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Forward converter with peak-current mode controller

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moonz:
Why stable operation of a forward converter with peak-current mode controller, the magnetization current should be less than the reflected current?

I figured out the forward converter, and I'm currently reading app-note(https://www.analog.com/media/en/technical-documentation/app-notes/designing-activeclamp-forward-converters-using-peakcurrentmode-controllers.pdf) where state:


--- Quote ---The converter is a peak current-mode controller that controls the peak current as seen by the current-sense resistor, which is the sum of the reflected load current and magnetizing current. It is necessary that the magnitude of reflected load current is always more than the magnetizing current for stable converter operation and well regulated output. This condition is always satisfied, if the magnitude of reflected load current at the input minimum is more than the magnetizing current. Hence, peak-to-peak primary magnetizing current is assumed to be half of the reflected load current at minimum input voltage
--- End quote ---

Do I understand correctly that it is needed in order for the feedback to 'know' what is really occurring in the load?

If the answer is yes, then how will the forward converter work with a slight load?

mtwieg:

--- Quote from: moonz on August 29, 2023, 07:29:58 am ---Why stable operation of a forward converter with peak-current mode controller, the magnetization current should be less than the reflected current?

I figured out the forward converter, and I'm currently reading app-note(https://www.analog.com/media/en/technical-documentation/app-notes/designing-activeclamp-forward-converters-using-peakcurrentmode-controllers.pdf) where state:


--- Quote ---The converter is a peak current-mode controller that controls the peak current as seen by the current-sense resistor, which is the sum of the reflected load current and magnetizing current. It is necessary that the magnitude of reflected load current is always more than the magnetizing current for stable converter operation and well regulated output. This condition is always satisfied, if the magnitude of reflected load current at the input minimum is more than the magnetizing current. Hence, peak-to-peak primary magnetizing current is assumed to be half of the reflected load current at minimum input voltage
--- End quote ---

Do I understand correctly that it is needed in order for the feedback to 'know' what is really occurring in the load?

If the answer is yes, then how will the forward converter work with a slight load?

--- End quote ---
I've designed peak CMC forward converters before and never observed this, and I can't see any reason why this would be true. In fact, magnetizing current effectively acts as a bit of slope compensation, helping against subharmonic oscillation.

Faringdon:
Why stable operation of a forward converter with peak-current mode controller, the magnetization current should be less than the reflected current?

It generally is...but yes in light load the magnetizing current peak may be more than the reflected...but who cares, you can still get regulation.

Turn the question round.......why do you want magnetizing current to be greater than reflected load current?
....or, what is it about magnetizing current that you dont like?

But generally, yes,  the load current is what you are trying to control, so why do you want much magnetizing current......magnetizing current is what you are forced to put up with.....due to V = Ldi/dt....but  as Mtwieg says...magnetizing current does give you a  nice bit of slope compensation if you need that.

moonz:

--- Quote ---why do you want magnetizing current to be greater than reflected load current?
--- End quote ---
I don't want  :). Magnetizing current is a detrimental effect for a forward converter, as it doesn't provide power to the load; instead, it increases power loss. Right?

I would like to continue discussing the topic of forward converters. Can someone please review my approach for selecting a core for the converter?

So... Input data:
Input voltage min: 10V
Out Power: 70W
Out Voltage: 54V
Maximum duty cycle(Dmax): 0.6 (I know, i's not typicaly for forward but for MAX17598 it allow)
Frequency switching: 1MHz

1) Calculate out current in load:
Iout=Pout/Vout = 1.3A

2) Calculate turns ratio:
n= Vout/(Vin*Dmax) = 9

3)  Calculate secondary current:
I skipped the inductor selection step, just say that ripple current are 0.33A. The secondary peak current Isec = Iout + Irip = 1.3+0.33 = 1.63A

4) Calculate primary current:
Ipri = Isec * n

5) Calculate magnetization inductance:
Suppose that the magnetization current should be less than 20% of the full load current. Imag = 0.2*Ipri
Then Lmag = 2uH (from V=L*di/dt)

For example choose B66285G0050X187


H = Ipri*Turns/le = 451 (for Al = 1520nH enought 1 turns in primary side) = Ipri/le = 586. Ipri iclude magnetization  current.
B = H * ue * Al = 0.35; It's less then 0.4 (saturation value),and this core is suitable for this design.

Is everything correct in my reasoning or did I make a mistake somewhere?

jonpaul:
PC200 is the power material

All others like N87 are not for power applications.

Magnetizing current is a natural result of any transformers primary inductance. As it is inductove and not resistive, there is no power loss just VARS.

j

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