1-25 volt power supply

[Kevin Weddle]

This might be an easier to use power supply than the other one.

View attachment 42066

 

[audioguru2]

It is almost the same as the other circuit.
Why does it have two opamps? One opamp has a gain of about 200,000 which is plenty of gain for it to be the error amplifier. One opamp will need to be a non-inverting amplifier.

Also, the simple circuit has nothing to limit the output current.

 

[Hero999]

The circuit will give poor regulation because the gain is low.

Use a single op-amp and put the pass transistor in the feedback loop.

View attachment 42068

 

[audioguru2]

Hero, you have the darlington base shored to its emitter.
It should be like this:

View attachment 42070

 

[Hero999]

Yes, you're right. The inverting input needs to be connected to the output so the Darlington pair is in the feedback loop.

 

[Kevin Weddle]

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

 

[audioguru2]

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.

 

[Kevin Weddle]

I know that circuit. The gain must be at least be 1. This circuit has a gain of 6 which has an obsolete voltage divider. Unless you want to get rid of something else.

 

[Hero999]

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.

View attachment 42072

 

[audioguru2]

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.

 

[Kevin Weddle]

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.

 

[audioguru2]

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.

 

[Kevin Weddle]

Okay nobody likes two opamps. Are you going to put the load here or there. A gain of 100 is better unless it's a power supply for microwaves.

 

[Hero999]

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.

 

[Kevin Weddle]

That's true. And you can set the gain at 10, 100, or 10,000 and still have the exact same ouput.

 

[audioguru2]

That's true. And you can set the gain at 10, 100, or 10,000 and still have the exact same ouput.

No Kevin.
The voltage reference IC is 2.5V. Then for 25.0V output the gain must be exactly 10, not 100 and not 10,000.

 

[Kevin Weddle]

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.

 

[Hero999]

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.

 

[Kevin Weddle]

Yes, maybe the transistors should be in the negative feedback loop of the opamp. But this circuit has to have similar function over much of the operating range.

I know the impedance of 200k looks higher than the transistor impedance. But the voltage is relatively the same.

View attachment 42109