A problem for the spice boys

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A problem for the spice boys
First of all let me say that I do not have or use Spice programs, so if what I am asking is too time consuming and troublesome, there are definately no hard feelings if you choose to ignore this post.
PROBLEM #1:
I have to filter and match a 100 MHZ HCMOS clock oscillator to a 50 0hm termination at a level of 0 DBM. Harmonic content should be <25 DBMc at the term. point. The oscillators maximum rise/fall time is 4 nSec. At 100Mhz the waveshape is a slight triangular shape with a moderate degree of truncation and 5volts PP (+18 DBM) , exactly as one would expect from this "half can " module.
I did a rough calculation on third harmonic content (this is the dominant one) with this wave shape and came up with approx.
20% of the fundamental. A quick scope check some what confirmed this.
PROBLEM #2:
My calculator died half way thru a proving analysis of the circuit design i came up with, and this is where I need help in confirming what I have.
WHERE I AM AT NOW:
I am capacitively coupling (no reactance here) the clock oscillator into a series 200 ohm resistor. The other end of this resistor feeds an LC Pi network and its purpose is to clean up the wave form and match it to a 50 ohm term.
PI filter input15 pf cap (100 ohm Xc) from resistor to grd.  150 nH inductor* (100 ohm XL) [this is my calculated value and also my problem area hence the asterisk], connected from cap./resistor node to another 15 pf cap of which the other end is grounded. The 50 ohm term. is across the last cap. Since this is wired right now on a ground plane in rather sloppy bread board fashion, there is a lot more lead lengths than would be in its final form (possibly 100 nH intermixed). The thing I do not understand at this point is that the PI network looks like it is still calling for more inductance in spite of all the extra added parasitic inductance of the sloppy circuit.
At this point, a very properly terminated scope check shows a fairly clean out put of + 4 DBM. The PI network is calculated to be a lo 'Q' network (maybe 10) resonating at 100 MHz and when optimized will probaly raise the level at the term point and require some adjustment of the series resistor to obtain the level desired here.
So if any one out there has a Spice program that can easily work out this type of circuit, I would certainly appreciate your input, espcially in regards to the inductor value as the other components seem to be more or less doing their job at this point. BTW the finished circuit will have no more than 1/16" lead lengths per component and calculated to be less than 15 nH total, so does not need to be entered into any answers at this point. The HCMOS out put impedance is assumed to be zero ohms out put so PI network input impedance is taken as 200 ohm.
PS  I did not intend for this post to be so long but felt I needed to supply all info I could in order to be more helpful.
PROBLEM #1:
I have to filter and match a 100 MHZ HCMOS clock oscillator to a 50 0hm termination at a level of 0 DBM. Harmonic content should be <25 DBMc at the term. point. The oscillators maximum rise/fall time is 4 nSec. At 100Mhz the waveshape is a slight triangular shape with a moderate degree of truncation and 5volts PP (+18 DBM) , exactly as one would expect from this "half can " module.
I did a rough calculation on third harmonic content (this is the dominant one) with this wave shape and came up with approx.
20% of the fundamental. A quick scope check some what confirmed this.
PROBLEM #2:
My calculator died half way thru a proving analysis of the circuit design i came up with, and this is where I need help in confirming what I have.
WHERE I AM AT NOW:
I am capacitively coupling (no reactance here) the clock oscillator into a series 200 ohm resistor. The other end of this resistor feeds an LC Pi network and its purpose is to clean up the wave form and match it to a 50 ohm term.
PI filter input15 pf cap (100 ohm Xc) from resistor to grd.  150 nH inductor* (100 ohm XL) [this is my calculated value and also my problem area hence the asterisk], connected from cap./resistor node to another 15 pf cap of which the other end is grounded. The 50 ohm term. is across the last cap. Since this is wired right now on a ground plane in rather sloppy bread board fashion, there is a lot more lead lengths than would be in its final form (possibly 100 nH intermixed). The thing I do not understand at this point is that the PI network looks like it is still calling for more inductance in spite of all the extra added parasitic inductance of the sloppy circuit.
At this point, a very properly terminated scope check shows a fairly clean out put of + 4 DBM. The PI network is calculated to be a lo 'Q' network (maybe 10) resonating at 100 MHz and when optimized will probaly raise the level at the term point and require some adjustment of the series resistor to obtain the level desired here.
So if any one out there has a Spice program that can easily work out this type of circuit, I would certainly appreciate your input, espcially in regards to the inductor value as the other components seem to be more or less doing their job at this point. BTW the finished circuit will have no more than 1/16" lead lengths per component and calculated to be less than 15 nH total, so does not need to be entered into any answers at this point. The HCMOS out put impedance is assumed to be zero ohms out put so PI network input impedance is taken as 200 ohm.
PS  I did not intend for this post to be so long but felt I needed to supply all info I could in order to be more helpful.

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Sorry, I don't have a spice program, just a couple of ideas.
"The thing I do not understand at this point is that the PI network looks like it is still calling for more inductance in spite of all the extra added parasitic inductance of the sloppy circuit." Due to breadboarding the capacitor leads' parasitic inductance decreases the effective capacitance of the capacitors; making it look like you need more inductance. (capacitor lead inductance is inductance in the wrong direction)
The series resistor and first capacitor of Pi are a voltage divider (feeding inductor of Pi) with less voltage at higher frequencies (RC filter). Increasing the value of the resistor will improve the filtering in addition to lowering the output amplitude.
"The thing I do not understand at this point is that the PI network looks like it is still calling for more inductance in spite of all the extra added parasitic inductance of the sloppy circuit." Due to breadboarding the capacitor leads' parasitic inductance decreases the effective capacitance of the capacitors; making it look like you need more inductance. (capacitor lead inductance is inductance in the wrong direction)
The series resistor and first capacitor of Pi are a voltage divider (feeding inductor of Pi) with less voltage at higher frequencies (RC filter). Increasing the value of the resistor will improve the filtering in addition to lowering the output amplitude.
Dale Y

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Good point Dale
Actually I had considered this to a certain extent and I guess if I had posted a picture it would have been worth a thousand words. Mechanically, each capacitor has one leg to ground about one inch long (each 40 nH and about 25 ohms XL) so that alone should reduce the effective Xc (100 ohm) by about 25 ohm leaving 75 ohms of Xc in each leg for an effective 20 pf. instead of 15 pf. Now the other 1" leads are totally in series with the inductor and feed points since they are soldered very near the capacitors body, so this adds 80 nH to the tank circuit ( 40 nH for each lead) so this has to be added to the total inductive leg of the tank circuit. Now this new effective capacitance of 10 Pf (20//20) across one leg of the tank would require an inductance of 250nH to resonate with it and by the calculations in this post it now has 240 nH ( coil of 160 nH +40+40) so it would seem that it should be resonant at 100 MHz as it stands on the breadboard. But some quick tests indicated that it wanted much more inductance to acheive resonance. Since my friggin calculator died on me , I am roughing these figures out in my head as I type so they are pretty loose but hopefully in the ball park.
As to the series resistor, it only for the purpose of setting the power level into the feed point of the tank circuit. Adding capacitance here will only change the resonant frequency and the tap in impedance. It won't behave as a simple RC filter.
I keep thinking I have made a mistake in my calculations somewhere along the way as it seems the inductor wants to be twice the value for proper resonance. Oh well, I will pick up some calculator batteries tomorrow and run thru this thing again. Hopefully some one with a spice program can input something by then. Thanks for your input as it helps me reinforce my thinking.
Actually I had considered this to a certain extent and I guess if I had posted a picture it would have been worth a thousand words. Mechanically, each capacitor has one leg to ground about one inch long (each 40 nH and about 25 ohms XL) so that alone should reduce the effective Xc (100 ohm) by about 25 ohm leaving 75 ohms of Xc in each leg for an effective 20 pf. instead of 15 pf. Now the other 1" leads are totally in series with the inductor and feed points since they are soldered very near the capacitors body, so this adds 80 nH to the tank circuit ( 40 nH for each lead) so this has to be added to the total inductive leg of the tank circuit. Now this new effective capacitance of 10 Pf (20//20) across one leg of the tank would require an inductance of 250nH to resonate with it and by the calculations in this post it now has 240 nH ( coil of 160 nH +40+40) so it would seem that it should be resonant at 100 MHz as it stands on the breadboard. But some quick tests indicated that it wanted much more inductance to acheive resonance. Since my friggin calculator died on me , I am roughing these figures out in my head as I type so they are pretty loose but hopefully in the ball park.
As to the series resistor, it only for the purpose of setting the power level into the feed point of the tank circuit. Adding capacitance here will only change the resonant frequency and the tap in impedance. It won't behave as a simple RC filter.
I keep thinking I have made a mistake in my calculations somewhere along the way as it seems the inductor wants to be twice the value for proper resonance. Oh well, I will pick up some calculator batteries tomorrow and run thru this thing again. Hopefully some one with a spice program can input something by then. Thanks for your input as it helps me reinforce my thinking.
Checking matching network design tables confirms your nominal values for the 100 ohm inductive and capacitive reactances to do the 200:50 ohm transformation for your topology but with a Q of 2. I agree that the inductance associated with lead lengths of discrete capacitors partly cancels their capacitive reactances. Try using near zero lead length 15 pf SMT caps (1206 is fine) tombstoned off the ground plane to act as terminal points for your original conventional leaded inductor, results should more closely conform unless inductor shunt capacitance parasitics or ground plane interactions with an unshielded inductor are excessive. Perhaps even if the inductor self resonant frequency is well above 100 MHz, there is still that distributed capacitance inside that is partly cancelling the inductive reactance and forcing selection of a much increased value of L to achieve a net 100 ohms inductive reactance.

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 Posts: 2276
 Joined: Wed Nov 24, 2004 1:01 am
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Sparkle
Thank you for confimation of values used. In my laziness and haste in bread boarding this filter, I did not want to follow proper RF design layout as once the components were in, they are no longer usable if they have to be removed. So in that respect I thought I would just cobble it up and predict performance by taking into consideration all of the parasitics the 'sloppy' circuit incorporated. But I guess I pushed it to far this time. It's a fact of life that RF design on paper rarely makes an exact match to real world construction and adjustments must then be made accordingly. I had calculated an SRF of 300+ MHz for the coil but this figure may be in error further complicating design. I am on my way out for some calculator batteries and will run thru the the numbers once more, rewind another inductor and then constuct the filter in proper RF fashion (something I probably should have done in the first place).Hopefully every thing will fall in place then.
Thank you for confimation of values used. In my laziness and haste in bread boarding this filter, I did not want to follow proper RF design layout as once the components were in, they are no longer usable if they have to be removed. So in that respect I thought I would just cobble it up and predict performance by taking into consideration all of the parasitics the 'sloppy' circuit incorporated. But I guess I pushed it to far this time. It's a fact of life that RF design on paper rarely makes an exact match to real world construction and adjustments must then be made accordingly. I had calculated an SRF of 300+ MHz for the coil but this figure may be in error further complicating design. I am on my way out for some calculator batteries and will run thru the the numbers once more, rewind another inductor and then constuct the filter in proper RF fashion (something I probably should have done in the first place).Hopefully every thing will fall in place then.
Hi there,
There is free software available online that can do network calculations
of any type like this. If you are interested i'll post a link and even
provide guidelines for how to set the equations up that can calculate
just about anything you want to know, including the frequency
response of the final network. Once you get the final equation
(which isnt terribly hard to do i guess) you can tweek values like
the inductor or change the guessed values of the series parasitic
L's and do the frequency response again all in about 10 seconds.
This would provide a valuable tool for doing this stuff i would think.
If you like, you can also get the equations into a form which would
allow you to shoot the input with the wave you saw on the scope
(dont know how much parallel cap that scope has yet though) and
see what the output should look like.
I wouldnt mind seeing a schematic of your circuit, including where
the oscillator connects and how the other components connect.
I think i got some of this info from your first post but it's always nice
to see a schematic that shows everything drawn in where it is connected.
Just to give an example of how you might set up the equ's for a
RLC network with components in order R, L, C, and loose end of C
connected to ground and output taken across C:
Vo(s)=E*zC/(R+zL+zC)
where you have defined zC and zL prevously as:
zL=s*L
zC=1/(s*C)
and you would plug in the values
L=100e9
C=15e12
or whatever they are.
You would then be able to plot a frequency response, change
L or C to whatever you wanted to, then do another frequency response.
You could then compare the frequency responses and see what
one you like better, just for example.
Some other things you could do:
Check input impedance
Check output impedance
Calculate losses
Calculate bandwidth, although you would get a good idea from
looking at the frequency response plots without actually calculating
a numerical value for this.
This sort of automates the design process and confirms that you
have the sort of circuit you want and the values are in the right
range. Depending on how well you convert real world to actual
circuit, you can come up with some really accurate stuff like this.
I think since you are taking parasitic elements into consideration
you should get very good results.
BTW, using the software is much faster than using a hand calculator.
I used to do the calculator thing a lot too, but then many years ago
switched to the computer for everything because it's not only faster
the equations seem to be easier to change and generate new ones
if needed. The calculator was also sort of slow for doing the frequency
response plots and time domain plots.
OH yeah that's another thing, you can convert to time domain and
see the response to pulses, etc. too.
Alternately, i'd be happy to post an equation for your circuit that
would allow you to do these calculations in a flash, if you dont want
to get too involved yourself.
LATER:
Here's a plot of output (with 1v input) vs frequency for the pi
network you described...two 15pf caps, one 150nH inductor.
200 ohms in series with input generator, 50 ohm output
termination.
BTW, the unloaded output impedance at 100MHz comes out to 46 ohms,
assuming zero impedance for the input source. The load on the source
would be a little over 200 ohms, so the source would have to be
capable of driving 200 ohms without a problem.
What this plot doesnt show is the third harmonic, which is down to
9 percent of the input, which puts your 3rd harmonic input of 20 percent
down to about 2 percent, so that's not too bad i guess.
There is free software available online that can do network calculations
of any type like this. If you are interested i'll post a link and even
provide guidelines for how to set the equations up that can calculate
just about anything you want to know, including the frequency
response of the final network. Once you get the final equation
(which isnt terribly hard to do i guess) you can tweek values like
the inductor or change the guessed values of the series parasitic
L's and do the frequency response again all in about 10 seconds.
This would provide a valuable tool for doing this stuff i would think.
If you like, you can also get the equations into a form which would
allow you to shoot the input with the wave you saw on the scope
(dont know how much parallel cap that scope has yet though) and
see what the output should look like.
I wouldnt mind seeing a schematic of your circuit, including where
the oscillator connects and how the other components connect.
I think i got some of this info from your first post but it's always nice
to see a schematic that shows everything drawn in where it is connected.
Just to give an example of how you might set up the equ's for a
RLC network with components in order R, L, C, and loose end of C
connected to ground and output taken across C:
Vo(s)=E*zC/(R+zL+zC)
where you have defined zC and zL prevously as:
zL=s*L
zC=1/(s*C)
and you would plug in the values
L=100e9
C=15e12
or whatever they are.
You would then be able to plot a frequency response, change
L or C to whatever you wanted to, then do another frequency response.
You could then compare the frequency responses and see what
one you like better, just for example.
Some other things you could do:
Check input impedance
Check output impedance
Calculate losses
Calculate bandwidth, although you would get a good idea from
looking at the frequency response plots without actually calculating
a numerical value for this.
This sort of automates the design process and confirms that you
have the sort of circuit you want and the values are in the right
range. Depending on how well you convert real world to actual
circuit, you can come up with some really accurate stuff like this.
I think since you are taking parasitic elements into consideration
you should get very good results.
BTW, using the software is much faster than using a hand calculator.
I used to do the calculator thing a lot too, but then many years ago
switched to the computer for everything because it's not only faster
the equations seem to be easier to change and generate new ones
if needed. The calculator was also sort of slow for doing the frequency
response plots and time domain plots.
OH yeah that's another thing, you can convert to time domain and
see the response to pulses, etc. too.
Alternately, i'd be happy to post an equation for your circuit that
would allow you to do these calculations in a flash, if you dont want
to get too involved yourself.
LATER:
Here's a plot of output (with 1v input) vs frequency for the pi
network you described...two 15pf caps, one 150nH inductor.
200 ohms in series with input generator, 50 ohm output
termination.
BTW, the unloaded output impedance at 100MHz comes out to 46 ohms,
assuming zero impedance for the input source. The load on the source
would be a little over 200 ohms, so the source would have to be
capable of driving 200 ohms without a problem.
What this plot doesnt show is the third harmonic, which is down to
9 percent of the input, which puts your 3rd harmonic input of 20 percent
down to about 2 percent, so that's not too bad i guess.
LEDs vs Bulbs, LEDs are winning.

 Posts: 2276
 Joined: Wed Nov 24, 2004 1:01 am
 Location: ASHTABULA,OHIO
 Contact:
MrAl
First let me say that was a very thorough and informative reply. As requested I am posting a schematic. I have recalculated values ,rewound a new coil and installed components in the circuit using proper RF construction techniques. Although there was some improvement, it is still not satisfactory. The problem with scope checking is that the passive probes have one or more resonant points that usually occur in the 30 80 MHz range. This is due to added inductance in the probes ground lead (even short ones) acting in series with the scopes input capacitance (nominal 10  15 pf for 10:1 probes).The spring clip probe ground attachment in lieu of the normal ground lead only offers minimal improvement.
There are 3 reliable probing types that are useful when entering the VHF region and above  the active probe which is very low capacitance ( 1 pf) and VERY expensive. Then the low impedance resistive divider probe (no capacitive compensation used) which is very simple and performs quite well in a 50 ohm transmission line environment but does present some circuit loading. I intend to design and build one of these in the near future. And lastly a high quality scope's true 50 ohm vertical input port. This is the one I use presently. In order to get a true picture of this circuits out put (or any 50 ohm load), I disconnect the circuit at that point and substitute the scopes internal load in place of it as I have done with this circuit. The end of the 50 ohm coax is tack soldered to make these measurments. Just telling you this so you know how I conducted my testing.
When the circuits output is connected to the scope only. the waveform is fairly pure ( I have a good eye for detecting sine wave distortion down to about 510%). So the filter looks like it is doing its job, however the level here is about +4 or + 5 DBM. I also used this method when the output was attached to its intended load and yes I know I was double loading the circuit with a consecutive 3 DB drop in the signal, but it is the only true way I have of viewing it. I also looked at that point with a 10:1 passive probe and in both cases the signal was quite distorted. So you can see that some of my test methods were not according to 'Hoyle' yet they offer some small degree of accuracy. In both cases adding capacity (1520 PF) to the values already in place greatly improved purity of waveform and closer to the level I am looking for.
Well if you had any doubts I hope the schem in figure #1 gives a clearer picture. Also from a purely intuitive stand point as viewed in Fig. #2, I see this as a lo Q resonant circuit loaded by 250 ohms and resonant at a much higher frequency than intended. Changing the caps to 33 pf each would resonate it at 100 MHz and in fact circuit tests seemed to back this up. I realize that this example is true only for unloaded caps due to the fact that parellel res. across them actually has the effect of lowering the resonant frequency. But in practice it did not seem to hold true  could it be that the load was an open circuit?
Now one final monkey wrench to throw into the works this circuit feeds the 50 ohm RF input of a doubly balanced mixer.The mixer thus far has been deenergized in the sense that no LO enegy has been injected, which in turn means that the internal bridge diodes may not be switching which means that the RF port may not be presenting a 50 ohm load yet Maybe I will be interested in your findings.
First let me say that was a very thorough and informative reply. As requested I am posting a schematic. I have recalculated values ,rewound a new coil and installed components in the circuit using proper RF construction techniques. Although there was some improvement, it is still not satisfactory. The problem with scope checking is that the passive probes have one or more resonant points that usually occur in the 30 80 MHz range. This is due to added inductance in the probes ground lead (even short ones) acting in series with the scopes input capacitance (nominal 10  15 pf for 10:1 probes).The spring clip probe ground attachment in lieu of the normal ground lead only offers minimal improvement.
There are 3 reliable probing types that are useful when entering the VHF region and above  the active probe which is very low capacitance ( 1 pf) and VERY expensive. Then the low impedance resistive divider probe (no capacitive compensation used) which is very simple and performs quite well in a 50 ohm transmission line environment but does present some circuit loading. I intend to design and build one of these in the near future. And lastly a high quality scope's true 50 ohm vertical input port. This is the one I use presently. In order to get a true picture of this circuits out put (or any 50 ohm load), I disconnect the circuit at that point and substitute the scopes internal load in place of it as I have done with this circuit. The end of the 50 ohm coax is tack soldered to make these measurments. Just telling you this so you know how I conducted my testing.
When the circuits output is connected to the scope only. the waveform is fairly pure ( I have a good eye for detecting sine wave distortion down to about 510%). So the filter looks like it is doing its job, however the level here is about +4 or + 5 DBM. I also used this method when the output was attached to its intended load and yes I know I was double loading the circuit with a consecutive 3 DB drop in the signal, but it is the only true way I have of viewing it. I also looked at that point with a 10:1 passive probe and in both cases the signal was quite distorted. So you can see that some of my test methods were not according to 'Hoyle' yet they offer some small degree of accuracy. In both cases adding capacity (1520 PF) to the values already in place greatly improved purity of waveform and closer to the level I am looking for.
Well if you had any doubts I hope the schem in figure #1 gives a clearer picture. Also from a purely intuitive stand point as viewed in Fig. #2, I see this as a lo Q resonant circuit loaded by 250 ohms and resonant at a much higher frequency than intended. Changing the caps to 33 pf each would resonate it at 100 MHz and in fact circuit tests seemed to back this up. I realize that this example is true only for unloaded caps due to the fact that parellel res. across them actually has the effect of lowering the resonant frequency. But in practice it did not seem to hold true  could it be that the load was an open circuit?
Now one final monkey wrench to throw into the works this circuit feeds the 50 ohm RF input of a doubly balanced mixer.The mixer thus far has been deenergized in the sense that no LO enegy has been injected, which in turn means that the internal bridge diodes may not be switching which means that the RF port may not be presenting a 50 ohm load yet Maybe I will be interested in your findings.

 Posts: 2276
 Joined: Wed Nov 24, 2004 1:01 am
 Location: ASHTABULA,OHIO
 Contact:
Most scopes offering 50 ohm inputs also have some internal capacitance of the input circuits across the 50 ohms, commonly 10 to 12 pf. You can use a Smith Chart and some simple cable dielectric delay wavelength calculations to see how a short length of interconnecting coax will transform this lump capacitance at the far end to look like various sizes of capacitive and inductive reactance at the near end where you hook in your circuit. Or, you can just try a series of 1/8 electrical wavelength incremental lengths of interconnecting coax to see if your results vary appreciably. I think about a foot and a half of cable will make the capacitor look like an inductor to your circuit under test, at the fundamental frequency. The harmonics with their shorter wavelengths will see totally different loading from the differently transformed scope input capacitance.Robert Reed wrote:...And lastly a high quality scope's true 50 ohm vertical input port. This is the one I use presently. In order to get a true picture of this circuits out put (or any 50 ohm load), I disconnect the circuit at that point and substitute the scopes internal load in place of it as I have done with this circuit. The end of the 50 ohm coax is tack soldered to make these measurments. Just telling you this so you know how I conducted my testing...
If you have sufficient voltage sensitivity in your scope to throw some away in an attenuator, making a simple uncompensated local 10:1 voltage divider or even better a pad attenuator will greatly isolate scope cable and input loading effects from future measurements. Without isolation, you could easily be trying to comprehend and optimize a circuit different from the intended schematic that does not exist when the filter is finally connected to drive the DBM mixer.
In industry, it is common to insert pad attenuators made from SMT resistors to drive and terminate DBMs which need and are spec'd with wideband 50 ohm terminations, but do not necessarily present signal level independent 50 ohm termination to the driving circuits. Try to avoid using axial lead spiral cut and trimmed film resistors at VHF and above. Generally they are ok for higher z bias applications but the inductive effect has a way of showing up at "interesting moments" particularly in the ohmicly lower valued gain setting, load and termination applications.
Hi again Robert,
I cant tell you if your load is presenting a constant 50 ohms to
the circuit or not. If you are not sure, connect a 50 ohm resistor
in parallel to the output, see what happens. The circuit will not
perform anywhere near expected without the right load, and if
that load varies over frequency too that could cause a lot of
problems like amplifying out of band frequencies more than
the 100MHz you are looking to work with, which of course
will look like distortion.
If you cant test this circuit properly then all i can think of at the
moment is that you should test it with your intended load and
see if it works the way you expect it to or not.
What i can tell you is that with two 16pf caps the inductor value
comes out to 158.71nH to resonate at 100MHz, and that the output
impedance is 49.7 ohms which is pretty close to 50.
Some series ESR in both L and both C's will decrease the resonate
point so reducing the L could help. Reducing by 1nH seems to
help so your value of 156nH isnt too bad really.
BTW, with no load on the output the network becomes
a band stop filter with center F around 100MHz, so that means
all frequencies other than 100MHz get more gain then the
100MHz part of the signal.
If you wish to cut back on the in band gain i guess you could increase
the value of the 200 ohm resistor. Is there any problem with changing
this to maybe 400 ohms or whatever is required? Does the input
load have to be around 200 ohms or can it be much higher?
I cant tell you if your load is presenting a constant 50 ohms to
the circuit or not. If you are not sure, connect a 50 ohm resistor
in parallel to the output, see what happens. The circuit will not
perform anywhere near expected without the right load, and if
that load varies over frequency too that could cause a lot of
problems like amplifying out of band frequencies more than
the 100MHz you are looking to work with, which of course
will look like distortion.
If you cant test this circuit properly then all i can think of at the
moment is that you should test it with your intended load and
see if it works the way you expect it to or not.
What i can tell you is that with two 16pf caps the inductor value
comes out to 158.71nH to resonate at 100MHz, and that the output
impedance is 49.7 ohms which is pretty close to 50.
Some series ESR in both L and both C's will decrease the resonate
point so reducing the L could help. Reducing by 1nH seems to
help so your value of 156nH isnt too bad really.
BTW, with no load on the output the network becomes
a band stop filter with center F around 100MHz, so that means
all frequencies other than 100MHz get more gain then the
100MHz part of the signal.
If you wish to cut back on the in band gain i guess you could increase
the value of the 200 ohm resistor. Is there any problem with changing
this to maybe 400 ohms or whatever is required? Does the input
load have to be around 200 ohms or can it be much higher?
LEDs vs Bulbs, LEDs are winning.

 Posts: 2276
 Joined: Wed Nov 24, 2004 1:01 am
 Location: ASHTABULA,OHIO
 Contact:
Hi Sparkle
"Most scopes offering 50 ohm inputs also have some internal capacitance of the input circuits across the 50 ohms, commonly 10 to 12 pf."
Not high end scopes like this one. When Tek says 50 ohms, it is 50 ohms and purely resistive. This is why it has it's own special port. I have checked and crosschecked this many times and it is the 'perfect'load.
"In industry, it is common to insert pad attenuators made from SMT resistors to drive and terminate DBMs"
Use them all the time where I can afford the loss and am using several at present in this circuit design. I was trying the filter for mainly harmonic reduction and the 200 ohm series resistor for level reduction. And yes I can be flexible with that value and as I see it now, i will be probaly increasing to 390 ohms for level setting.
"Try to avoid using axial lead spiral cut and trimmed film resistors at VHF"
Thats a good point and up to a couple of years ago, I would have completely ageed with you. I avoided film resistors like the plague for RF, using strictly carbon comp resistors. Then came one project ,I had depleted my carb comp supply and tried some film resistors in its place. Resultthey performed just fine and had to handle upwards of 200 MHz. But one saving grace here I never use resistors higher than 300 ohms in VHF signal paths. In that realm, high ohmage resistors are not even resistors any more as I am sure you are well aware of this fact judging by the high level of your replys. The port I am feeding on this mixer wants to be terminated in 50 ohms. Up to this point, I have not been injecting any LO power which is planned for +7DBM. Do you think the lack of that would affect the input 'Z' at the RF port (the one I am currently having a problem with)?
"Most scopes offering 50 ohm inputs also have some internal capacitance of the input circuits across the 50 ohms, commonly 10 to 12 pf."
Not high end scopes like this one. When Tek says 50 ohms, it is 50 ohms and purely resistive. This is why it has it's own special port. I have checked and crosschecked this many times and it is the 'perfect'load.
"In industry, it is common to insert pad attenuators made from SMT resistors to drive and terminate DBMs"
Use them all the time where I can afford the loss and am using several at present in this circuit design. I was trying the filter for mainly harmonic reduction and the 200 ohm series resistor for level reduction. And yes I can be flexible with that value and as I see it now, i will be probaly increasing to 390 ohms for level setting.
"Try to avoid using axial lead spiral cut and trimmed film resistors at VHF"
Thats a good point and up to a couple of years ago, I would have completely ageed with you. I avoided film resistors like the plague for RF, using strictly carbon comp resistors. Then came one project ,I had depleted my carb comp supply and tried some film resistors in its place. Resultthey performed just fine and had to handle upwards of 200 MHz. But one saving grace here I never use resistors higher than 300 ohms in VHF signal paths. In that realm, high ohmage resistors are not even resistors any more as I am sure you are well aware of this fact judging by the high level of your replys. The port I am feeding on this mixer wants to be terminated in 50 ohms. Up to this point, I have not been injecting any LO power which is planned for +7DBM. Do you think the lack of that would affect the input 'Z' at the RF port (the one I am currently having a problem with)?

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Hello MrAl
I can increase the 200 ohm resistor and probably will need to just to reduce the power level at that point, maybe to 390 ohms. At this point I am unsure of the input 'Z' of the mixers RF input, only that the Manu suggests feeding it from a 50 ohm source, so I am assuming its also a 50 ohm input.Problems see tail end of my last reply. Its a difficult point to measure as a passive probe not only loads but is quite inaccurate at this frequency and they are even worse when measuring low impedance nodes. 50 ohm scope port direct is very faithful but I am double loading the node I am measuring (if its 50 ohm). This normally would be very predictable, but now I don't even know the impedance here to make a prediction. I think what I will do is leave the circuit as it stands (since everyone agrees it is set up right). then I will inject LO power and do my testing at the mixer IF port, since that is the bottom line anyway.
I can increase the 200 ohm resistor and probably will need to just to reduce the power level at that point, maybe to 390 ohms. At this point I am unsure of the input 'Z' of the mixers RF input, only that the Manu suggests feeding it from a 50 ohm source, so I am assuming its also a 50 ohm input.Problems see tail end of my last reply. Its a difficult point to measure as a passive probe not only loads but is quite inaccurate at this frequency and they are even worse when measuring low impedance nodes. 50 ohm scope port direct is very faithful but I am double loading the node I am measuring (if its 50 ohm). This normally would be very predictable, but now I don't even know the impedance here to make a prediction. I think what I will do is leave the circuit as it stands (since everyone agrees it is set up right). then I will inject LO power and do my testing at the mixer IF port, since that is the bottom line anyway.
My guess: Looking at a MiniCircuits DBM internal schematic as an example, and assuming absolutely perfect balance, here are two mechanisms of RF port mismatch power level sensitivity.Robert Reed wrote:... The port I am feeding on this mixer wants to be terminated in 50 ohms. Up to this point, I have not been injecting any LO power which is planned for +7DBM. Do you think the lack of that would affect the input 'Z' at the RF port (the one I am currently having a problem with)?
First, there are two diode drops to turn on if only the RF is driven. I'm uncertain if the RF port transformer turns ratio is the same as the LO transformer, but if so the diodes would turn on with less than +7dBm RF drive. There is a Conversion Loss vs LO Drive chart for folks thinking of starved LO operation in the How to Select a Mixer tutorial section of the Fat Book that indicates a conversion loss of 8.5 dB at 0 dBm LO drive (as opposed to a conversion loss of 5 dB for normal op) for what looks like a +7dBm style DBM. When a pair of diodes turn on from RF drive alone (this is not normal mixer operation, but certainly the norm for applying a mixer to phase detector application) the energy is just circulating and either reflected back up the port or turning into heat, to the extent things are really balanced. Below this level of RF driven turn (below perhaps 0 dBm drive, see also the Mini Circuits comment on drive required for phase detector application), the RF port has got to be heading towards becoming an open mismatch because that RF port transformer is driving diode shunt capacitance and magnetizing current/transformer losses only, no diode series resistance losses. So my guess is that with increasing RF only drive at somewhere between 10dBm and 0dBm, the RF port transitions from somewhat open to a fair but not great match.
Second, with proper +7 dBm LO drive, the diodes turn on with no RF applied at all, and if any RF shows up at the RF port it periodically unbalances the currents established by the LO, which are up to this point circulating (perhaps reflecting some back the LO port) and turning into heat again to the extent things are really balanced. The unbalance directs some RF energy (minus conversion loss) to the IF port, which we assume is perfectly terminated at all frequencies somehow. Conservation of energy suggests that RF port will "know" at least to a small extent, if that LO is driven, because if in the LO case the IF gets the mixer product energy and in the no LO case the IF gets next to no energy at all. So the RF port load would seem to be increased by somewhat more than the energy associated with a 6 dB typical conversion loss when LO drive was applied. The LO drive gives the RF port energy somewhere additional to go, the IF termination. The converted energy plus all other diode and transformer losses hopefully yield a good match at the RF port.
Summary: 1. Apply low then increasing RF port drive alone, no LO, RF port "knows" when the diodes turn on because it transitions from a predominantly capacitive open mismatch to a fair match as RF drive approaches the level normally associated with using a mixer as a phase detector. 2. Apply any normal RF port drive level, RF port "knows" when LO is present because it can then send energy to the IF port, in addition to the energy absorbed in a somewhat matched RF port (transformer and diode losses) that is established when there is LO diode drive, and it becomes a good match.
Hi again Robert,
As i was saying, i cant really help too much with what input Z your
load is, but another idea, although a little far out, is this:
Connect 50 scope in series with load (presume 50 ohms).
Connect 100 ohm carbon resistor in parallel with output of network.
Doing this brings the total load Z back to 50 ohms but still allows
connection of the scope. Of course this means attenuation for the
load and for the scope too, so you'll have to do the math to see
how reduced the scope waveform will be because of this. At least
in theory this works If you see 1v on scope that means 2v
total output for example, provided the load really is 50 ohms.
If the manual says 50 ohms then i guess that's all we have to go
on unless you can find a signal standard to test it with.
So anyway, i take it you want 1mW into your load. If i did my own
math right, the following will work well (assuming 50 ohm load):
R1=420 ohms
C1=C2=37.6pf
L1=103.3nH
This fixes the center frequency very close to 100MHz and
the attenuation to get 1mW into 50 ohms, and sets the output
impedance again to very close to 50 ohms. The only other
assumption here is that the source is 5vpp and contains 5vpp
fundamental. The third harmonic still gets significantly attenuated
also.
As before, after adding a small ESR to all caps and inductors and a
little series L to both caps, reducing the inductor value slightly brings
the center frequency back to 100MHz. I didnt recheck Zout however
which i guess would also be interesting after the change.
Yes, i was saying test with the real load because that's the end all of
all end alls to the project. If it works ok, if not, it's not ok
Although that's not always as informative as we would like, sometimes
it's the only way.
Oh yeah, one thing i forgot to mention...
There is going to be offset dc in the output. This of course will be
less than the input but still there.
Another circuit that reduces the offset somewhat and provides
matching as above is this:
R1=750 ohms
C1=10.6pf
C2=27.3pf
L1=269nH
This has a little better frequency selection i think and still
matches everything else well, plus reduces dc in the output
better.
As i was saying, i cant really help too much with what input Z your
load is, but another idea, although a little far out, is this:
Connect 50 scope in series with load (presume 50 ohms).
Connect 100 ohm carbon resistor in parallel with output of network.
Doing this brings the total load Z back to 50 ohms but still allows
connection of the scope. Of course this means attenuation for the
load and for the scope too, so you'll have to do the math to see
how reduced the scope waveform will be because of this. At least
in theory this works If you see 1v on scope that means 2v
total output for example, provided the load really is 50 ohms.
If the manual says 50 ohms then i guess that's all we have to go
on unless you can find a signal standard to test it with.
So anyway, i take it you want 1mW into your load. If i did my own
math right, the following will work well (assuming 50 ohm load):
R1=420 ohms
C1=C2=37.6pf
L1=103.3nH
This fixes the center frequency very close to 100MHz and
the attenuation to get 1mW into 50 ohms, and sets the output
impedance again to very close to 50 ohms. The only other
assumption here is that the source is 5vpp and contains 5vpp
fundamental. The third harmonic still gets significantly attenuated
also.
As before, after adding a small ESR to all caps and inductors and a
little series L to both caps, reducing the inductor value slightly brings
the center frequency back to 100MHz. I didnt recheck Zout however
which i guess would also be interesting after the change.
Yes, i was saying test with the real load because that's the end all of
all end alls to the project. If it works ok, if not, it's not ok
Although that's not always as informative as we would like, sometimes
it's the only way.
Oh yeah, one thing i forgot to mention...
There is going to be offset dc in the output. This of course will be
less than the input but still there.
Another circuit that reduces the offset somewhat and provides
matching as above is this:
R1=750 ohms
C1=10.6pf
C2=27.3pf
L1=269nH
This has a little better frequency selection i think and still
matches everything else well, plus reduces dc in the output
better.
LEDs vs Bulbs, LEDs are winning.
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