Synchronous boost.. or not?

2kW synchronous boost.. or not?

Hello,

I'm designing a high power non isolated boost converter (24/48V, 2kW). For simplicity and reliability, I prefer to avoid a synchronous topology.

As the rms current of the inductor will be around 50Amps, I found it hard to get a suitable Schottky diode. Actually, I only have the option of putting two 30A common cathode (single chip) diodes in parallel. Even so, there is a very little safety margin (60A vs 50A).

Anyway, reading the datasheet of the power Mosfet (IRFP4568) used for the boost switch, I found out that its body diode looks quite sturdy (170/680A continuous/peak) and its reverse recovery time is about 120ns.

Do you think it's feasible to use the Mosfet body diode as freewheel diode in this particular situation (power range and topology)?

What about the reverse recovery current (8A)? Will it flow from the output to ground (through the active switch) or it's just for a short enough duration to worry about? Will a Schottky diode be more efficient in this situation?

Well, my ultimate question was supposed to be:

Is it feasible to choose a synchronous topology (50Amp, hard switched) or should I go with one of the solutions above?

Thank you very much for your time.

If the inductor RMS current is 50A, then if using a freewheel diode (a normal one) would dissipate around 40W, that's a pretty high loss, taking into account it could be much lower. Using a schottky diode would decrease that loss to around 20W (1% of the maximum output power). Using the synchronous FET, that loss could reach around 15W without counting the rise/fall losses.

About using diodes, what frequency are your working at? If 2 diodes in parallel are faster than the single FET body diode, use the faster solution whenever the time transitions become a problem. If "speed" is not a matter here, use the one that will provide you a bigger safety margin. It will be better at long terms.

A synchronous boost will only improve efficiency. That can be seen in ultra low power or maximum efficiency circuits, where it can reach about 98%, and a diode could drag it to 90%. But as you are working with very high currents there's almost not big difference (in power loss) by using synch or freewheeling, so if I were you, to avoid complications, I'd choose the non-synchronous.

Thank you very much for your comments. I further read about it and it looks like the freewheeling diode reverse recovery is quite a big problem in (high power) buck/boost topologies.

There are many practical solutions to mitigate this issue (some sort of active snubbers and such) but, once again, I'll rather keep it simple.

BrunoARG said:

But as you are working with very high currents there's almost not big difference (in power loss) by using synch or freewheeling

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I wish you were right.. Actually, the forward voltage of any diode being almost constant, the disipated power (V*I) greatly depends on current. But, in this particular situation, even an active switch (Mosfet) will not improve the efficiency by more than a few percents.

If I'll have to add some degree of complexity to this circuit, I'd better go with an interleaved topology instead. That way, I'll keep a reasonable safety margin (25A of 60A) by using those dual Schottky diodes.

Hi,

I only have the option of putting two 30A common cathode (single chip) diodes in parallel. Even so, there is a very little safety margin (60A vs 50A).

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You don´t say which diode you are talking about.

But I assume the current often isn´t the problem. I rather think the dissipated power is the problem.
If so, then using two 30A diodes in one package doesn´t help.

About efficiency.
I don´t think that an interleaved topology helps. It´s the same as with paralleling diodes. It changes total power dissipation only marginally.
Even duty cycle has no big influence.

Only reducing rectifier forward voltage helps --> this leads to synchronous rectifiers.

Klaus

KlausST said:

You don´t say which diode you are talking about.

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Sorry, I'm taking about 60CPQ150 (IR/Vishay).

KlausST said:

But I assume the current often isn´t the problem. I rather think the dissipated power is the problem. If so, then using two 30A diodes in one package doesn´t help.

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According to the math above (which is correct), we're talking about 20W dissipated power. I guess it's not a problem even with a passive heatsink.

KlausST said:

I don´t think that an interleaved topology helps. It´s the same as with paralleling diodes. It changes total power dissipation only marginally.

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With an interleaved topolgy, you can decrease the dissipation power of the active switch (during conduction) by a factor of 4 (Rdson * I/2 * I/2). Of course, the dissipated power of the freewheeling diode will be reduce by a factor of two but you could bet on it being more accurate rather than two paralleled diodes.

KlausST said:

Even duty cycle has no big influence.

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How come? The average dissipated power of the freewheeling diode depends on the duty cycle.

KlausST said:

Only reducing rectifier forward voltage helps --> this leads to synchronous rectifiers.

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Moreover, synchronous + interleaved decrease the conduction power loss by a factor of 4.

(to be more specific: it decrease the power loss by a factor of 4 _per device_)

Following your suggestions, I've decided to go with the synchronous topology. Now I just need to optimize the gate control timmings (dead time) to completely avoid body diode conduction for the high side (freewheeling) switch.

Another more off beat way to do this (not saying it is better) would be to use a push pull topology and a centre tapped inductor (not a choke).

The "off" side rises to twice the input voltage, and a pair of diodes should then give you pretty much pure dc at twice the dc input voltage.
Any required dc output choke and filter capacitor could be made quite small.

Only real disadvantage is there is no easy way to regulate the output voltage, but you would not have the massive rising current ramps you get with a boost converter.

Hi,

With an interleaved topolgy, you can decrease the dissipation power of the active switch (during conduction) by a factor of 4 (Rdson * I/2 * I/2). Of course, the dissipated power of the freewheeling diode will be reduce by a factor of two but you could bet on it being more accurate rather than two paralleled diodes.

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But in my post above i didn't talk about mosfet power dissipation, but diode dissipation. Especially catch diode...and paralleling diodes.

Duty cycle:
If you need 6A output current.
And the diode duty cycle is 50%, then the average current needs to be 12A
And if duty cycle is 10%, then current will be 60A.
Power dissipation is: I x Vf x dc. " I x dc" is about constant (even for paralleled diodes) and thus power dissipation mostely depends on the marginal variation of Vf

But your formula is true for Mosfets conductive dissipation. Here total power dissipation is independent of (interleaved) mode.
Using two Mosfets will reduce power dissipation .... interleaved mode or not.

Klaus

KlausST said:

Duty cycle:
If you need 6A output current.
And the diode duty cycle is 50%, then the average current needs to be 12A
And if duty cycle is 10%, then current will be 60A.

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Sorry, Klaus, you were mostly right but I was thinking of a situation like this:

For a boost converter, Pin = Vin x I (inductor current). Accordingly, Pout = Vout x I x (1 - D), where D is the duty cycle. That means the freewheeling diode is only conducting during this (1-D) percent of the switching period.

So, if my requirements were for a higher output voltage (24/96V instead of 24/48V), the freewheeling diode (50Amps) stress would have been halved.

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@Warpspeed:

I thought that we agreed long time ago that we'll never-ever use unclamped (push-pull) topologies again.

Warpspeed said:

Only real disadvantage is there is no easy way to regulate the output voltage, but you would not have the massive rising current ramps you get with a boost converter.

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I'm using this circuit for a battery charger and I hear that pulse charging might actually help with desulphation! (j/k)

Hi,

So, if my requirements were for a higher output voltage (24/96V instead of 24/48V), the freewheeling diode (50Amps) stress would have been halved.

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for sure .. because average output current is halved.

You could go for 500V (or even more) output voltage, then the current drops even more. Reducing diode loss.

This is why we use high voltage transmission lines: Less current, less power loss.

Klaus

with your current though....i woudl go for paralleled (interleaved) boost converters.........presumably they are current regulated, so there will not be able to hog the current, as you will regulate the current from each booster.