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.
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.
การแปล กรุณารอสักครู่..
