OP Amp circuit question

crutschow said:

It would make sense if the circuit worked as you think it does. But it doesn't. Do the simulation, don't just talk about it.

As soon as the output tries to rise about the transistor's base voltage it will start to apply voltage to the 10k resistor so as to keep the voltage at the base voltage value (in other words clamping it at that level). This will continue until the transistor saturates. Only when the transistor saturates, when the voltage at the 10k resistor reaches the reference voltage level, will the output then start to rise with a new gain value based upon the parallel value of the two feedback resistors.

Make sense to you?

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Carl,

Then it is not a clamping circuit, the gain changes as i had said all along, but it does not clamp the output to any set value.
So you see why you can not remove the 10k resistor and have it work the very same way? As i said several times, with the 10k resistor it is not a clamp, but without the 10k resistor it then becomes more of a clamp (10k shorted).
The new gain will vary from whatever 43k provides to whatever 43k in parallel with 10k will provide as the transistor conducts more and more. That's definitely not a clamping action, and it's not when the transistor saturates either. The output is free to rise more as the input signal level is increased.

I dont see how you can say there is any clamping action, that's what you are saying that i dont see, that's the only thing. Explain how the output goes to some set level and doesnt rise significantly (aside from running out of supply voltage of course).

Roff said:

MrAl, the transistor is active (unsaturated) until about 500uA flows through the 10k collector resistor. When the transistor is active, the 10k resistor has no effect.
With the output at 5V, 116uA (5V/43k) will be flowing through the 43k resistor . This means that the source resistor has to be providing (500uA +116uA)=616uA before the transistor saturates. This would require that the input voltage would have to be more negative than -6.16V in the posted simulations.

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How can you say that a 10k in parallel with 40k does not have any effect? The input voltage is not set (yet) until we know the true value of the output impedance (or resistor value) of the previous stage.

MrAl said:

How can you say that a 10k in parallel with 40k does not have any effect? The input voltage is not set (yet) until we know the true value of the output impedance (or resistor value) of the previous stage.

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That's true. My comments were in the context of the simulations that Carl and I posted. However, if the input current is sufficient to saturate the transistor, there will still be "flat spots" in the output where the transistor isn't saturated.
I made some arbitrary assumptions about vref, the source resistance, and the input voltage, but I believe the results will hold in the general sense as those values are changed.

MrAl said:

Carl,

Then it is not a clamping circuit, the gain changes as i had said all along, but it does not clamp the output to any set value.
So you see why you can not remove the 10k resistor and have it work the very same way? As i said several times, with the 10k resistor it is not a clamp, but without the 10k resistor it then becomes more of a clamp (10k shorted).
The new gain will vary from whatever 43k provides to whatever 43k in parallel with 10k will provide as the transistor conducts more and more. That's definitely not a clamping action, and it's not when the transistor saturates either. The output is free to rise more as the input signal level is increased.

I dont see how you can say there is any clamping action, that's what you are saying that i dont see, that's the only thing. Explain how the output goes to some set level and doesn't rise significantly (aside from running out of supply voltage of course).

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As the transistor starts to conducts it will clamp the output at the Vref value. There is no variable gain at this point and the 10k will not have any effect until the transistor saturates. It's only after the transistor saturates that the value of the resistor value comes into play. Just do the damn simulation and go argue with that.

Just to satisfy my curiosity, I designed a couple of precision feedback clamps. I'm posting them here because I thought others might find them useful, or at least interesting.
Response time for the simple clamp is primarily a function of op amp saturation recovery time and slew rate.
Response time for the "improved" clamp is a function of slew rate and GBW product. The op amps don't saturate in this version.

crutschow said:

As the transistor starts to conducts it will clamp the output at the Vref value. There is no variable gain at this point and the 10k will not have any effect until the transistor saturates. It's only after the transistor saturates that the value of the resistor value comes into play. Just do the damn simulation and go argue with that.

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He he Ok Carl i'll do as you suggest and do a simulation today some time. I'll get back here a little later. I have a feeling maybe the reason why we are seeing this differently is because we are assuming different operating points for this circuit and testing under different limits. That's one of the reasons i wanted to see the OP response to what the circuit is really intended to be used for. I still feel it would be better to comment after obtaining that information, but i also see this circuit now as something that might be interesting in its own right, used possibly in a different manner than originally intended.
Thanks for your comments and i'll be back a little later hopefully...

crutschow said:

MrAl, as Roff noted the 10k resistor (R4 in the original schematic, sorry for the confusion with two R4's) has no significant effect on the clamp. It does not provide variable gain. You can remove the resistor and the clamp will operate just the same. Your thought experiment is in error. Run the simulation if you don't believe.

And of course, Roff's comment about the base-emitter breakdown is correct.

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Hello again Carl,

No variable gain you say? Well maybe then you would like to argue with the simulation because after seeing the simulation apparently i dont have to

As you can PLAINLY see from the first diagram below, the gain changes, and it changes three times. It starts at one gain A1 and changes to A2 as i predicted. There is a short portion where it is flatter but nonetheless still a gain A3 and not a clamp, although you may wish to call that one tiny section where Vin is just right a "clamp" if you like.
This is exactly why i had suggested it may be a sensor (or transducer) amplifier because it can be used to correct a transducer output curve if it was innately nonlinear.

Convinced yet?

If you still are not, then check out the second drawing that shows the response with the 10k resistor shorted out. It clearly looks more like a clamp circuit now just as i had suggested previously. It's a very different response without the 10k there. This also shows why i suggested removing the 10k resistor if they wanted a better actual clamp circuit.

MrAl said:

As you can PLAINLY see from the first diagram below, the gain changes, and it changes three times. It starts at one gain A1 and changes to A2 as i predicted. There is a short portion where it is flatter but nonetheless still a gain A3 and not a clamp, although you may wish to call that one tiny section where Vin is just right a "clamp" if you like.

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The difference in our simulations is that I used a 10k input resistor and you used a 1k, which gives 10 times higher gain and a 10 times shorter clamp region. I always stated that the 10k resistor had no effect until the transistor saturates. But between the time the transistor turns on and the transistor saturates, the circuit clamps the output.

So we have no real disagreement as to the operation of the circuit, just one of interpretation.

Hi again Carl,


Very well said ! And i agree fully

I only wish that the OP would come back and fill us in on how they wanted to use the circuit. It's starting to look like they will never come back.

MrAl said:

Hi again Carl,

I only wish that the OP would come back and fill us in on how they wanted to use the circuit. It's starting to look like they will never come back.

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Yes, he may have left the building. It seems to be a common problem on these forums. The person who started the thread is never heard from again while the rest of us are left yammering at a blank wall.

Op amp is most important basic IC

Shyamal said:

Op amp is most important basic IC

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Perhaps in the analog world.

Shyamal said:

Op amp is most important basic IC

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This borders on spam, considering the signature and the lack of relevance to the thread.

Hi,

I had that same feeling myself