Using gate drive IR2110

Thread Starter

mtabadi2524

Joined May 2, 2021
9
Dear Sir/Madam,
I am trying to use the gate drive IR2110 which has two independent high and low sides for driving mosfet and IGBT. At the first stage, I made the following circuit on the breadboard. Without any command to HIN and LIN, I measured approximately 13 V at the source port of mosfet (between mosfet and Resistance 10K) and approximately 14 V at the gate port of mosfet. I found that the mosfet was ON and the current was flowing through it without any command at HIN. How can I solve this problem.
Best regards.
 

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Sensacell

Joined Jun 19, 2012
3,432
The problem with this configuration is that the bootstrap capacitor must charge through the load, not a low-side FET.

Depending on the load, you might have problems with the gate drive voltage being too low.
 

Thread Starter

mtabadi2524

Joined May 2, 2021
9
The problem with this configuration is that the bootstrap capacitor must charge through the load, not a low-side FET.

Depending on the load, you might have problems with the gate drive voltage being too low.
How should I modify the circuit to resolve such problem?
 

Thread Starter

mtabadi2524

Joined May 2, 2021
9
Add the low-side FET. You do not specify what the load is, if it's purely resistive, you could use a much smaller FET, since it's only carrying the bootstrap current.
Thanks for useful comment, I added another mosfet in the low side, as shown in attached file. It blocks the current flowing through both mosfets when HIN and LIN are connected to GND.
I connected HIN, LIN, SD and VDD and VSS to arduino UNO microcontroller. I observed the following cases:
1. If SD=0 and HIN=LIN=1 (PWM with duty 1), then the current flows through both mosfets.
2. If SD=0 and HIN=LIN=0, then no current flows through mosfets.
3. If SD=0 and HIN=0 and LIN=1 (PWM with duty 1), then the current flows through both mosfets.
Therefore, the mosfet on the high side is ON regardless to status of HIN. How can I solve such problem?
 

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Sensacell

Joined Jun 19, 2012
3,432
Your low-side FET is not solving the problem- you must provide a low impedance path to ground, so the bootstrap cap can charge via this path- ditch the 10K resistor.
You must be sure that the signals driving the high and low FETs have no overlap, so they are never on at the same time.

Look at the schematic, trace the path of current flow that charges the bootstrap cap.
 

Thread Starter

mtabadi2524

Joined May 2, 2021
9
Your low-side FET is not solving the problem- you must provide a low impedance path to ground, so the bootstrap cap can charge via this path- ditch the 10K resistor.
You must be sure that the signals driving the high and low FETs have no overlap, so they are never on at the same time.

Look at the schematic, trace the path of current flow that charges the bootstrap cap.
Thanks for your useful comments. I modified the circuit as shown in the attached file. It works very well and IR2110 turns the high side gate off and on. I have found that a diode should been considered parallel to resistance 10 ohm (in the input of gate). Why should such diode be considered in the circuit?
 

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Thread Starter

mtabadi2524

Joined May 2, 2021
9
If you had studied the blog on the IR2110 I posted, it has it in there, it is to allow a faster turn off time for Q1 & Q2
Thanks for useful link and explanation. I modified the circuit as shown in the attached file. I also added two resistors R3 and R4 (1K Ohm) similar to the circuits presented in the link. Please let me know the role of these resistors on the performance of circuit.1.jpg
 
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