BJT in opamp feedback

  • #1
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Main Question or Discussion Point

I can't understand an opamp circuit whenever there is an active element in the feedback loop.
For example, in the current monitor ckt(attached). I don't understand how it works.
I can't figure out how the opamp offset and BJT gain would affect the output.
This ckt is from national semi AN31.
 

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Answers and Replies

  • #2
uart
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You picked a bad circuit to start with, it makes no sense. It has no obvious input or output and the OPAMP feedback is positive. The opamp output will sit at positive saturation levels. I can see nothing at all interesting happening with that circuit.
 
  • #3
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You picked a bad circuit to start with, it makes no sense. It has no obvious input or output and the OPAMP feedback is positive. The opamp output will sit at positive saturation levels. I can see nothing at all interesting happening with that circuit.
uart, its a current monitor ckt.

The output is at R6. The current is measured across R4, R5 (0.1 ohm) resistors.
Vout = ILoad*0.1 ohms*R6/R1

For example, let's say load resistance is 22.5 ohms, Vcc is also 22.5v.
So, the current flowing into the load will be 1Amp.

Vout at R6 will be

1*0.1*10 = 1volt.

Vout gives the equivalent voltage value corresponding to the load current. It's a high side current monitor.
 
  • #4
uart
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It's a high side current monitor.
Yes I figured that's what it was supposed to be, but that is not what it is (it doesn't work).

In that circuit the opamp non-inverting input is about 22.3 volts, the inverting input is about 20.5 volts, the opamp output is at it's positive saturation level (probably approx 21 volts) and the BJT is in heavy saturation.

If you want someone to explain how an opamp circuit with BJT operates then you need to choose a circuit that works correctly.
 
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  • #5
uart
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Ok likephysics, I've just gone over that circuit and it may work as advertized if you interchange the +ive and -ive inputs to the op-amp. I know that might seem counter-intuitive to you but you really need to do that to get -ive feedback.

Yes the circuit will then work, provided that the opamp has a suitable common mode range (aka input voltage range). In this case the opamp needs to function with input voltages within about 0.1 volts of the positive rail (most opamp wont handle this, I'll have to check the data sheet for the LM107).

Edit. Yes ok this opamp is designed to operate with input voltages very close to the positive rail, so yes it should work if you interchange "+ and -" inputs as I said above.
 
  • #6
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Yes I figured that's what it was supposed to be, but that is not what it is (it doesn't work).

In that circuit the opamp non-inverting input is about 22.3 volts, the inverting input is about 20.5 volts, the opamp output is at it's positive saturation level (probably approx 21 volts) and the BJT is in heavy saturation.

If you want someone to explain how an opamp circuit with BJT operates then you need to choose a circuit that works correctly.
uart, this circuit works. I have it on the breadboard working for about a week now. Instead of the L107, I have used OPA2137 and LM358.
With LM358 the +Ve powersupply was 24V.
 
  • #7
uart
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uart, this circuit works. I have it on the breadboard working for about a week now. Instead of the L107, I have used OPA2137 and LM358.
With LM358 the +Ve powersupply was 24V.
Then you most certainly have the opamp + and - inputs connected opposite to what you have shown on the schematic. Go double check your breadboard, I guarantee it.
 
  • #8
uart
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I have used OPA2137 and LM358. With LM358 the +Ve powersupply was 24V.
The OPA2137 will work fine in this circuit as it's common mode input range extends all the way up to the positive rail. It is a very good choice for this circuit.

The LM358 is a poor choice here as the opamp inputs should stay about 1.5 volts clear of (below) the positive rail, a condition that is clearly not meet by this circuit. Regardless or whether it works or not it is operating outside it's design specifications. It is not suitable for this circuit with the resistor values as shown.
 
  • #9
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Then you most certainly have the opamp + and - inputs connected opposite to what you have shown on the schematic. Go double check your breadboard, I guarantee it.
uart, it works as is without interchaning + and - terminals.
With the + and - interchanged, it still is a current monitor. But that's not what I have on the breadboard.
If you can explain either one, It will still be useful.
 
  • #10
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The OPA2137 will work fine in this circuit as it's common mode input range extends all the way up to the positive rail. It is a very good choice for this circuit.

The LM358 is a poor choice here as the opamp inputs should stay about 1.5 volts clear of (below) the positive rail, a condition that is clearly not meet by this circuit. Regardless or whether it works or not it is operating outside it's design specifications. It is not suitable for this circuit with the resistor values as shown.
I already mentioned with the LM358, the +ve supply is 24V.
Voltage to the load is 22.5v.
 
  • #11
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  • #12
uart
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uart, it works as is without interchaning + and - terminals.
No it does not. As I said, go double check your breadboard and double check your opamp pinout. You certainly have it reversed to what you've shown. It CAN NOT work with the + and - opamp input as shown in your schematic.

If you can explain either one, It will still be useful.
- As the load current increases the voltage at the opamp INVERTING input (junction of R5 and RL) falls.

- This causes the opamp output voltage to increase and hence the base (and emitter) current of Q1 to increase.

- Since the BJT is operating in it's active region the collector current is approximately equal to the emitter current, so Ic increases and hence the voltage at the NON-INVERTING opamp input decrease. This provides the negative feedback

- With the negative feedback operating as outlined above the circuit reaches balance when the voltage drop across R1 is equal to the voltage drop across R4+R5.

- At this point the collector current is equal to 10^-4 times the load current and the emitter current is approximately the same.

- Hence the output voltage, which is 10^4 times the emitter current is numerically equal to the load current (that it, the trans-resistance gain is unity).
 
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  • #13
uart
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I already mentioned with the LM358, the +ve supply is 24V.
Voltage to the load is 22.5v.
Yes you mentioned that the +ive voltage was 24volt but you DIDN'T mention that the voltage to the load remained at only 22.5 volts. Obviously that allows it to operate within the correct common mode input voltage range.
 
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  • #14
uart
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What, that's a completely different circuit! It uses a pnp (darlington) transistor instead of an npn as you have shown. And if you look you'll see that the positions of the collector and emitter are reversed in that circuit compared to the one which you posted previously. That makes a LOT of difference.

Obviously before you can understand how opamp circuits with BJT's work you need to understand the basic operation of the BJT including such things as the difference between a PNP and and NPN transistor and the difference between the collector and the emitter ok.
 
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  • #15
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What, that's a completely different circuit! It uses a pnp (darlington) transistor instead of an npn as you have shown. And if you look you'll see that the positions of the collector and emitter are reversed in that circuit compared to the one which you posted previously. That makes a LOT of difference.

Obviously before you can understand how opamp circuits with BJT's work you need to understand the basic operation of the BJT including such things as the difference between a PNP and and NPN transistor and the difference between the collector and the emitter ok.
Did you happen to look at the LM358 datasheet?
 
  • #16
uart
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Did you happen to look at the LM358 datasheet?
Yes, on page 13 it has the circuit you originally posted, BUT WITH THE + AND - INPUTS TO THE OP AMP THE OTHER WAY AROUND - EXACTLY AS I'VE BEEN TRYING TO TELL YOU ALL ALONG

Yes I'm shouting. Yes my nickname is Mr Grumpy, but you wasted a lot of my time by not posting the correct circuit to begin with and then arguing the point when you were wrong. Sorry but I'm done here.
 
  • #17
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Sorry for wasting your time.
 
  • #18
uart
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Ok sorry for shouting and dont feel bad. :redface: It's just that I could have answered the question in just a few minutes with the right circuit diagram. My reply #12 explains how the circuit on pg13 of the LM358 datasheet works.
 

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