Can anyone make the new circuit function same as the old circuit? (SEE ATTACHMENT) Dear Sir's and Madams; and the Q and A colume of NUTS and VOLTS magazine.
I have a question about a circuit that should have gone to TJ BYERS but need someone eltz to answer my question?
It's part of a bigger circuit only a cutout shown here, see the attachment. There is a older circuit and I want to updated to a newer circuit and function the same way as the older circuit. I used two IC's (1) a LM311N (2) a LM555
I tryed to convert the circuit over to the new one, can anyone tell me if I did it right.
When S2 is closed the relay is actived, and stays on until the voltage on D5 gets higher enough then the relay is cutout for a moment, but inmedally turns on the relay again. It stays on until once again the rising voltage gets high enough turns on Q2 which turns off Q1 and the relay drops out for a short moment, Q1 turns on again which actives the relay again. The LM311N is used as a compartor, and the LM555 timer is used as a thrshold switch or Schmitt (Schmidt) trigger and relay driver.
I am replacing the old circuit with the new circuit. And I want to use the new circuit. And I want the OLD circuit to work as the NEW circuit. This whould have gone to TJ BYERS but now I need another bright mind. The 5K trim-pot, is to provide a adjustment window.
Craig Kendrick Sellen
Carbondale PA
Perhaps you can tell us a little more about what your circuit is
supposed to do. I guess you want to build the circuit using
ic's instead of transistors, but then i have to ask why the
added 555 timer ic? It seems like the first circuit could be done
with just an op amp and a single transistor. The transistor would
be used to drive the relay (or what looks like a relay).
Using a single transistor is a cheap and simple way to drive a relay.
Craig
I don't understand, if the old circuit works and you do not want the new cicuit t function any differently - why change it ?
There wouldn't even be much savings in board 'real estate'.
Craig,
Tell me what you expect your new circuit do. How long should the relay be ON and OFF? what is the purpose of S1 and S2? Are you sure the LM555 can drive the relay? If it is a heavy relay you'll need a transistor to drive it. Oh, and what is the peak value of the sawtooth?
I think you are going in the right direction but unless you can answer the questions, I can't help you.
Good luck
Paul J. Weijers
I've fixed the schematic so "old" isn't sideways and "new" isn't upside down:
It really would help if Craig would say what the expected behavior is, more info about the input signal, and spec of the relay. Anyway, here's some observations on the circuit -
There is no hysterisis(sp?) or latching in either old or new circuit, so switch S2 "Drop" is redundant. Also, depending on the source of the input, the output could "chatter", wearing out the relay in a short time.
The setup of Q4 and Q5 in Darlington configuration implies that Q5 is a low gain power transistor. That means the 555 can't drive the load directly as shown in "new".
The "heart" of a 555 is a comparitor, so the "new" circuit could be made with EITHER U1 OR U2, using both is redundant. If trying to add a minimum "on" time, use a 555 and driver transistor. Or, just the 311N and a driver transistor for better trigger level adjustment.
Craig, are you still there? There are 4 of us (at least) willing to help, but specific info is missing.
THANKS TO ALL OF YOU FOR YOUR HELP. I POSTED A NEW MODIFIED CIRCUIT, IS THIS BETTER? JUST CHECK THIS OUT PLEASE. THE ONLY QUESTION I HAVE IS DO I HAVE THE + AND - ON THE COMPARATOR LM311N RIGHT? OR DO THAY NEED TO BE REVERSED? I OMITED THE 555. AND REPLACED IT WITH A PNP TRANSISTOR. QUESTION DOE'S ANYONE KNOW OF A GOOD AND STILL AVAILABLE PNP DARLINGTON TRANSISTOR?
CRAIG KENDRICK SELLEN
The inverting input on the original circuit's differential amplifier is Q1 so it looks like you did not reverse the inputs on the comparitor. The circuit looks like it should work and you wouldn't need the darlington unless the relay had a particularly high coil current (>200mA rating of the 2n3907).
Im not sure how much output current (to base of PNP)) the LM311 can deliver (the datasheet lacked this spec) but there are lots of other op amps that will do this job just fine. With sufficient output current you can just use a bigger PNP but the darlington gets you higher gain and allows you to turn on the output transistor with less base current. I believe you can just use two PNP transistors if you can't find what you want in a single 3 pin darlington package. You still need a bigger (2nd) PNP if you want more current for the relay.
I revised the circuit again and reduced parts count. I replaced the LM311N with a conventional LM741 op-amp. The pot. is to adjust the thrushold or trip level, while the + input goes to the slowly rising voltage in the positive direction. See circuit diagram. Inregards to the relays, I plan to use DIP type relays which don't use much current, maybe 50mA each. Ether that I need a good and still in production Darlington PNP transistor. Rather then mess with two transistors.
Did you at least simulate this circuit to see if it would work?
With that diode there the transistor can only get 14.3 volts to
try to turn it off. This might not be enough and the relay might
stay on. Try connecting a diode in series with the transistor
emitter also, so the base only has to reach 14.3 to turn off
rather than the full 15v. The relay should work ok at 14v.
Also, the way it is set up with the 1k and 470 ohm resistors,
the 470 ohm resistor will always keep the transistor turned on.
Try making the 1k ohm resistor zero ohms and make the
470 ohm resistor 2.2k ohms.
These fixes should ensure that the transistor turns on and
off properly.
It would be good for you to take another look at this circuit with
these things in mind i think. It really looks like there are a few things
that are fundamentally wrong with this newest circuit.
Dear, MrAI
I will slid the diode down to the junction of S2 (drop) and base of the PNP transistor, in other words in place of the 1K. And I will omit the 1K resistor. The 470 ohm resistor is a limiting resistor, so when the drop switch is closed the relays will pull in and stay on until pin 6 of the comparator trips positive, cutting off the transistor & relays. The transistor is normally off, until S2 is closed, or pin 6 goes positive. I can omit the 1K because there is no current flowing through the transistor and op-amp, & diode when it goes positive, & S2 is open.So the op-amp woun't be damaged. I want a 470 ohm resistor to be sure the transistor conducts the current to drive the relays, and a LED. So I think a 2.2K may be a little low to do this. Unless it has a high HFE, Im thinging using a 2N3906 or a 2N4403.
I thought you said the relay pulls in with 50ma. 2.2k would be
good enough for that even with a HFE as low as 20.
Since you are using a diode in place of the 1k, that's good, but
to ensure the transistor turns off you might also need a diode in
series with the emitter, otherwise it may stay on or turn
on slightly. We havent even talked about the upper output
limit of the op amp yet, which will also be a problem if it
doesnt have a rail to rail output (ie output may only go up to
10 volts instead of the actual 15v supply voltage). These
things could easily keep the transistor turned on more than
you would like it to be. In fact, looking at the specs for the LM741
it shows the output swing to be only +/- 16 volts for a supply
voltage of +/- 20 volts. This is going to be a problem.
If you keep the 470 ohm resistor your op amp will have to put out
roughly 32ma, which is quit a bit for a lot of op amps, and at
that current you probably wont get full output voltage even if
it's a rail to rail device.
Here's an example of this new circuit with a 2.2k resistor and the
load is 300 ohms (50ma at 15v relay coil) and using the spice
models:
when the op amp puts out 4v (wont get to 0v either but that's ok)
the load gets 49.698ma.
when the op amp puts out +11 volts (hoping to turn the relay off)
the load gets 49.711ma.
Notice the current is HIGHER when the op amp tries to turn the
relay off with this circuit.
Thus, the relay stays on, and stays on
with full current. It doesnt matter that much though even if
the current goes a little lower, because it wont change enough to
make a difference. It would go lower in some cases, but
certainly not enough to turn the relay off.
Now lets say we found an op amp that actually can put out 15v,
the full supply voltage. We still have the transistor base emitter
voltage drop to deal with, which means the base is at approx 14.3
volts. Since the diode also drops about 0.7 volts this means only
14.3 volts (approx) can reach the base through the diode, which
means the transistor can remain partially on. Again using the
spice models, the actual current in the load is 7.2ma, which may
or may not be enough to keep the relay on, but remember we
only got this far assuming we had a rail to rail output op amp,
which we dont have.
Just so you know, with a 470 ohm resistor instead of the 2.2k resistor
the current remains as high as 47.6ma, which is certainly enough
to keep the relay turned on. This is why i recommended a trial
value of 2.2k instead of 470. And this is also with a rail to rail op
amp, which again the 741 is not.
Now lets put a 1N4001 diode in series with the transistor emitter
and change the other diode (that replaces the 1k resistor) to
a 1N4001 diode also.
With the op amp putting out 4v, the relay gets 47.3ma.
With the op amp putting out 15v, the relay gets 64na (64 nanoamps)
which means it finally turns off.
Problem is, the 741 cant put out 15v, it might only put out 11v for
certain conditions, which means we have to look at this circuit with
the op amp putting out only 11v again:
With the op amp putting out 11v, the relay again gets 47.3ma, which
means using the 741 op amp the relay will not turn off, even using
the 2.2k resistor instead of the 470 ohm as well as the extra diode
in series with the emitter.
Ok, so what does work?
What does work is putting three diodes in series with the emitter
and using a resistor divider on the output of the op amp (before
the diode) to 'step up' the output voltage a little. This requires
two resistors of values about 250 ohms and 2.5k ohms.
This means the relay turns on with 42ma and off with 664ua (microamps)
but this is still not as good of a design as the previous circuit.
What else might work is to use a LM339 comparator, which can
be used with a pullup resistor to turn the transistor off, but even
this isnt that good of an idea because whatever value you use for
the 470 ohm resistor (which also has to satisfy the relay current
requirement) the value of the pullup resistor will end up being too
low for the LM339 to handle working with +15 volts.
What else might work is a rail to rail output op amp, if using the 2.2k
ohm resistor AND the extra diode in series with the emitter.
This is the best bet if you absolutely must use this circuit.
I really think you were much better off using the previous circuit,
which although a little more complex (more parts) it looked like
it could work without adding a second transistor.
Dear MrAI;;;
see revised circuit below. Is this OK with you. Which resistors from the output of op-amp does the 250 ohms and 2.5K? Which one is grounded, and which one goes to op-amp? I just had a thort, how bout a ZENER diode in the emitter of the transistor, with the anode to the emitter. With a rating of 4.7V or maybe 3.3V, since the 3 diodes has a rating of 1.8V drop?
The 250 ohm and the 2.5k ohm resistor divider i talked about actually
connect as you have it except the 250 ohm resistor goes to +15v
not to ground. The 2.5k ohm connects to the output of the op amp,
and the other end to the 250 ohm resistor. The cathode of the
diode (in series with the base) connects to the junction of the
250 ohm and 2.5k ohm resistors. In this way, the voltage driving
the diode can be higher than the output of the op amp, and in
combination with the three diodes in series with the emitter can
turn the transistor off.
A zener in series with the emitter? Well, remember that whatever
drop is here also drops the 15v supply for the relay. The lower
the less voltage the relay gets. The question then becomes how
low of a voltage can the relay work at.
I still dont like this design as much as your previous design though,
where the op amp did everything and no need for a bunch of series
diodes.
Let me take a minute to make sure i understand how you want your
circuit to work...
When both switches are open, the relay is not energized.
When the one switch closes, the relay turns on and stays on but if the
other switch is closed too then the external pulsing signal turns
the relay on and off at some low frequency like 1Hz or lower,
and when this pulse goes high the relay turns off.
Is this correct?
Also, do you have any LM358 ic's around? They are a little bit better
and they work down to ground too.
Also, do you have any NPN transistors around?
The LM311 might be used by connecting the one switch in series with
pin 1 (from pin 1 to ground). The other switch in series with the
external pulse. The output would drive the PNP transistor base with
a 1k pullup resistor. Also, a 10k pulldown on the non-inverting input.
This would eliminate some parts, but this circuit
would have to be breadboarded to make sure the LM311 works ok with
pin 1 not connected to ground, as they didnt have any designs like
that on the data sheet.
ADDED LATER: This circuit does in fact work using the spice model
downloaded from the National site. There are no diodes required for
this circuit. Note also that the pinout is different depending on the
package used. Pin 1 here is the pin always labled 'ground' on the
data sheet regardless of package.
I would still breadboard this circuit however to make sure it works
as expected.
HELLO MrAI;
Im posting another circuit using LM311N. The values will have to be droped in. Rember it uses a PNP. The only thing I didn't think of is maybe it needs a pullup resistor to V+ to pin 7. (see schematic below) What I need is when S2 is closed the relays pull in regardless of LM311N. But when diode is fordward bias and conducts, pin 7 goes positive or high and cuts off the PNP and relays. Until the diode is reverised bias again.
If i understand you correctly what you need to do is connect
switch S2 in series with pin 1 and ground of the LM311.
Then, short the two resistors you have drawn there.
Then, connect a pullup resistor of 1k (2.2k prob work ok too).
What this does is the following...
With both switches open, the relay is off.
(At this point if S1 were to close nothing would happen).
When switch S2 is closed, the relay turns on and stays on
(S1 is still open) because the output of LM311 is low.
Now if S1 is closed the output of the LM311 goes high and low
pulsing with the frequency of the external pulse wave. This
pulses the relay on and off.
If at any time S2 is opened the relay turns off and stays off
even if S1 is closed.
