Thread Starter

flaviotmz5

Joined Aug 23, 2021
14
I'm trying to understand how to drive the MOSFET's in the two-switch forward, I designed a 200W DC-DC converter, 50V input 100V output but I was looking for using an MCU and IC in this case I have an IR2110 high side low side for the gate drive, I don't know if it is possible to use it? and if it is how can I adapt to the circuit?
The duty cycle, in this case, is fixed at 0.4 (100kHz)

two_switch_forward.png
 

Ian0

Joined Aug 7, 2020
4,875
In theory you can use the IR2110 (just connect HIN to LIN as both FETs switch on at the same time), but it has a problem in charging up the bootstrap capacitor because the charge current is blocked by the lower diode.
If you can get it started up it will run, as the cap will charge when the lower diode conducts
I've seen versions with two auxiliary windings - one to produce the normal supply for the 2110 and another to produce the floating supply for the top FET.
I've also made a 3-transistor version, using another MOSFET instead of the lower diode, which eliminates the capacitor charging problem.
 

Thread Starter

flaviotmz5

Joined Aug 23, 2021
14
In theory you can use the IR2110 (just connect HIN to LIN as both FETs switch on at the same time), but it has a problem in charging up the bootstrap capacitor because the charge current is blocked by the lower diode.
If you can get it started up it will run, as the cap will charge when the lower diode conducts
I've seen versions with two auxiliary windings - one to produce the normal supply for the 2110 and another to produce the floating supply for the top FET.
I've also made a 3-transistor version, using another MOSFET instead of the lower diode, which eliminates the capacitor charging problem.
Sorry, I'm a newbie in this bootstrap thing, the capacitor is a huge problem? You've said that it will run if I started up, but how do I do this?, I'm looking for the simplest solution
 

Ian0

Joined Aug 7, 2020
4,875
Sorry, I'm a newbie in this bootstrap thing, the capacitor is a huge problem? You've said that it will run if I started up, but how do I do this?,
That's the problem. There's a good application note on the subject if only I could find it again. I'll post if I remember where I saw it.

The problem arises because we need to get the gate voltage of the top FET to a voltage about 10V above the positive supply.
It is done with the bootstrap capacitor. In the normal half-bridge it's easy. When the bottom transistor is ON, the capacitor is charged through the diode from the 12V supply. When the voltage needs to rise, the bootstrap capacitor provides the gate voltage.
In the two-transistor circuits, there's a diode blocking the current.
Have a look at the LM5015 (I know it's not powerful enough for your application). The capacitor charging problem is mentioned on page 10 of this application note.
Have you considered the asymmetric half-bridge instead? It doesn't have this problem.
 

Thread Starter

flaviotmz5

Joined Aug 23, 2021
14
That's the problem. There's a good application note on the subject if only I could find it again. I'll post if I remember where I saw it.

The problem arises because we need to get the gate voltage of the top FET to a voltage about 10V above the positive supply.
It is done with the bootstrap capacitor. In the normal half-bridge it's easy. When the bottom transistor is ON, the capacitor is charged through the diode from the 12V supply. When the voltage needs to rise, the bootstrap capacitor provides the gate voltage.
In the two-transistor circuits, there's a diode blocking the current.
Have a look at the LM5015 (I know it's not powerful enough for your application). The capacitor charging problem is mentioned on page 10 of this application note.
Have you considered the asymmetric half-bridge instead? It doesn't have this problem.
Some mates suggested that I use an optoisolated gate drive single-ended made by them, so that doesn't involve bootstrapping, do you think this will solve my problem?
 

Ian0

Joined Aug 7, 2020
4,875
Some mates suggested that I use an optoisolated gate drive single-ended made by them, so that doesn't involve bootstrapping, do you think this will solve my problem?
You'd have to post the circuit for me to be certain, but doesn't it just make it into a single-ended forward-converter?
If both FETs are N-channel then either you need a bootstrap circuit or an auxiliary supply.
 

Thread Starter

flaviotmz5

Joined Aug 23, 2021
14
You'd have to post the circuit for me to be certain, but doesn't it just make it into a single-ended forward-converter?
If both FETs are N-channel then either you need a bootstrap circuit or an auxiliary supply.
I'm still wait for them to send me, when I get it I'll post.

But It's something similar to this IC 2SC0108T
 

Thread Starter

flaviotmz5

Joined Aug 23, 2021
14
I'm still wait for them to send me, when I get it I'll post.

But It's something similar to this IC 2SC0108T
You'd have to post the circuit for me to be certain, but doesn't it just make it into a single-ended forward-converter?
If both FETs are N-channel then either you need a bootstrap circuit or an auxiliary supply.
Here it is the gate driver that I mentioned

1636584334315.png
 

Thread Starter

flaviotmz5

Joined Aug 23, 2021
14
You'd have to post the circuit for me to be certain, but doesn't it just make it into a single-ended forward-converter?
If both FETs are N-channel then either you need a bootstrap circuit or an auxiliary supply.
I think it is an auxiliary supply in this case right?
 
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