I'm trying to design a flyback converter to power my stuffs, I need about 3 A of current at 24 V.
At first I though this was straightforward. However, after reading "Design Guidelines for Off-line Flyback Converters"(AN4137) of Onsemi, I have found that it isn't that easy. Capacitors have max ripple current specification which mean I could risk blowing them if I just use a single of them. Therefore I think I should use some of them in parallel to distribute the ripple current (I think so). According to the appnote,
I calculated the output peak voltage ( about 88 V in my design) so using 100 V rated caps should be fine. Another problem arises : I don't have much 100V-or-larger electrolytic capacitors.
After thinking awhile, I remembered that there is someone who uses "capacitance multiplier" in there power supply design.
Will it do the trick ? Instead of going to the store and buying some caps, should I use it ?
By the way, please help me correct grammar error, I am aan English learner, not a native.
Thanks !
Best Answer
A "capacitance multiplier" is definitely a valid way to reduce ripple in a power supply.
It is basically a single-pole R-C low-pass filter followed by an emitter follower as a buffer. Instead of shunting the ripple voltage to ground through a capacitor (creating a ripple current problem), the transistor simply blocks it.
Of course, this does not come without a cost. First of all, you need to make sure that the output voltage is less than the minimum input voltage -- i.e., the lowest voltage that the ripple reaches. This means that you need to add something like a zener diode or a resistor between the transistor base and ground.
This voltage difference needs to be large enough to keep the transistor conducting at all times, which also means that it needs to be large enough to supply the required base current through the resistor in your diagram. Suppose you're willing to live with a minimum voltage drop across the transistor of 2 V. If the transistor has a current gain of 50, that means you need 60 mA of base current for an output current of 3 A. Since the minimum drop across the resistor is 1.3 V, this means that the resistor can be no larger than 1.3 V / 60 mA = 22 Ω
Any voltage difference between the instantaneous input voltage and the output voltage gets dissipated in the transistor at whatever the load current is. For example, if the ripple is 3 Vp-p on top of a minimum drop of 2 V, the average voltage across the transistor is at least 3.5 V. At a load current of 3A, this becomes a minimum of 10.5 W of wasted power. This represents a significant drop in overall efficiency, which may negate the benefits of using a switching converter in the first place.
There are several confusing aspects to your question, however. If your desired output is 24 V, why do you have 80-100 V at the rectifier? There's a huge mismatch somewhere in your basic design. Also, you talk about a flyback converter, but your diagram shows a transformer connected directly to the mains. Which is it?