Transistor driven by one voltage that switches another voltage?

DGElder

Joined Apr 3, 2016
351
You may be measuring two different things. It appears the simulator is measuring the voltage across the 10 ohm R, the top with respect to the bottom. So the bottom is arbitrarily called 0V. Is that correct? But is that how you are measuring with the O-scope or are you measuring with respect to ground. Explain exactly how you are measuring the output with your scope. How many probes, are they 10X and have you compensated them, are you fastening the ground clip to your circuit -if so where, is the scope set for AC or DC coupling, what input impedance is selected for the scope, is the scope or any equipment floating or plugged in directly to the mains?

Can you show the output and input signals at the same time and give a better scope shot so we can see the voltages, and where the signal is with respect to ground, and compare input to output waveforms?
 

wayneh

Joined Sep 9, 2010
17,496
I think you need a path to ground when Q4 (the bottom one) is not conducting. The voltages out of the simulator may be a bit of an artifact of the component models. With nearly zero current, I don't think the voltages mean much.
 

Thread Starter

electronice123

Joined Oct 10, 2008
339
I think you need a path to ground when Q4 (the bottom one) is not conducting. The voltages out of the simulator may be a bit of an artifact of the component models. With nearly zero current, I don't think the voltages mean much.
So should I add a ground between the two Q4 transistors?
What value or does it matter?
 

Thread Starter

electronice123

Joined Oct 10, 2008
339
You may be measuring two different things. It appears the simulator is measuring the voltage across the 10 ohm R, the top with respect to the bottom. So the bottom is arbitrarily called 0V. Is that correct? But is that how you are measuring with the O-scope or are you measuring with respect to ground. Explain exactly how you are measuring the output with your scope. How many probes, are they 10X and have you compensated them, are you fastening the ground clip to your circuit -if so where, is the scope set for AC or DC coupling, what input impedance is selected for the scope, is the scope or any equipment floating or plugged in directly to the mains?

Can you show the output and input signals at the same time and give a better scope shot so we can see the voltages, and where the signal is with respect to ground, and compare input to output waveforms?

Yeas I was measuring directly across the load resistance R8 in both the simulation and the breadboard design using a single probe with the ground clip on emitter of the lower Q4 transistor. Probes are 10X and were compensated. Scope set to DC coupling.

I am considering starting over completely as I can't get it to work right no matter what I do.

How would you guys go about it? like I said I'm trying to apply a positive square wave across the load which voltage changes every x number of pulses with both the maximum and minimum voltage adjustable (Like my scope shot shows in the simulation).
 

DGElder

Joined Apr 3, 2016
351
Yeas I was measuring directly across the load resistance R8 in both the simulation and the breadboard design using a single probe with the ground clip on emitter of the lower Q4 transistor. Probes are 10X and were compensated. Scope set to DC coupling.

.
As I suspected, you are not measuring the same thing in the sim and breadboard. You are measuring the voltage across R8 and Q4 on the breadboard, so of course your signal does not match the simulation voltage across R8 alone. Put a probe on each side of R8 and use the Math function on the scope to see the difference between the two signals.

Conversely, look at the sim waveform when you measure across R8 and Q4 like you did on the breadboard.
 

Thread Starter

electronice123

Joined Oct 10, 2008
339
new dwg.png I rebuilt the circuit on the breadboard and it does exactly what I need it to....Everything is the same as my previous schematic....except this time I don't have the transistors connected to the circuit at all.
The question I have now- how do I connect the transistors so that the lower frequency signal will cause the high frequency signal's voltage to be reduced when the lower frequency signal goes low?

How do I arrange and connect the transistors to allow me to do this?

V LEVE.png

In the scope shot you can see the yellow signal has an adjustable voltage during it's low period as well.
Like I mentioned previously, I want to produce a positive square wave pulse train across the load at one voltage level for x number of pulses then the same positive square wave pulse train at a lower voltage for x number of pulses.
 
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Alec_t

Joined Sep 17, 2013
14,280
The use of Darlington transistors as shown means that the output pulses will be a volt or more less than the voltage V3 is set to. You could overcome that by including the Darlington in the opamp feedback loop. I don't see the purpose of Q4?
 

Thread Starter

electronice123

Joined Oct 10, 2008
339
The use of Darlington transistors as shown means that the output pulses will be a volt or more less than the voltage V3 is set to. You could overcome that by including the Darlington in the opamp feedback loop. I don't see the purpose of Q4?

Thanks for the help....Sorry I labeled both transistors Q4. I'm assuming it's the upper one that is not necessary?

I need to do more reading on op amp feedback. I'm guessing that I need to add a resistance between the Q4 emitter and the inverting input of the op amp to bring the output pulses voltage higher?
 

Thread Starter

electronice123

Joined Oct 10, 2008
339
Is this closer to what you want, with output amplitude rising in sympathy with V2 voltage?
View attachment 115450
That looks good- but are you showing the voltage increasing slowly in a ramp by adjusting the pot or does the circuit do that automatically?

I want the voltage to stay where it is unless I adjust the pot (to change voltage across the load when V2 goes low) or unless I adjust the V3 voltage.
 

Alec_t

Joined Sep 17, 2013
14,280
are you showing the voltage increasing slowly in a ramp by adjusting the pot or does the circuit do that automatically?
The ramp is the result of V2 changing slowly. If V2 is kept constant then the waveform peaks are constant too. The pot sets the ratio of the tall peaks to the shorter peaks. N.B my voltage sources are numbered differently from yours. Your V3 is my +V (which is kept constant). Voltage to my load is varied by the opamp/Darlington.
 
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Thread Starter

electronice123

Joined Oct 10, 2008
339
The ramp is the result of V2 changing slowly. If V2 is kept constant then the waveform peaks are constant too. The pot sets the ratio of the tall peaks to the shorter peaks. N.B my voltage sources are numbered differently from yours. Your V3 is my +V (which is kept constant). Voltage to my load is varied by the opamp/Darlington.
The circuit works but I'm having one problem. The output of the Op Amp is not a clean square wave with short rise and fall times. Instead the rise and fall times are longer and it's causing the waveform across the load and at the base of the darlington to look more like a ramp waveform with the top cutoff.Untitled.png

What do I need to do with the Op Amp to get clean square waves out of it?

Oscilloscope is measuring across the resistor with the ground clip on the ground side.


Wow, has it really been so long that I forgot about Op Amp slew rates?
Thanks Alec-t
 
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