0
\$\begingroup\$

Can I treat 2 secondary coils with dual voltage output as one with double voltage for the purpose of calculations?

This seems to be OK for all formulas based on output power as technically the secondary IS a single coil with tap in the middle.

However for LT converters that use flyback in primary as feedback the formulas for feedback resistor(s) are based on Vout and turns ratio. How do I apply these?

UPDATE:

Allow me to clarify the question. Let's say I used all the right formulas and made myself a nice stable converter that puts out +50V at 100mA from isolated secondary coil and uses flyback voltage on primary as feedback.

Now, what happens if I tap the middle of secondary coil and convert this into dual voltage supply? Will I get a nice stable converter with dual +/-25V at 100mA outputs (minus drop on second diode, of course)? Without any changes in the feedback circuit or re-calculating the whole thing from scratch?

I am about 95% sure tapping output coil will not do anything to the functioning of converter, because the transferred power did not change. Except for that feedback resistor that according to LT8302 datasheet is calculated as:
Rfb = Rref * Nps * (Vout + Vf) / Vref
The Vout part of it is confusing, since now I don't have 50V, I have 2x25V.

\$\endgroup\$
1
  • \$\begingroup\$ I would suggest using one of the secondaries for the feedback circuit, then the other one use it for the voltage doubler. \$\endgroup\$
    – drtechno
    May 29, 2018 at 11:59

1 Answer 1

0
\$\begingroup\$

Can I treat 2 secondary coils with dual voltage output as one with double voltage for the purpose of calculations?

With a flyback converter, energy is stored in the primary and released to the secondary at a rate dictated by the switching frequency. As most Flyback converters are DCM types, the output voltage produced depends entirely on the load resistor, the duty cycle, the input voltage, the frequency and the primary inductance: -

enter image description here

Picture source.

Regulation of the output voltage is performed by using feedback and this counters the effects of a varying load by applying greater or less duty cycle thus, it controls output voltage by increasing or decreasing the energy transferred per cycle.

However, for flyback converters with multiple outputs you have to be more careful. If one load suddenly takes less power, the other output voltage will rise and, in order to keep that voltage regulated, the energy throughput is reduced and the first load (that took less power) may become significantly reduced in voltage due to regulation feedback occuring on the 2nd load.

It's certainly worth modelling your load variations in something like micro-cap or LTSpice to see what happens.

However for LT converters that use flyback in primary as feedback the formulas for feedback resistor(s) are based on Vout and turns ratio. How do I apply these?

Assuming that method works fine for a single output (and it does but isn't quite as good as direct feedback) then you still have the same basic problem of what happens when loading imbalances can occur.

\$\endgroup\$
6
  • \$\begingroup\$ Can you say "power in/out"? Because if that P is actually Pin then your formula can be rearranged as Lp=(Vin^2*D^2)/(2*Pin*Fsw), which looks a lot like the one I was using for calculating primary inductance (DCM mode). Anyway, as it happens you've misunderstood my question, probably because it is much simpler (stupider?) than you expected. I am updating the question to make it clearer. \$\endgroup\$
    – Maple
    May 29, 2018 at 17:40
  • \$\begingroup\$ I look forward to that update. \$\endgroup\$
    – Andy aka
    May 29, 2018 at 17:43
  • \$\begingroup\$ I see the edited question and my answer still stands. Maybe you didn't understand my answer? \$\endgroup\$
    – Andy aka
    May 29, 2018 at 18:06
  • \$\begingroup\$ Absolutely possible that I didn't. When I look at your formulas I see the calculation for single coil Vout. What I am asking about is if your calculations will still stand if I split the coil to get 2x (Vout/2). While there was P in the formula it was all clear to me, because Pout stays the same. However you got rid of it in the process and that is where it got confusing. \$\endgroup\$
    – Maple
    May 29, 2018 at 18:25
  • \$\begingroup\$ I'd encourage you to make a simple sim circuit if you are able. \$\endgroup\$
    – Andy aka
    May 29, 2018 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.