IR2112 connected to buck converter

[mah]

i used the attached circuit with atmega 8 to drive dc buck converter and it was working very well , then i increased the input voltage on the drain to 20v and the input current was 4 A (from a power supply ) suddenly the output on Ho of IR2112 was zero when i removed the converter the output on Ho was 7.6 v dc constant when the volt on drain Vd was 9.6 . and when i connected power resistor to the source of the mosfet Ho was zero ,did i lost the driver ?
also I noticed that when I increased Vd to 20 volt the output on IR2112( Ho) increased too ,the pwm was amplified to 25 v on the Mosfet gate how could the IRfz48 mosfet survive while Vgs from (10-20 v )?

 

[ronsimpson]

Your picture is of a IR2110 but you text is of a IR2112.
Most people don't under high side gate drivers.
1) The bottom side driver:Sits on ground and drives the bottom MOSFET. The output voltage is approximately 0 or 12 volts. It connects to the bottom MOSFET's G and S.
2)The top side driver: It does not sit in ground! It connects to the MOSFET's G and S. (some times there is a source resistor) It gets its power from +12V through a diode. Current goes through D1 and charges up C1,2 and through R5 load to ground. If R5 is missing the MOSFET driver can not function. (C1,2 will not charge up to 11.3V) The MOSFET must be off for 90% of the time so C1,2 can charge up!
3)To turn on the the MOSFET: The high side driver puts 11 volts across the gate to source. Now the FET turns on and the source goes up to 20 volts. (load supply) Notice the bottom of the high side drives also goes to 20 volts. (While the gate is at 31 volts the MOSFET has only 11 volts across G to S. ) You asked how can the MOSFET survive with high voltage on the gate. The FET only cares what is from G to S. It does not know that your meter is at ground.
4) The top side driver can not get power while the FET is on. The driver is living from power stored in C1,2. This power will run out after a while. The FET needs to be off 10% of the time to get power back into C1,2.

A high side driver is hard to trouble shoot even if you know how it works. Hope this helps you.

 

[mah]

thank you for your perfect explanation however i still don't know why did i loose the driver ?! ah i used ir2112 instead of ir2110 they are the same .
why does the voltage on Ho increase with increasing load supply voltage ?
please take into account that there was buck converter before i loose the driver ( mosfet - toroidal inductor- diode - capacitor and resistor)

 

[ronsimpson]

why does the voltage on Ho increase with increasing load supply voltage ?

The high side driver rides on the source of the mosfet. When the mosfet is on (closed) the source of the mosfet goes up with the supply voltage.

please take into account that there was buck converter before i loose the driver ( mosfet - toroidal inductor- diode - capacitor and resistor)

No I can't take into account anything because you have not given me a correct/complete schematic.

why does the voltage on Ho increase with increasing load supply voltage ?

You enter the elevator at floor 1. The up button is 3 feet above your feet. You ride the elevator to the 9th floor. Your friend back at floor 1 yells up "why are the buttons 93 feet up". You yell back down "they are 3 feet up."

I hope this helps. Thanks for the like.

 

[shortbus=]

Your picture is of a IR2110 but you text is of a IR2112.
Most people don't under high side gate drivers.
1) The bottom side driver:Sits on ground and drives the bottom MOSFET. The output voltage is approximately 0 or 12 volts. It connects to the bottom MOSFET's G and S.
2)The top side driver: It does not sit in ground! It connects to the MOSFET's G and S. (some times there is a source resistor) It gets its power from +12V through a diode. Current goes through D1 and charges up C1,2 and through R5 load to ground. If R5 is missing the MOSFET driver can not function. (C1,2 will not charge up to 11.3V) The MOSFET must be off for 90% of the time so C1,2 can charge up!
3)To turn on the the MOSFET: The high side driver puts 11 volts across the gate to source. Now the FET turns on and the source goes up to 20 volts. (load supply) Notice the bottom of the high side drives also goes to 20 volts. (While the gate is at 31 volts the MOSFET has only 11 volts across G to S. ) You asked how can the MOSFET survive with high voltage on the gate. The FET only cares what is from G to S. It does not know that your meter is at ground.
4) The top side driver can not get power while the FET is on. The driver is living from power stored in C1,2. This power will run out after a while. The FET needs to be off 10% of the time to get power back into C1,2.

A high side driver is hard to trouble shoot even if you know how it works. Hope this helps you.

View attachment 84626

That was me with the like. And I have some questions about this.

1.Does your marked up drawing mean that with a resistor, source to ground, you don't need a low side mosfet to charge/recharge the bootstrap cap? Always thought a low side mosfet was needed.

2. What would the value of the resistor need to be? Or formula to figure the value?

Thanks for this, I learn something new all the time.

 

[ronsimpson]

1.Does your marked up drawing mean that with a resistor, source to ground, you don't need a low side mosfet to charge/recharge the bootstrap cap? Always thought a low side mosfet was needed.

You schematic shows a load resistor. That works well.
You said you use a buck PWM and diode. That works well. The inductors slams down into the diode.
You don't need a low side MOSFET. You need something that pulls the source of the MOSFET down near ground. (pulls the bottom of the high side driver down)

The high side driver used some amount of power. (see data sheet) At 50% duty cycle you must pull down so that twice that much current enters C1,2. If the driver is on 75% of the time and off 25% then all the power charge C1,2 only has 25% of the time. (charging current is 4x) That is why these drivers have a hard time at 90% on and 10% off.

 

[shortbus=]

You schematic shows a load resistor. That works well.
You said you use a buck PWM and diode. That works well. The inductors slams down into the diode.
You don't need a low side MOSFET. You need something that pulls the source of the MOSFET down near ground. (pulls the bottom of the high side driver down)

The high side driver used some amount of power. (see data sheet) At 50% duty cycle you must pull down so that twice that much current enters C1,2. If the driver is on 75% of the time and off 25% then all the power charge C1,2 only has 25% of the time. (charging current is 4x) That is why these drivers have a hard time at 90% on and 10% off.

Hi Ron, I think your confusing me with the OP. I was asking about any circuit, with a PWM or pulsed high side mosfet instead of this particular circuit. Sorry to highjack this but didn't know how to start my own new thread and have you answer.

Will a resistor to ground, from the VS terminal of the driver, recharge the bootstrap cap, when the H out is not conducting?

 

[ronsimpson]

Will a resistor to ground, from the VS terminal of the driver, recharge the bootstrap cap, when the H out is not conducting?

The resistor to ground from VS is the load. With out a load the high side driver will not work. You can add a resistor across the load so the circuit will have a load even if the real load goes away.

 

[mah]

this is my complete circuit with buck converter before i lost signal from the driver

 

[kubeek]

What was the duty cyle when it happened?

 

[ronsimpson]

What happened to pins 9 and 13?

 

[mah]

What happened to pins 9 and 13?

pin9 is connected to +5v ,pin13 is ground