The parts list sounds good. What's not as good is when you write, "I have the DMM on and switched to 20K in the Ohm section." You'll want your meter set to read voltage. A 20V range should be adequate for these measurements.
Only use the Ohm section to measure a circuit that has no power applied. Otherwise, damage to the meter could result, or an error will be added to the measurements at the least.
With the circuit plugged in, the voltage on the capacitor should be close to the voltage measured on the green and blue wires (about 0.7V difference). You should also find that voltage if you measure across the collectors & emitters of the transistors.
When you write, "I unplug the device from COM1. Placing the black and red leads at the same point the DMM counts up from minus figures to zero. (I gather this is the capacitor charging up taking power from the DMM??), that could actually be the capacitor discharging into the DMM.
It sounds like the IR diodes are connected correctly. For diodes with a clear body, you can see inside and there will be two metal sections. The larger of the two is the negative connection. The diode test function on the meter is good for measuring when there is no power to the circuit. Your readings look good. If power is applied to the circuit, use the voltage setting on the meter. The reading you get will be the same as that obtained with the diode test. The diode test simply applies power from the meter. A diode will have a forward-bias voltage that permits conduction and will be constant in spite of the amount of current flow.
See if you can measure the voltage across both the IR diodes with the circuit operating. It should go from zero volts at idle to about 3.5 volts when DTR changes to positive. You can also monitor the voltage across the transistor collector/emitter. It should drop significantly when the circuit is activated. The meter display may change quickly.
IR remote circuit running out of steam?!
Re: IR remote circuit running out of steam?!
I think it would be possible to use a 9-volt battery in place of the serial port connection. This might be helpful when taking voltage readings of the circuit because it get rid of the brief pulses and allows time for measurements. Just connect the + of the battery to ground and the - to the DTR input to charge the capacitor. Then switch the battery connections to simulate the control pulse. The visible-light LED or an infrared detector is necessary since the IR won't be modulated.
Here's my analysis of the circuit operation, using +/- 9V as DTR voltage. Corrections & additions are welcomed.
When DTR is negative to ground, D1 blocks current and D2 allows the capacitor to charge. The transistors are both OFF which prevents the other diodes from conducting.
When DTR goes positive, D2 blocks current and D1 conducts (note that the "bottom line" where the negative side of the cap is connected is now about -18V potential from DTR). When 0.7 volts is developed across the lower 10k resistor, the NPN transistor is biased ON. The emitter of the NPN is negative with respect to ground, so this completes the path through the two silicon diodes. When they conduct, it puts a negative voltage to the base of the PNP transistor. The PNP transistor turns on, completes the current path, and the capacitor discharges through the IR diodes. When the capacitor discharges below about 3.5 volts, it can no longer overcome the voltage drop across the IR diodes and will cease conduction.
Here's my analysis of the circuit operation, using +/- 9V as DTR voltage. Corrections & additions are welcomed.
When DTR is negative to ground, D1 blocks current and D2 allows the capacitor to charge. The transistors are both OFF which prevents the other diodes from conducting.
When DTR goes positive, D2 blocks current and D1 conducts (note that the "bottom line" where the negative side of the cap is connected is now about -18V potential from DTR). When 0.7 volts is developed across the lower 10k resistor, the NPN transistor is biased ON. The emitter of the NPN is negative with respect to ground, so this completes the path through the two silicon diodes. When they conduct, it puts a negative voltage to the base of the PNP transistor. The PNP transistor turns on, completes the current path, and the capacitor discharges through the IR diodes. When the capacitor discharges below about 3.5 volts, it can no longer overcome the voltage drop across the IR diodes and will cease conduction.
Re: IR remote circuit running out of steam?!
Hi Sofaspud<p>Ok (thank you for pointing out my errors) i have the DMM switched to 20V.
A reading across the green and blue wires is -10.22
Oh this is with the black lead from the DMM on the green wire. When the leads are swaped over its the possitive of the reading. Not sure which color lead should go where especially as you talk about minus figures. I dont want to give the wrong readings!!<p>The reading across the Capacitor is 9.88 thats a difference of 0.34V. Is that still ok??
The reading across the PNP is .31
The reading across the NPN is 9.34
The reading across both the IR LED is 0.14 and yes when the circuit is activated i do read a significant drop in voltage.<p>Can i ask just to reasure myself. The NPN & PNP transistors have a flat side to them. Looking at the schematic the flat side of the NPN will face the IR LED. The flat side of the PNP faces the silicon diodes??? Im sure ive got it right, but just checking.<p>I will have to buy a 9V battery to test your analysis.<p>Many thanks
Mike
A reading across the green and blue wires is -10.22
Oh this is with the black lead from the DMM on the green wire. When the leads are swaped over its the possitive of the reading. Not sure which color lead should go where especially as you talk about minus figures. I dont want to give the wrong readings!!<p>The reading across the Capacitor is 9.88 thats a difference of 0.34V. Is that still ok??
The reading across the PNP is .31
The reading across the NPN is 9.34
The reading across both the IR LED is 0.14 and yes when the circuit is activated i do read a significant drop in voltage.<p>Can i ask just to reasure myself. The NPN & PNP transistors have a flat side to them. Looking at the schematic the flat side of the NPN will face the IR LED. The flat side of the PNP faces the silicon diodes??? Im sure ive got it right, but just checking.<p>I will have to buy a 9V battery to test your analysis.<p>Many thanks
Mike
Re: IR remote circuit running out of steam?!
Here is i hope a clear pictorial representation of what my device looks like.
Re: IR remote circuit running out of steam?!
You have it correct with the black probe on the green wire. The placement of the meter leads isn't of prime importance. The absolute value of voltage is all that is needed, if the status of the DTR line is kept in mind.
The 9.88 voltage on the capacitor is good. I'm not sure why it is only a 0.34 drop from the DTR voltage, but circuit operation shouldn't be adversely affected.
The PNP 0.31 voltage doesn't sound right. I'm thinking it should read close to the 9.34 you found on the NPN.
The voltage measured across and IR diodes should be close to zero or close to 3.5 volts. Measured separately, you should get the 1.80 and 1.77 that you measured with the diode test of the DMM.
Looking at the flat side of the 2N3904 & 2N3906, the legs are Emitter-Base-Collector.<p>I see a layout of the circuit was posted while I was typing. I'm taking a lunch break but I'll look it over afterwards.<p>[ May 15, 2005: Message edited by: sofaspud ]</p>
The 9.88 voltage on the capacitor is good. I'm not sure why it is only a 0.34 drop from the DTR voltage, but circuit operation shouldn't be adversely affected.
The PNP 0.31 voltage doesn't sound right. I'm thinking it should read close to the 9.34 you found on the NPN.
The voltage measured across and IR diodes should be close to zero or close to 3.5 volts. Measured separately, you should get the 1.80 and 1.77 that you measured with the diode test of the DMM.
Looking at the flat side of the 2N3904 & 2N3906, the legs are Emitter-Base-Collector.<p>I see a layout of the circuit was posted while I was typing. I'm taking a lunch break but I'll look it over afterwards.<p>[ May 15, 2005: Message edited by: sofaspud ]</p>
Re: IR remote circuit running out of steam?!
I've looked at the layout and don't see any errors. The transistors are oriented correctly. I would double-check the measurements for the PNP transistor. With the capacitor charged, the voltage across the collector-emitter of the PNP should be nearer the capacitor charge.
You can remove the PNP from the circuit and test it with the diode checker of the DMM, or the transistor hFE test if the meter has one.
You can remove the PNP from the circuit and test it with the diode checker of the DMM, or the transistor hFE test if the meter has one.
Re: IR remote circuit running out of steam?!
Hi so that Sofaspud or anyone else, does not start to think that i have given up on this, i am posting this. I have bought a 9V battery and will attempt to get back to mesuring voltages maybe this weekend.<p>Many thanks
Mike<p>[ May 26, 2005: Message edited by: scag ]</p>
Mike
<p>[ May 26, 2005: Message edited by: scag ]</p>Re: IR remote circuit running out of steam?!
Hi SofaSpud, I hope you are still about. Here goes..
I have wired my 9V battery to the right GND & DTR (TXD).
Ok once all the readings settle down.
Reading across the
1K is 0.00
D2 is 0.32
2.2K is 0.00
Cap is 9.4
D1 is 0.32
(1st) 10K is 0.00
(2md) 10K is 0.00
3.9K is 0.00
(1st) Silicon Diodes is 0.00
(2nd) Silicon Diodes is 0.00
22 Ohm is 0.00<p>swaped over the terminals on the battery
Reading across the
1K is 0..45
D2 is 6.42 (decreasing)
2.2K is 0.00
Cap is 2.70 (decreasing)
D1 is 0.53
(1st) 10K is 5.30
(2md) 10K is 0.65
3.9K is 0.00
(1st) Silicon Diodes is 0.00
(2nd) Silicon Diodes is 0.00
22 Ohm is 0.00<p>I gather this looks bad??
Are the lack of change in the readings of some of the components due to the changes happening quickly that i can not get too quick enough??<p>Ok i think this has beaten me. I have no idea what to do next. Could i send it to you??
Many thanks
Mike
I have wired my 9V battery to the right GND & DTR (TXD).
Ok once all the readings settle down.
Reading across the
1K is 0.00
D2 is 0.32
2.2K is 0.00
Cap is 9.4
D1 is 0.32
(1st) 10K is 0.00
(2md) 10K is 0.00
3.9K is 0.00
(1st) Silicon Diodes is 0.00
(2nd) Silicon Diodes is 0.00
22 Ohm is 0.00<p>swaped over the terminals on the battery
Reading across the
1K is 0..45
D2 is 6.42 (decreasing)
2.2K is 0.00
Cap is 2.70 (decreasing)
D1 is 0.53
(1st) 10K is 5.30
(2md) 10K is 0.65
3.9K is 0.00
(1st) Silicon Diodes is 0.00
(2nd) Silicon Diodes is 0.00
22 Ohm is 0.00<p>I gather this looks bad??
Are the lack of change in the readings of some of the components due to the changes happening quickly that i can not get too quick enough??<p>Ok i think this has beaten me. I have no idea what to do next. Could i send it to you??
Many thanks
Mike
Re: IR remote circuit running out of steam?!
Hi Mike,
I think your readings looks quite normal. This design is not made for continous reverse powering, and will not output anything when you wait too long.<p>I've been looking through the thread and I wonder why you are removing one of the IR-diodes all the time? Is it to reduce the power consumption?<p>When you does this you still keep the same output current on the IR-diode, but reduces the min. voltage the circuit need. You also most certainly turn off the indicator LED at the same time. (This is depending on the FWD voltage of the LED)<p>The whole idea behind this circuit is that the active period is much less than inactive. You may compare it to PWM, or inversed PWM. The shorter the duty cycle, the higher the voltage over the supply capacitor.<p>If you want somebody to check it out, you may send it to me, I'm not that far away. <p>TOK
I think your readings looks quite normal. This design is not made for continous reverse powering, and will not output anything when you wait too long.<p>I've been looking through the thread and I wonder why you are removing one of the IR-diodes all the time? Is it to reduce the power consumption?<p>When you does this you still keep the same output current on the IR-diode, but reduces the min. voltage the circuit need. You also most certainly turn off the indicator LED at the same time. (This is depending on the FWD voltage of the LED)<p>The whole idea behind this circuit is that the active period is much less than inactive. You may compare it to PWM, or inversed PWM. The shorter the duty cycle, the higher the voltage over the supply capacitor.<p>If you want somebody to check it out, you may send it to me, I'm not that far away.
<p>TOKGorgon the Caretaker - Character in a childrens TV-show from 1968.
Re: IR remote circuit running out of steam?!
Hi Gorgon<p>The idea of removing one of the IR LED was due to my original problem, some people questioned why there was 2 IR LED. So to see if the circuit actually needed 2 of them i took one out. The device stoped working. I bought all components again, put it together, it worked. I took the IR LED out just to see if by accident it was something else that had stoped it all working. It stoped working and failed to start again when i put it back.
My last attempt to piecing it all back together was that i bought all the pieces again replaced one component at a time and still nothing.<p>Sorry you lost me with 'PWM, or inversed PWM'<p>Many thanks for your offer to have a look at it, give me a week or so before i decide what to do if thats ok with you.<p>Many thanks
Mike
My last attempt to piecing it all back together was that i bought all the pieces again replaced one component at a time and still nothing.<p>Sorry you lost me with 'PWM, or inversed PWM'<p>Many thanks for your offer to have a look at it, give me a week or so before i decide what to do if thats ok with you.<p>Many thanks
Mike
Re: IR remote circuit running out of steam?!
Yes, I'm still about.
Many of the voltages will indeed change quickly. It's the nature of the circuit and something that just needs to be worked around.
What stands out most to me is the lack of voltage across the two silicon diodes in each set of measurements. The description calls these the "only critical components" but they don't seem to be operating. They are important to the operation of the PNP transistor.
The description calls the transistor configuration a pulse current amplifier, which is true enough. I prefer to see the PNP transistor as a switch. When the GND line is positive relative to DTR, the output IR LEDs are forward biased. No current flows because the PNP transistor is turned off. The positive pulse to DTR is what turns on the PNP and allows current flow through the IR LEDs.
Connect the 9-volt battery with the + at GND and the - at DTR. If you place your meter across the collector & emitter of the PNP, you should measure near the battery voltage across them. Leaving the meter leads on the transistor, switch the battery connections at GND and DTR. The voltage reading on the meter should abruptly drop to near zero and stay there, letting you know that the PNP has turned on.
If you don't have clips on your meter leads to allow you to keep contact on the transistor leads, use some small wire to tightly wrap around the leads and tack the other ends to your circuit with a bit of solder.
It isn't necessary to send me your circuit. I'll breadboard the thing and compare with yours.
Many of the voltages will indeed change quickly. It's the nature of the circuit and something that just needs to be worked around.
What stands out most to me is the lack of voltage across the two silicon diodes in each set of measurements. The description calls these the "only critical components" but they don't seem to be operating. They are important to the operation of the PNP transistor.
The description calls the transistor configuration a pulse current amplifier, which is true enough. I prefer to see the PNP transistor as a switch. When the GND line is positive relative to DTR, the output IR LEDs are forward biased. No current flows because the PNP transistor is turned off. The positive pulse to DTR is what turns on the PNP and allows current flow through the IR LEDs.
Connect the 9-volt battery with the + at GND and the - at DTR. If you place your meter across the collector & emitter of the PNP, you should measure near the battery voltage across them. Leaving the meter leads on the transistor, switch the battery connections at GND and DTR. The voltage reading on the meter should abruptly drop to near zero and stay there, letting you know that the PNP has turned on.
If you don't have clips on your meter leads to allow you to keep contact on the transistor leads, use some small wire to tightly wrap around the leads and tack the other ends to your circuit with a bit of solder.
It isn't necessary to send me your circuit. I'll breadboard the thing and compare with yours.
Re: IR remote circuit running out of steam?!
Hi Mike,
I aggree with Sofaspud, I don't think you need to send the circuit. I took a review of your last measurements and didn't find the voltages over the two transistors. I also noticed from your veroboard that you have removed the indicator LED. (By the way, you have connected together all the leads on the top row?)<p>One addition to the schematic I would have done, is a 10k resistor from the base of the PNP to the GND, across the 2 SI-diodes. This will turn of the PNP properly and faster. Also reducing powerloss.<p>One puzzling thing is that the circuit die when you use 1 IR-diode. This indicates that the PNP transistor(or/and the IR-diode) is destroyed. Two things may do this; a shorted 22-ohm resistor or a break in the 2 Si-diodes. This will disable the current control, and more or less short the capacitor through the transistor and IR-diode.<p>I think a trip to the basement tomorrow is in place. <p>TOK
I aggree with Sofaspud, I don't think you need to send the circuit. I took a review of your last measurements and didn't find the voltages over the two transistors. I also noticed from your veroboard that you have removed the indicator LED. (By the way, you have connected together all the leads on the top row?)<p>One addition to the schematic I would have done, is a 10k resistor from the base of the PNP to the GND, across the 2 SI-diodes. This will turn of the PNP properly and faster. Also reducing powerloss.<p>One puzzling thing is that the circuit die when you use 1 IR-diode. This indicates that the PNP transistor(or/and the IR-diode) is destroyed. Two things may do this; a shorted 22-ohm resistor or a break in the 2 Si-diodes. This will disable the current control, and more or less short the capacitor through the transistor and IR-diode.<p>I think a trip to the basement tomorrow is in place.
<p>TOKGorgon the Caretaker - Character in a childrens TV-show from 1968.
Re: IR remote circuit running out of steam?!
Hi Gorgon
Ooops
Yes I have connected together all the leads on the top row. Sorry missed that out on the graphics. I have now changed it.<p>Ok i have not lost all heart now, thank you both! So i need to do what Sofaspud advises and maybe need to look at using different Si Diodes.<p>The Si Diodes that i have been using are 1N4148 from Maplin here in Wales, though i was told about another type of Si Diodes 1N914. Just looking on the website there is a slight difference between the two, the reverse current?? Is this any help??<p> Signal Diodes Signal Diodes<p>Type Construction Case PIV IF Reverse Current
Style V mA IR (µA V)
1N914 Si whiskerless D035 100 75 25na 25V
1N4148 Si whiskerless D035 100 75 25nA 20V<p>Many thanks
Mike
Ooops
Yes I have connected together all the leads on the top row. Sorry missed that out on the graphics. I have now changed it.<p>Ok i have not lost all heart now, thank you both! So i need to do what Sofaspud advises and maybe need to look at using different Si Diodes.<p>The Si Diodes that i have been using are 1N4148 from Maplin here in Wales, though i was told about another type of Si Diodes 1N914. Just looking on the website there is a slight difference between the two, the reverse current?? Is this any help??<p> Signal Diodes Signal Diodes<p>Type Construction Case PIV IF Reverse Current
Style V mA IR (µA V)
1N914 Si whiskerless D035 100 75 25na 25V
1N4148 Si whiskerless D035 100 75 25nA 20V<p>Many thanks
Mike
Re: IR remote circuit running out of steam?!
Hi Mike,
I've looked at the data for the two diodes. You may use both, but the 1N914 has lower max. current. I don't think the type of diodes has anything to do with your problem. 1N4148 is a good choice to use.<p>I'll get back to you after I've tried to build yor circuit.<p>TOK<p>[ May 31, 2005: Message edited by: Gorgon ]</p>
I've looked at the data for the two diodes. You may use both, but the 1N914 has lower max. current. I don't think the type of diodes has anything to do with your problem. 1N4148 is a good choice to use.<p>I'll get back to you after I've tried to build yor circuit.<p>TOK<p>[ May 31, 2005: Message edited by: Gorgon ]</p>
Gorgon the Caretaker - Character in a childrens TV-show from 1968.
Re: IR remote circuit running out of steam?!
For testing purposes you may want to use two batteries, one for power supply and one for a DC signal. This will avoid the problem of the storage capacitor discharging during measurements.<p>The power battery would be connected across the storage capacitor with the positive terminal to the positive electrode of the capacitor (the green wires on your drawing) and the negative terminal to the negative end of the capacitor (the connections across the top of your drawing).<p>The battery used to supply a signal would be connected between the positive capacitor terminal and your input signal lead. In one position (positive to capacitor, negative to input) it will simulate a negative input signal. In the other position (positive to input, negative to capacitor) it will simulate a positive input signal.<p>By using two batteries you will be measuring steady voltages, with no necessity for making sure that the storage capacitor is reqharged.
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