SEPIC LED driver...where inject the signal for Loop gain & phase measurement

Hello,

I have done a coupled inductor SEPIC LED driver, which regulates its own input current. We wish to do a stability analysis using an AP300 frequency analyser, so that we can get the gain and phase Bode plots for the converter.

My question is, where should we inject the perturbation signal?, (i.e. the test signal which is to sweep over the frequencies of interest in order to get the Bode Plot?)

Schematic.
https://i46.tinypic.com/jttquc.jpg

Vin is a 6V battery.

The load is any LED lamp which could be 5V up to 40V.

The average input current is set by applying a control voltage into the non-inverting input of the U7 opamp. (677mA per volt applied).

Now, in order to get the Bode plots of gain and phase of the sepic, we need to inject a signal into it, as described in the following web page)

**broken link removed**

Here is the schematic with the loop injection resistor added (R32)…

SEPIC schematic with loop injection resistor (R32)
https://i48.tinypic.com/2lo6hdl.jpg

…The criteria for the loop injection point is that one side of the injection place must be high impedance and the other side must be low impedance. (The output impedance of the opamp buffer, U17, provides the low impedance side).

So do you think that R32 is a good place for the voltage injection, for the purposes of getting the loop gain and phase?

Your circuit is not at all what is in the web page. They are regulating output voltage and you are regulating input current.

Why do you regulate input current? Which is almost the same as regulating input power. If you have a 6V LED running at 677mA and you changed to a 12V LED it will run at 338mA. And a 24V LED will run at 169mA. Why not regulate the LED current?

I think your loop gain will change when you change what type of LED. 6V or 24V or 40V

I think you should inject at a point between U17 & R16.

I think you should inject at a point between U17 & R16.

Click to expand...

..thanks, do you mean as shown above?....you dont think the 1K resistor (R16) is too low value to be considered a high impedance?

Why not regulate the LED current?

Click to expand...

.....because we want to regulate the output power no matter what voltage luminaire is attached....as you say, it regulates input power, which is almost the same as regulating output power , as we take into account the efficiency via the micro that sets the current via its DAC.

How would you assess the magnitude of the injection signal, as you know, if too big, it will be ineffective, if too small, it'll be swallowed by the noise.

You could also inject between R13 & U17. R13 is not low but U17 is very high.

I would inject a signal about 10%. So the sign wave will effect the current about 67mA. I think 1% is ok but maybe too close to the noise.

You could also inject between R13 & U17. R13 is not low but U17 is very high.

Click to expand...

OK thanks, but if the injection resistor is 22R, then isn't the 1K8 value of R13 too high?....i thought the resistance at the "Low impedance" end of the injection resistor had to have an impedance much lower than the injection resistor?

Also, i presume i would have to monitor Vout whilst injecting?, to ensure that it wasnt changing too much, indicating a too big test signal is being injected?

Also, i presume i would have to monitor Vout whilst injecting?, to ensure that it wasnt changing too much, indicating a too big test signal is being injected?

Click to expand...

You are not regulating Vout. You are regulation Iin!

R18 (1.8k) sits on ground. I would lift the ground end and inject the signal there. If the generator could push/pull on R18 a little the signal will get into the error amplifier.

A friend of mine would open U17 pin3. Then add another amplifier with a gain of 1. The output of this new amplifier will have the same voltage as R18 but with a low impedance. Now add your transformer from the low impedance output to U17 pin3.

R18 (1.8k) sits on ground. I would lift the ground end and inject the signal there.

Click to expand...

..it sounds good, but this doesnt break the feedback path does it?

The extra amplifier sounds the best idea, but needs more room on the PCB.....but i reckon that is the golden solution.
Why would it be an amplifier witha gain of 1, -couldnt it just be a follower , or is that what you mean anyway?

The fact that you suggest these other solutions suggests to me that you think that R16 (1K) is simply not a high enough impedance for the injection point to be the R32 resistor (22R)?

Does not break the feedback. Just adding a gain of 1 should not effect things much.

An emitter follower drops the voltage by 0.7 volts. That is a problem. Also a emitter follower only pulls up not down.

We had a 2 opamp board made. The first opamp is to buffer the signal. Then comes the injection transformer and then another buffer just in case we needed to drive a low impedance next stage.

This type of testing is good but I don't do it often. I usually just do a step response. I inject a step into the reference or the error amplifier. I am looking for a step up and down of 5% and I want a fast response with no hint of ringing. By varying the square wave frequency I can see how the amplifier is working. This is looking at "time" and not "frequency" but can tell you the same information.

sorry i did not mean emitter follower, i meant an opamp follower.

Does not break the feedback

Click to expand...

....but i thought when injecting a voltage into a feedback loop to measure it, you had to actually break the feedback path?....indeed many other of your suggestions do involve actually breaking the feedback path.

This type of testing is good but I don't do it often. I usually just do a step response.

Click to expand...

..OK but what do you think of the article "Transient response and loop gains of power supplies" by Dr Ray Ridley.?...it can be found on the web at switchingpowermagazine.com but you have to log in etc. ( i have attached it)

This article says looking at the transient response is not good enough, here is a short exerpt....
"
Many old-timers in the industry claim that
they can see all the characteristics necessary
by just looking at the transient response, and
that there is no need to make loop measurements
at all. This misconception can often lead
to expensive errors in design, long and expensive
time delays in product development and
instability in the field.
Because I hear this claim often, I decided to
set up some test cases. Through these test cases,
I will illustrate how much trouble it can cause.
"

"Breaking the error amplifier loop " does not mean breaking open. If the loop is open the supply will run away and break something. The loop must continue to work. You need to add in a small signal with out changing the DC of the loop. I ofter AC couple the signal in using capacitors not a transformer.

Dr Ray Ridley: I am a "old-timers" that makes supplies by experiences. If this was my first or tenth supply I would use a network analyzer.

A friend of mine would open U17 pin3. Then add another amplifier with a gain of 1. The output of this new amplifier will have the same voltage as R18 but with a low impedance. Now add your transformer from the low impedance output to U17 pin3.

Click to expand...

This sounds good....very low impedance at one side of the injection resistor, and very high impedance at the other side,..however, doesn't one side of the injection resistor have to be the parameter being controlled (in this case the input current).....

and indeed thats impossible to arrange, so maybe its just not possible to use a frequency analyser to measure the gain and phase margins of this sepic led driver.?

Usually, the AP300 style frequency analyser technique for getting the bode plot, is done with an output voltage regulated smps, and the injection resistor is indeed connected to the vout (the parameter under control)
...in your above quoted method, we are not connected directly to the parameter under control (the input current), so is this method still valid for getting bode plots of this sepic led driver?

You are regulating the voltage at the out put of the current sensors.