op-amp with unregulated supply

wakoko79 · Jun 14, 2012

Hi.. My question is about op-amps with unregulated supply.. the op-amp will be used on a linear power supply.. I haven't done it but I have doubts. Will the performance of the op-amp be influenced by a fluctuating supply?

For an example:
https://ludens.cl/Electron/Ps20/Ps20.html

Are there specific words that I have to look for when choosing op-amps for this application?

Thanks

ericgibbs · Jun 14, 2012

hi,
Look at the 'boxed' section of the 741 d/s, note the Supply Voltage Rejection values.

E.

wakoko79 · Jun 14, 2012

ericgibbs said:
hi,
Look at the 'boxed' section of the 741 d/s, note the Supply Voltage Rejection values.

E.

Thanks=)

I didn't know something like this exist until now.

unclejed613 · Jun 14, 2012

so the answer is "yes it will have an effect", but the PSRR will determine how much of an effect. on some op amps, the PSRR can be 100db or more (meaning any power supply ripple will be 100db smaller than the signal you are amplifying)

ronsimpson · Jun 14, 2012

unclejed613 said:
so the answer is "yes it will have an effect", but the PSRR will determine how much of an effect. on some op amps, the PSRR can be 100db or more (meaning any power supply ripple will be 100db smaller than the signal you are amplifying)

The PSRR is very frequency dependent. At DC it is good (100db). At 100khz it gone to hell. I think it starts down at about 400hz (can't find data now).

wakoko79 · Jun 15, 2012

power supply question

I found this site:
https://ludens.cl/Electron/Ps20/Ps20.html

I tried simulating this circuit for 24V with a max output of 6A. It was pretty good.

I already understand how the op-amp part works. But I'm having some question about Q2's operation.
Is this really a LINEAR power supply?
About Q2, why is it a Darlington?

Aren't the pass transistors operated only at cut-off and saturation?

Please explain this to me. The site is outdated and there is no Q&A part there so I'm asking help from you guys now.

Thanks again!

ronsimpson · Jun 15, 2012

wakoko79 said:
I already understand how the op-amp part works. But I'm having some question about Q2's operation.
Is this really a LINEAR power supply?
About Q2, why is it a Darlington?
Aren't the pass transistors operated only at cut-off and saturation?

The 741 can only output mAs. Q2, darlington is used to boost the current. the 2N3055s need a large amount of base current.
This is a linear power supply.
None of the transistors are in saturation.

wakoko79 · Jun 18, 2012

ronsimpson said:
The 741 can only output mAs. Q2, darlington is used to boost the current. the 2N3055s need a large amount of base current.
This is a linear power supply.
None of the transistors are in saturation.

Thank you sir.
Ahh, so the base currents are high enough. I didn't notice it since I used a comparator instead:
View attachment 64990

But still, (in the original circuit) the pass transistors are configured as CE amp. But the op-amp that is controlling the base driver (Q2) only has 2 states (saturation right?) high or low. So how would it drive the pass transistors in the linear region? This is why I saw the circuit earlier operating only in cut-off or saturation.

ronsimpson · Jun 18, 2012

wakoko79 said:
But still, (in the original circuit) the pass transistors are configured as CE amp. But the op-amp that is controlling the base driver (Q2) only has 2 states (saturation right?) high or low. So how would it drive the pass transistors in the linear region? This is why I saw the circuit earlier operating only in cut-off or saturation.

I don't think the amplifier is in saturation (on/off).
From the link you gave:

How it works
U1 provides a regulated reference voltage that's always 5V below the positive rail. U2 compares this reference voltage to a sample from the output (ground against positive rail) and drives a Darlington transistor connected as emitter follower, which in turn drives the four pass transistors connected in common emitter configuration. Four resistors equalize the current through the transistors, and one of these resistors does double duty by serving as current sense resistor. If the current through this resistor exceeds about 6A, then Q1 will start conducting, swamping the drive from U2 to the negative rail and thus limiting the output current.

No parts were added to control frequency response, loop damping, etc. All trust was placed on the 741's rather low frequency response and high stability, combined with a 1000µF capacitor across the output. In practice this has proven to work well enough, but purists may want to experiment with the loop response and add some compensation capacitor.

Normally in a power supply a capacitor is from the output of the amplifier to the (-) input to slow down the response. The last paragraph above talks about the 741 is so slow this is not needed. I think the 741 and all the transistors are in linear mode. The silicon is so slow that the time from ON to OFF and from OFF to ON is so long that the transistors spend most (if not all) of their time in the land between ON and OFF.

unclejed613 · Jun 18, 2012

ronsimpson said:
The PSRR is very frequency dependent. At DC it is good (100db). At 100khz it gone to hell. I think it starts down at about 400hz (can't find data now).

usually that information is shown in a graph in a data sheet. the turnover frequency for PSRR (i.e. where it starts dropping) is usually the same as the turnover point for the open loop gain.

wakoko79 · Jun 18, 2012

ronsimpson said:
I don't think the amplifier is in saturation (on/off).
From the link you gave:

Normally in a power supply a capacitor is from the output of the amplifier to the (-) input to slow down the response. The last paragraph above talks about the 741 is so slow this is not needed. I think the 741 and all the transistors are in linear mode. The silicon is so slow that the time from ON to OFF and from OFF to ON is so long that the transistors spend most (if not all) of their time in the land between ON and OFF.

So the op-amp needs to be slow? But my idea is the opposite. I think the op-amp should be fast so that it can react to the voltage fluctuations. Then the silicon (pass transistors) should be slow so that, like what you said, it would stay on the linear region.

The original author also said that the 0.1 Ohms are there as equalizing resistors. But will that really be enough to assure that thermal runaway wont happen? In the old thread I started (https://www.electro-tech-online.com/threads/simple-linear-power-supply-design.126416/) I added current limit circuitry, which is not very good since it just passes the current load from the pass transistor back to the lm317 which has internal current limit. The circuit now indeed has current sense, but only from one leg of the pass transistors though.

op-amp with unregulated supply

wakoko79

New Member

ericgibbs

Well-Known Member

wakoko79

New Member

unclejed613

Well-Known Member

ronsimpson

Well-Known Member

wakoko79

New Member

ronsimpson

Well-Known Member

wakoko79

New Member

ronsimpson

Well-Known Member

unclejed613

Well-Known Member

wakoko79

New Member

Similar threads

op-amp with unregulated supply

New Member

Well-Known Member

New Member

Well-Known Member

Well-Known Member

New Member

Well-Known Member

New Member

Well-Known Member

Well-Known Member

New Member

Similar threads

Privacy & Transparency

Privacy & Transparency