What is this circuit for?

[ENE]

Hello,
What is this circuit working for?

 

[willem2]

I'm pretty sure this circuit has no practical use whatever. There will be no feedback if the output ever gets positive

 

[ENE]

I have used op amp and comprator like 741 and L339.
I don't understand the circuit Q npn.

 

[f95toli]

I have used op amp and comprator like 741 and L339.
I don't understand the circuit Q npn.

It is a transistor. But I agree with willem2 that the circuit looks very odd and does not serve a practical purpose, I can't see why you would put a transistor in the feedback loop of an op-amp; especially not with the base grounded(!)
Hence, I don't think the circuit will do anything useful but that does not mean that you can't answer the question.

 

[Averagesupernova]

For you naysayers it is most likely a log amplifier.

 

[lychette]

If a load is connected between V_out and -15V then the circuit acts as a constant current (5mA) source for resistive loads of up to 3kΩ.
This configuration is the basis of constant current sources using op amps.

 

[NascentOxygen]

It looks like it could serve a number of functions.
(a) Over a range of positive input, the output voltage will be a logarithmic function of the input voltage [at fixed temperature].

(b) Or if you maintain the input voltage fixed, the output will be a linear function of temperature of the transistor probe.

(c) Or it could simply be an analog comparator, with output of -0.7V or +15V.

 

[ENE]

So, what is the answer?

 

[ENE]

How to calculate it?

 

[NascentOxygen]

I didn't realize that you are asking for help with the question in your post, I thought you were asking something else, viz., what might this circuit be used for.

In operation, either the transistor will be functioning as a transistor, or it won't be. If it is not functioning as a transistor then it is "off". For the transistor to be functioning as a transistor, describe the voltages around the circuit you will definitely expect to see.

 

[ENE]

In operation, either the transistor will be functioning as a transistor, or it won't be. If it is not functioning as a transistor then it is "off".

let the transistor be off so open circuit for feed back.
so, +ve pin to 0V
and -Ve pin to 5V
that mean it should be -15V?
how to do?

 

[NascentOxygen]

let the transistor be off so open circuit for feed back.
so, +ve pin to 0V
and -Ve pin to 5V
that mean it should be -15V?

Now using the result that V_out = -15V does this confirm your premise that Q is "off"?

 

[ENE]

NO, the question same as above i am guessing that it will be -15V when BJT is open.
no idea when it will work.

 

[NascentOxygen]

If V_out = -15V will this turn the transistor "off"? Look at the voltages that this type of transistor requires for it to be on or off.

 

[ENE]

The BJT is npn it will get more -ve with respect to base for turing on it want 0.7V

 

[NascentOxygen]

The BJT is npn it will get more -ve with respect to base for turing on it want 0.7V

We need a yes or no answer to the question: Is it possible in this circuit for V_out = -15V to be keeping the transistor off?

If these 2 conditions can coexist then that is how it must be. If these conditions cannot coexist, then you need to make a different initial assumption.

 

[lychette]

If you look at the input side of the amplifier the 5V DC with 1kΩ resistor will make a current of 1mA flow towards the inverting input. The current cannot flow into the op-amp and will therefore flow by the feedback link towards the output. It will be the collector current of the transistor.
Can you see what the output voltage will be?

 

[NascentOxygen]

If a load is connected between V_out and -15V then the circuit acts as a constant current (5mA) source for resistive loads of up to 3kΩ.

This is not correct.

 

[rude man]

For you naysayers it is most likely a log amplifier.

Yes, but not a good one.
V_out = V_T ln{(V_in/R_in)/i_s} assuming high beta of transistor. But i_s = saturation current which varies with temperature & device to device within the same family. V_T = kT/q, k = bolzmann constant, T = kelvin temp., q = electronic charge. It's a poor indicator of temperature also since again i_s = i_s(T).