Electronics > Projects, Designs, and Technical Stuff

DIY Injection Transformer for Power Supply Control Loop Response Measurements

<< < (4/11) > >>

2N3055:

--- Quote from: AcHmed99 on October 10, 2016, 05:08:59 pm ---
--- Quote from: 2N3055 on October 10, 2016, 04:51:04 pm --- AcHmed99,

I think you didn't quite understand how it's done..

You don't inject current into PSU output.. You take PSU, you break feedback loop and insert injection transformer in series with feedback loop.. So it gets amplified by error amplifier, not suppressed......

--- End quote ---

The injection is done at the output top of the Rdiv string through what is commonly refered to as an injection resistor 22 oHms or so.

Huh? The opamp has high rejection at low frequency so it will attenuate (by a factor of 40-100dB) the disturbance which is what is shown in Basso book and in the literature with FRAs. If the opamp is actually amplifying the disturbance at low frequency (DC) then your supply has problems.

--- End quote ---

I thank Jay_Diddy_B for explaining it, my English is failing me sometimes (most of the time  :-DD).. I wanted to say that injected signal gets superimposed on top of normal feedback voltage (vector added to be more precise).. As Jay_Diddy_B explained, result is modulation of output, using feedback amp and pass transistors (let's presume linear PSU for a moment) as amp with unity gain..
 For audio enthusiasts A class amp  :-DD

Floyo:

--- Quote from: diyaudio on October 10, 2016, 05:09:19 pm ---   

--- End quote ---

I think the specific core I used is a TDK RM-8 in H5C2 material, the data can be found in this pdf https://product.tdk.com/info/en/catalog/datasheets/ferrite_mz_tl_rm_en.pdf
the AL is pretty high. I don't know exactly how many turns it has, but going from the 18H primary inductance and the specced AL that comes to around 1000 turns. The secondary being about 26db down in the plot gives a pri/sec ratio of around 1/19.5 so that makes the secondary around 50 turns. Wire size is "really thin" and airgap is "0". I hope that helps a bit, I didn't arrive at this transformer by any maths, just experimentation, and luckily the 1000 turn primary was already done when I found the thing in my junk bin :).

2N3055:
The attenuation part..

What you said is perfect if you take low impedance signal generator and just modulate output.. Regulator "fights the change" and will negate your disturbance, and will suppress it as a part of regulation.

If you inject into the loop, you simply modulate  feedback signal , and by measuring frequency and phase response to that signal (minus DC of PSU) you are basically measuring it for PSU loop..

I apologize if there is misunderstanding...

Jay_Diddy_B:
Hi group,

I have put together an LTspice simulation to illustrate how this works. By studying an op-amp circuit we easily determine if we have the correct answer.



In the model I have used an op-amp to amplify a 10V reference to 20V a gain of two. I have included a network analyzer in the model, connected between nodes A and B. The LTspice will measure the loop gain, in the time domain, the same way as if I used my HP3577A in the lab. There is no need for an isolation transformer, because in SPICE the voltage source is floating.

If we start at node B and go clockwise around the loop. R1 and R2 are a divider, divide by 2 or -6dB. It has an output impedance of R1 in parallel with R2, R1//R2, = 10k.

The divider resistance combined with R3, 40k, has a gain of 4x or +12db. The total gain in resulting from the resistors is -6dB + 12dB = 6dB.

There is a low pass filter caused by C2 and R3

F= 1/2 x Pi x 1600pF x 40,000 = 2.5kHz (-3dB)


There is a zero at the frequency caused by R3 and C1

F= 1/2 x Pi x 0.16uF x 40,000 = 25 Hz (+3dB)

The results from LTspice are:



If we look at the relative size of the signals at A and B we see:



The units are dBV, that is dB relative to 1V rms.

And if we look at them in the time domain, at one frequency, 316 Hz we see:



The vector sum of the signals A and B is always equal to the injected signal.

If this works for this example, it will work for more complex circuits too, where it is difficult to know the right answer. :D


I have attached the LTspice model.

The model could be expanded by adding the transformer model between the source Vac1 and points A and B.

Regards,


Jay_Diddy_B


MagicSmoker:

--- Quote from: 2N3055 on October 10, 2016, 04:51:04 pm --- AcHmed99,

I think you didn't quite understand how it's done..

You don't inject current into PSU output.. You take PSU, you break feedback loop and insert injection transformer in series with feedback loop.. So it gets amplified by error amplifier, not suppressed......

--- End quote ---

Yes! Thanks to you and Jay_Diddy_B for explaining this while I was off doing chargeable work!

Navigation

[0] Message Index

[#] Next page

[*] Previous page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod