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Is this loop injection resistor OK for SMPS feedback loop measurement?

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Flyback

Well-Known Member
Hello,
In the below schematic, is R22 sufficient as an injection resistor?, -so that the gain and phase of the feedback loop can be measured with a frequency analyser?

The following article states that the impedance seen looking from either side of the injection resistor should be high one way and low the other.

Page 1, second text block down...
**broken link removed**
Presumably, R10 being 1k , can be considered much greater than the 50 Ohms of the sense resistor , R6.?
 

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  • SEPIC LED driver 50mA.pdf
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Hi,

That seems to look ok. The main idea is to be able to add the control signal to the feedback signal, and have the control signal remain known which would not happen if it could change relative to the actual output.
So in other words, the left side should be completely changeable through the controlled test signal, while the right side should be not changeable due to the controlled test signal directly. It will of course be changeable indirectly due to the added feedback signal, but that's what we want. We just dont want it to be able to change due to the injected signal directly or else the measurement of the injected signal relative to the test procedure is not accurate (of course it will be accurate relative to itself but alone that's not enough).
 
Actually i've seen that my above has a mistake, it fails to account for R11 and C7.

So is the following change OK....? (new schem and sim attached)

... is R14 going to be OK for use as a loop injection resistor for feedback loop measurement? (attached new , adjusted schematic & LTspice simulation).
Also, the opamp, U2 , is included purely for ensuring that the feedback loop measurement can be successful. (low impedance of opamp output is needed)
That is, R14 and U2 (opamp) are added so that the gain and phase of the feedback loop can be measured with a frequency analyser.

Is this going to be ok for feedback loop measurement (gain and phase margin)?
 

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  • SEPIC LED driver 50mA -loopinje.pdf
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  • SEPIC LED driver 50mA -loopinject.txt
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Hi,

[See below "EDIT" before replying to this post]

Are you saying that you added an op amp to the circuit along with a small "injection" resistor?
I am not sure you can get away with that, even though the op amp is a unity gain amplifier. That's because the network function is changed when you add a unity gain amplifier between two successive stages of any network made of resistors and capacitors. It depends partly on the values and the frequencies of interest but looking at the 5.1k resistor and the 22k resistor, 5.1k is too large of a percentage of 22k (about 25 percent roughly) which means adding a unity gain stage probably changes the dynamics too much. It would also depend on the 220nf capacitor but that sounds large enough where we would still think about comparing the two resistors directly.
In other words, the response of the entire circuit could be very different with the unity gain amplifier in there. If it were not for that 5.1k resistor and 220nf capacitor (part of the 22k network) i would not be so hesitant, but because the three make up a network i would not feel comfortable at all using that op amp like that. Now if it was on the other side (right at the 50 current sense resistor) i might not feel as bad about it because 50 ohms (sense resistor, not injection resistor) is a small percentage of 5.1k and 22k.

The whole idea is to be able to inject a signal that "rides" on the normal signal without altering the circuit too much. The more you alter the circuit the less accurate the results are, and in this case, a unity gain amplifier probably modifies the circuit way too much. Not only does it modify the network transfer function because of the unity gain isolation property, it also introduces other frequency altering components which may or may not affect it as much but still has to be considered.
You could test though, doing it both ways and see if you can measure any difference in the simulation.

In case this doesnt make enough sense, to understand this concept better you could look at a passive two stage RC low pass filter. Look at the response when the two stages have equal values for the resistors and capacitors, then add a unity gain amplifier BETWEEN the two stages and notice how the response changes drastically. The impedance presented to the input of the second stage gets much lower, and the impedance presented to the output of the first stage gets much higher, which changes the circuit drastically.

EDIT:
I just double checked and i see that the 5.1k resistor is not 5.1k it is 5k1, which means it is a 5.1 Ohm resistor. Somehow i read that resistor value wrong with the first look.
This presents a different problem because 5.1 Ohm is a very small value. This means that the unity gain amplifier will not change the dynamics as much as if it was really a 5.1k resistor.
This also means that the ideal way to drive this is with a zero ohm source injected between the 50 ohm sense resistor and the 5k1 network. That would probably mean a low impedance transformer with the secondary in series with the 50 ohm sense resistor and the 5k1 network.

If you do it any other way you risk altering the results, but lucky for us today we have the simulators so you could test it both ways and that should tell you if the unity gain amp is ok to use that way. In the simulation you can use a very very small value resistor or even just a sine source in series with the 50 ohm sense resistor and the 5k1 resistor network. Try it both ways and see if you can measure any difference in the gain and phase.

This turned out to be a little more difficult because we dont really have a high impedance on the left side of the 50 ohm sense resistor unfortunately. We need a source with an impedance much less than 5.1 Ohms to do this right.
 
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thanks, i'm still going though your thorough answer, though I have simulated with and without the opamp buffer and it makes no difference to the loop response, as seen by doing transient tests in the simulator.
I had intended to put the secondary of the isolation transformer that comes with the AP300 frequency analyser across the 50R injection resistor, R14, and just inject the signal like that?

Also, if I write "5K1", then to me that means 5.1K = 5100R
 
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EDIT:
I just double checked and i see that the 5.1k resistor is not 5.1k it is 5k1, which means it is a 5.1 Ohm resistor. Somehow i read that resistor value wrong with the first look.

I agree, 5k1 or 5K1 both mean 5100 Ω . 5R1 means 5.1 Ω.

0100 is too late to be working. ;)

John
 
Thanks, by the way, you know who replied me by email and pointed out the problem with my loop injection resistor site of the top post?.....Dr Ray Ridley himself!!!!
-the reply he gave me was very brief, I assume he is a very busy Guy so was amazed he got back to my private email to him.
 
Hi,

You guys are certainly right, so i was right the first time when i thought it was 5100 Ohms. I dont know why i thought it was 5.1 Ohms all of a sudden <smile>.

So with that resistor being 5100 Ohms that means the op amp unity gain amp *could* have an effect, but if it does not then that's just fine. A simulation using various frequencies will tell the tale (not just one test frequency).
 
Hi Flyback,

Yes i see that, but that is different than the way you did it in your circuit so there comes a question of whether or not it will alter the circuit too much. The way they did it was they put the amp right on the sense resistor, while you put the amplifier downstream after some other components. That's a world of difference. If you get lucky and it works out the same then all is well, but without knowing that ahead of time you have to assume that it is not the same. Theoretically though it is definitely not the same, just as their illustration is not *exactly* the same, but they both may work if the ratio of the impedances is favorable.

A unity gain amplifier changes the impedance of both the preceding stage as well as the following stage. That can easily change the frequency response both the phase and amplitude. That's why it is so important to do it a certain way only.

Note that in their circuit it is not exact either. They are banking on the sense resistor being small as compared to the circuit to the left of the amplifier (on the output of that amplifier). If certain conditions like this are not met it will provide different results. But again since we have the option to simulate, that's a good test i think as long as we test with various frequencies and note that they all give a similar response no matter how we implement the test. In the previous circuit the presence of the capacitor will mean that lower frequency responses may match pretty well but higher frequency responses may not due to the fact that the 5100 Ohm resistor starts to become more significant then.

So the bottom line is test it, but test it completely before assuming it will provide the same response.
You know what the best method is (zero impedance source in series with the output where the output already has a low impedance and the impedance of the feedback network to the left of that is relatively high). So test using a real zero ohm source and then with the op amp over several ranges of frequencies. That should be a pretty good test to do.

If i get a chance i'll show you a simulation of how much a unity gain amplifier can change things in a previously all passive network.
 
Hello,
The Ridley literature states that the injection needs to be in the loop, and that is the case here. Ill dig out the Ridley literature. As long as the injection point, if considered to be a the site of loop breakage, it breaks the entire loop(s), then its ok.

Figure 8 of the following shows how their are various places where the loop can be broken in order to get the feedback loop measurement...
**broken link removed**
 
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Hi,

Yes but we were not talking about 'breaking' the loop alone, we were talking about breaking the loop AND using a unity gain amplifier. So i dont see what that pdf has to do with using a unity gain amplifier in the loop.

What i was saying before is if the unity gain amplifier changes any impedance too much then it is not a good idea unless you can add something to the circuit to restore the original impedances as presented to both parts of the circuit (left side and right side of the unity gain amplifier).
I'll see if i can dig something up on this myself as soon as possible.

This is really a simple concept though that in it's more basic form is just about inserting things in a feedback circuit that were not there in the original design. If we insert a resistor for example we change the DC output voltage. If we insert a resistor and capacitor we change the DC and AC response. So we have to be careful what we insert in the loop, that's all this really is about.

I'll try to get back here soon with a circuit and illustration.

LATER:
The diagram shows the responses for two different circuits. The top section is one circuit, the bottom section the other.
The top graph in each section shows the amplitude responses (blue), the bottom graph in each shows the phase responses (red).
So there are two amplitude responses for each circuit and two phase responses for each circuit.

One amplitude and one phase response is for the circuit without a unity gain amplifier inserted, and the other amplitude and phase response is for the same circuit with the unity gain amplifier inserted.

As can be seen, the top circuit did not change to significantly, although it did in fact change. But the bottom circuit changed quite significantly.

So the conclusion is that sometimes the circuit changes significantly and sometimes it does not change significantly although it does change somewhat. It all depends on the impedances involved. But because it *could* change, we have to assume that it does change significantly unless we can prove that it does not change at all.
 

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