Power supply spice analysis question

Thread Starter

coinmaster

Joined Dec 24, 2015
502
My understanding of feedback theory and instability is limited because my experience with them is limited.
I was designing a power supply in spice to improve on my current one and I think it's time I fully understand what spice is telling me here.
This is the spice reading of the old supply

This is the spice reading of the new one

The output impedance is improved but I don't understand what the phase reading is telling me. What does it mean exactly? Phase of what? Why/where does it matter?
 

crutschow

Joined Mar 14, 2008
23,291
Why are you concerned about the output impedance versus frequency in your power supply?
Do you have a requirement for a fast responding supply?

The phase is normally shown between the input (source) and output voltages.
In this case, I'm not sure what it represents.
 

Thread Starter

coinmaster

Joined Dec 24, 2015
502
It's for an audio supply. I want the impedance to be as low as possible between 20hz 20khz to eliminate it as a source of possible distortion. Of course the performance of the previous power supply was likely more than sufficient for practical purposes but it was a 200v supply using a 20v opamp on the high side. Too many time's have I had to repair this circuit so I built a low side version instead. which just happened to show better performance impedance-wise.
I just don't understand the meaning of the phase reading.
The phase is normally shown between the input (source) and output voltages.
In this case, I'm not sure what it represents.
I hooked them both up to a current source 300ma idle 100ma current swings and measured the the output of the regulator using Vout/ILoad to get that result. What does the phase mean here?
 

crutschow

Joined Mar 14, 2008
23,291
It's for an audio supply. I want the impedance to be as low as possible between 20hz 20khz
Normally that's achieved with a large output filter capacitor.

If you want to check the stability of the supply, then use a voltage source and plot the output voltage vs. frequency.
For stability you want the phase shift to be much less than 180° (good phase margin) when the gain crosses 0dB (gain of 1).
Any significant peaking in gain also indicates possible instability.
 

Thread Starter

coinmaster

Joined Dec 24, 2015
502
Normally that's achieved with a large output filter capacitor.
Indeed but they will never touch the size, performance, or cost efficiency of a good regulator.
If you want to check the stability of the supply
Is that the only use of the phase measurement?
then use a voltage source and plot the output voltage vs. frequency.
Not sure I follow, do you mean attach a voltage source to the output of the regulator?
For stability you want the phase shift to be much less than 180° (good phase margin)
So by this logic the result of the second picture I posted is much more stable than the first?
when the gain crosses 0dB (gain of 1).
Gain of what? What sort or gain are we talking about?
 
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Thread Starter

coinmaster

Joined Dec 24, 2015
502
Oh okay so as long as gain doesn't cross 0db and 180 degree phase shift at any frequency it is stable right?
As an example this measurement is stable to 1Mhz?
 
Last edited:

crutschow

Joined Mar 14, 2008
23,291
As an example this measurement is stable to 1Mhz?
No.
Your supply is showing a gain of -180dB (Av = 1E-9), which is essentially no gain (the same as if the supply was not powered).
That's likely because the supply is biased off with no load.
In order to do the test, the power supply must be supplying power to a load with all active elements conducting current.
An AC voltage source is added at the amplifier input with an AC voltage of one.

Normally the gain will be much greater than 1 (0dB) giving a positive dB number.
A negative dB number is a gain less than 1.

If you are simulating in LTspice and upload the .asc file here, I can try to run it and check the stability.
 

Thread Starter

coinmaster

Joined Dec 24, 2015
502
The output of the supply is connected to a 300ma current source. It's definitely not unloaded.
Although I put the AC1 on the current source not on the input.
Putting AC1 on the input doesn't really change it much though.
 

crutschow

Joined Mar 14, 2008
23,291
Here's an open loop Bode Plot simulation.
I added C7 and R12 to open the AC loop (but allow DC to pass to stabilize the DC bias values).
I then injected an AC signal using V2 through C8.

The response shows a 35db gain peak at about 109Hz with a sharp change in the phase-shift.
Even though the phase doesn't go to 180°, this large gain peak and fast phase change indicates there is a possible instability in the circuit.
So compensation should be added/changed to minimize that peak.

Since that design is a rather odd configuration, I'm not sure where the problem is but it may be related to the values of R6, R7, R8, C4, and C6.

upload_2018-4-6_13-57-3.png
 

Thread Starter

coinmaster

Joined Dec 24, 2015
502
So your supposed to cancel out the feedback loop and feed AC1 into the opamps input? How are you supposed to get the performance characteristics if you shunt the AC feedback into ground?
 

crutschow

Joined Mar 14, 2008
23,291
So your supposed to cancel out the feedback loop and feed AC1 into the opamps input? How are you supposed to get the performance characteristics if you shunt the AC feedback into ground?
As I previously stated, it's an "open-loop" test to determine the response of the loop.
Open-loop means you put a break in the AC feedback.

You can also do a closed-loop test by putting the AC source in series with one of the op amp inputs without breaking the loop, but that generally doesn't tell you as much about the loop stability margin.
 
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