What does this op amp/transistor circuit do?

Question

Consider this circuit:

^{simulate this circuit – Schematic created using CircuitLab}

It's not clear to me what this circuit does. Here's what I think I understand:

• Depending on the position of the pot, the voltage at the non-inverting input is between 9V and 10V
• Because of negative feedback, the voltage at the collector of the transistor is also between 9V and 10V (or is it? It's not clear how the second power supply input is affecting the feedback cycle).
• The current through the load will be at most 10 A (perhaps this is a current source?).

I'm a beginner and these things are very confusing to me. In addition to a theoretical understanding, I'm also interested in how real (ie. non-perfect) op amps might behave differently and the tradeoffs involved in picking the right op amp for this circuit.

Answer 1

The circuit is broken because it mistakenly implements positive feedback, where the intent is to have negative feedback. The more resistance you dial in with the potentiometer, to raise the voltage on the + input, the more the MJE transistor conducts. This causes current to increase through R3, which lowers the voltage at the bottom of R3. This lowered voltage is fed back to the - input of the op-amp, and has the effect of increasing the output!

In effect, it's a double negative: R3 is the load resistor of an inverting gain stage, and this inverted output is fed to the inverting input of the op-amp. Inverted output fed to an inverting input is positive feedback.

The proper approach to take the feedback signal simply from the output of the op-amp (classic voltage follower/buffer). This buffer then simply drives the base of the transistor, implementing a classic emitter-follower to increase the current driving ability. R3 is not necessary.

Alternatively, the feedback signal can be taken from the node where the transistor's emitter meets the load (top of the load). That topology will then eliminate the VBE voltage drop, since it moves the VBE drop into the feedback loop. The consequence is that the load voltage will then closely follow the voltage on the op-amp's + terminal, rather than be a VBE drop lower.

Concretely, here is a version of the circuit modified with the above feedback topology, and also cleaned up to simulate. I got rid of the series resistor from the potentiometer so we can vary the voltage from 0 to 10V (by varying the "k" parameter of the pot from 0 to 1). The role of the load is played by R4.

^{simulate this circuit – Schematic created using CircuitLab}

If we move the feedback line to the original location, then the DC simulation shows that the circuit latches up with the transistor open.

Even if R3 is made substantially larger, the hypothesized positive feedback action doesn't happen precisely as described, but rather as follows: When power is applied, no current flows through R3, and so the - input of the op-amp is held at the 10V power rail. The + input cannot rise above this, and so the op-amp output is driven low. In this manner, the positive feedback keeps the transistor in a cut-off state. If we turn the potentiometer all the way up, it's more interesting: then the inputs are both nominally at 10V. The actual behavior will depend on their precise values, which are determined by bias currents and leakage through the cutoff transistor.

Answer 2

In order to analyse an op-amp circuit as an amplifier, certain assumptions must be made. One of those assumption is that negative feedback is present.

But, in this circuit, negative feedback is not present.

To see this, imagine that, for some reason (noise or some other disturbance), the voltage at the output of the op-amp increases (place an up arrow at the output of the op-amp).

Now, if negative feedback is present, the effect of this disturbance will be to make the voltage at the output of the op-amp to decrease. If this is not so, you do not have negative feedback.

So, trace this disturbance 'round the loop. Since the output of the op-amp is connected to the base of the transistor, the increase in base voltage will result in a decrease in collector voltage (place a down arrow at the collector).

The inverting input of the op-amp is connected to the collector (place a down arrow at the inverting input of the op-amp).

But, since this is the inverting input, a decrease in voltage there results in an increase in the op-amp output voltage.

In summary, a disturbance that increases the output voltage of the op-amp will result in feedback that tends to amplify that disturbance. This is the sign of positive feedback.

So, this circuit is not stable; any disturbance will be amplified rather than attenuated. I don't believe this is the desired operation. It's likely that the inverting input of the op-amp should be connected to the emitter of the transistor.

Answer 3

Looking at the circuit, I would think that by swapping the + and - inputs of the opamp you get a current source. Haven't tried the circuit though!

To analyse stability, use a thought experiment: what happens when I raise the voltage on the plus input. As a result, the output of the opamp will be raised as well. If this raise leads to a raise on the - input, the circuit can be stable, as the difference between the inputs is reduced by the action of the opamp.

In the original circuit, a raise on the + input leads to a raise on the output, which leads to more current through the transistor, which leads to more drop over R3, which leads to a drop in the - input. That circuit is unstable. Swap the inputs, and you have a nice current source.

For currents near zero, the opamp must be capable of handling inputs near the rail voltage, not all opamps are capable of this.

Answer 4

The circuit as drawn is a threshold detector.

It is component specific as there is a race condition at power on that might go either direction depending on the specific components and possibly even the layout. To make it start more reliably, put a small capacitor across BC.

The circuit has no reset as drawn so it can only be triggered once per power cycle. A N.C. switch in the feedback would allow resetting.

If the load is more than about an amp, the resistance has to be adjusted.

Answer 5

Circuit is a VCCS, load current is derived from the OP Amp ref input Voltage divided by the Sense Resistor(R4) value, I.E. assume the input ref voltage is 5.0v (pot R1 is in mid position) then the load current is 5.0v/100ohm = 5.0mA.