Understanding feedback circuit

[gkmaia]

I got a flyback power supply that outputs 8V.

Circuit seems very simple with just a MOS on the primary to oscillate and a diode on the secondary to rectify.

My question is. I want to bring it down from 8V to 6.5V.

Looking at the secondary (market in red) I can see a Zener A2-11 with Vz 9.7~10.1v

I am wondering. If I were to replace this Zener with a 8v Zener it would reduce the voltage output?

Maybe other option would be to unwind the transformer secondary a few turns to reduce the voltage output?

Am I right on my assumptions? Any advise?

[fzabkar]

AFAICS, the voltage sensing happens on the primary side. The 10V Zener might be an overvoltage crowbar :?

What are the markings on D16 and D17? Is one of them a Zener?

Edit: I can't make sense of your circuit. Could it be something like Fig 7 in the following datasheet?

If so, then you could wind additional turns in phase and in series with the secondary control winding. The more turns you add, the lower will be the output voltage. However, a better approach would be to add an optocoupler and TL431 to provide much better regulation. There are examples in the datasheet.

http://pdf.datasheetcatalog.com/datasheet_pdf/power-integrations-inc/TOP100YAI_to_TOP103YAI.pdf

[gkmaia]

I've updated the schematic

The schematic you send based on TOP100 is very similar to the example MIP2E7D which is the MOS used on my adapter.

Looking at my PCB there is a slot for an optocoupler and seems they designed it to support multiple topographies. I could try that, but that would be quite a big task and I am trying to avoid it.

I was trying to find a way to get this done without refactoring the circuit too much. 

[fzabkar]

Your circuit looks very similar to the TOP100Y application circuit. However, you have drawn the transformer taps incorrectly.

ISTM that the upper secondary winding should be connected to taps 3 and 4, not 2 and 3.

The primary winding (bottom) should be connected to taps 3 and 4. The lower secondary winding (the control winding) should connect to taps 1 and 2.

MIP2E7D, Panasonic, IPD, 700V, 3A, TO220, 100kHz, 40W - 60W:
http://pdf.datasheetcatalog.com/datasheets/480/467388_DS.pdf
http://datasheet.elcodis.com/pdf2/118/25/1182521/mip2e4d.pdf

Edit: The bottom end of the control winding connects to the 0V reference. If we open this end of the winding and insert two silicon diodes in series with the winding, then the control voltage (at the winding) will be 6V - 1.2V = 4.8V. This should then reduce the output voltage by the same ratio, ie Vout = 8 x (4.8 / 6) = 6.4V. Of course these diodes need to be biased in the same direction as D3 in the application circuit. Small signal diodes should suffice (the current draw is very low).


Edit #2: My suggestion would increase Vout, not decrease it. Sorry. However, my original suggestion regarding additional turns for the control winding should work.

[gkmaia]

Ok, so one last question them...

That seems like a very simple one transistor oscillator flyback that hands all the voltage regulation to the DC/DC transformer.

Is that right?

[T3sl4co1l]

D15 may be a zener diode, which sets primary side voltage.  Secondary side regulation then depends on leakage in the transformer.

D17 is a TVS, like in the above circuit.  It may be higher voltage rated; 100-200V is typical.

I don't think it's a plain MOSFET, as there's no bias path to the gate, and there's usually a current limiting turn-off transistor as well.  (Or if a BJT is used, base current can provide turnoff action.)  In that case, D15 must be for control power and feedback, and doesn't need to be a zener.

Tim

[fzabkar]

https://na.industrial.panasonic.com/products/semiconductors/power-devices/ac-dc-converter-power-supply-ic-ipd

"Panasonic IPD (Intelligent Power Devices) are AC-DC converter ICs that simplify power supply design."

[gkmaia]

As per the image bellow. D22 is a Zener 11v.

Does it leaks if voltage raises above 11v as a way to protect from voltage surges?

What would happen I replaced it with a Zener 7V for example?

[ogden]

If you look at the circuit bellow. It is from a TOP100.

R1 and R2 split the current to U2 but both receive the same voltage. Vr2 allows for anything above 6.2 to leak out.

So that makes me think what would happen if I change VR2 and then allow the voltage to leak under 6.2v. Something like 5.1v. I imagine it will have an affect on my output voltage.

Is my logic correct?

For circuit you are showing - you may change feedback circuit to alter output voltage indeed. Unfortunately your supply does not have optocoupler feedback. It's already said - voltage sensing happens on primary side. Such "primary sensing" supplies does not regulate well, so I would drop idea of modifying supply but just add linear regulator like LM317.