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1-25 volt power supply

Kevin Weddle

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Yes the circuit can be done. The problem is the gain has to be at least one. But not more.

There is no problem with our fixed circuit. The output of the circuit is connected directly to the inverting input of the opamp so the output of the circuit has a voltage exactly the same as the voltage at the non-inverting input of the opamp. Then the gain is exactly 1.
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Your circuit uses more components and is inferior.

It doesn't have to have a gain of 1. It's possible to use a higher gain.

See the circuit attached which has a gain of 10, enabling a more accurate 2.5V reference to be used, rather than a zener diode. It also has current limiting and a capacitor is connected to the output to improve the transient response.

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Hero99, nobody can read that dumb circuit.

No Kevin. It is a perfectly normal power supply circuit with good voltage regulation and it has current limiting.

1) Rz powers the LM385-2.5 which is a good 2.5V voltage reference IC.
2) The opamp drives a TIP121 darlington transistor rated at 5A as a follower.
3) R1 and R2 are a voltage divider providing negative feedback to the opamp-darlington so that their voltage gain is (180k/20k) +1= 10 times so that the output is 25.0V.
4) TR2 senses the output current in Rs and begins to turn off the darlington transistor when the output current exceeds about 0.7V/0.47 ohms= 1.49A to limit the output current.
5) C1 filters high frequency transients caused by the load.
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Yes, a gain of more than 1 is more desireable as the load current changes by a lot. A gain of 10 is very fine.

If the darlington output transistor was not included in the negative feedback loop of the opamp then its output voltage world drop about 1.4V when the load current is increased from almost nothing to 1.5A. The current limiter reduces the output voltage an additional 0.7V for a total reduction of 1.4V + 0.7V= 2.1V.

But the opamp has an internal voltage gain of about 200,000 which is reduced to 20,000 since the total gain is 10 for correcting the voltage error. So the 2.1V error is reduced to 2.1V/20,000= 0.1mV if the circuit wiring has no resistance.

The 2.5V voltage reference was selected since it provides excellent voltage regulation, much better than a zener diode. The opamp and darlington transistor can simply amplify it up to 25V with a gain of 10.
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There's nothing special about an op-amp. It's just a differential amplifier with a very high gain.

In the circuit posted above, all the op-amp does is try to keep the voltage on the inverting and non-inverting inputs equal by adjusting its output voltage. The voltage across R2 will be nearly equal to the voltage on the LM358-2.5 which is 2.5V. R2 forms a potential divider with R1 with a division factor of 9 so the voltage across R1 is 9 times higher than R2. The voltage accorss RL is the same as the voltage across R1+R2.

Putting two op-amps in the same feedback loop is a bad idea.

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  • 1 month later...

No Hero99, that gain of 10 circuit is the same as the original project. Something simple. It is not the sort of circuit that is used.
It isn't the same. In my circuit, the output transistor is inside the negative feedback loop of the operational amplifier. In your circuit, there's negative feedback before the output transistor.

In a power supply circuit, the power element (i.e. output transistor) should always be inside the feedback loop of the error amplifier.
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  • 2 weeks later...

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