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Gate Driver ICs Forum Discussions

Not applicable
I AM USING IR 2110 TO TURN THE MOSFET IR 540 N USING HIGH SIDE CONFIGURATION MY MOSFET IS NOT TURNING ON TO CHARGE A BATTERY PLEASE HELP
VDD IS +5v ,VCC is 12+VOLT

I AM POSTING THE SCHEMATIC OF CIRCUIT I HAVE USED2682.attach
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1 Solution
tengD
Moderator
Moderator 5 sign-ins 10 replies posted 5 replies posted
Moderator
Zuhaib,

As I already mentioned, "Only when you solved the charging issue for Cbs, then the circuit can work properly. " Actually there's no charging loop for Cbs as in your circuit.

Normally the charging loop is from Vcc --> bootstrap diode --> Cbs-->Vs--> low side mosfet--> COM; but in your application, Vcc=12V, Vs is connected to the battery, it is 14.5V, and low side Mosfet is OFF. So the charging loop doesn't work. So there's no enough voltage on Cbs, then HO will keep low, the circuit doesn't work as you expected.

For your application, I would suggest you to use our smart high side switch, for example, BTS441, here is the link for the datasheet:
https://www.infineon.com/dgdl/Infineon-BTS441TG-DS-v01_21-EN.pdf?fileId=5546d4625a888733015aa9afd699...

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11 Replies
tengD
Moderator
Moderator 5 sign-ins 10 replies posted 5 replies posted
Moderator
zuhaib,

Yes, this circuit can't work properly as you expected. As it is for charging the battery, so the Source of the Mosfet is connected to the battery, the voltage on this pin is 12V as shown on the pic.
And for the HO of the IR2110, the power supply is the voltage between Vb and Vs, but there's no connection for Vs, then how to charging the capacitor between Vb and Vs? -- This is the key in the application. Only when you solved the charging issue for Cbs, then the circuit can work properly.
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srhim1971
Employee
10 replies posted 5 replies posted First question asked
Employee
Hello Zuhaib,

It's a bit difficult to read the details in the photo snapshot that you provided, but we assume that the serial resistor is 10 Ohms. You used the additional diode to do the discharging faster, correct?
You can check your circuit against the typical one here http://www.infineon.com/dgdl/ir2110.pdf?fileId=5546d462533600a4015355c80333167e on page 1 .

-You don't have a cap at Vss and Vdd.
-The MOSFET needs to be connected with Pin 5. This line is missing in your circuit; therefore, the MOSFET has no potential.
-The cap at the MOSFET drain to ground is missing.
-Please consider that cable or connectors can have influences. Wires should be quite short.

Please check with an oscilloscope if the gate charge and the timing are ok.

Check here:
http://www.infineon.com/dgdl/an-1092.pdf?fileId=5546d462533600a401535595bc541044 and see chapter 2.1

For timing, see page 3 (Ton,toff,etc) and figure 4 see page 6. Ton/off is typical 120 & 94 ns.

As for details for the gate charge please see here:

http://www.infineon.com/dgdl/an-944.pdf?fileId=5546d462533600a40153559eb9841190
http://www.infineon.com/dgdl/an-937.pdf?fileId=5546d462533600a40153559ea1481181

Also, consider this idea: Here you can use a simulation to check your circuit, see here. Then you can compare the simulation against your osci-pic
s.
http://www.infineon.com/cms/en/product/power/gate-driver-ics/level-shift-gate-drivers/high-and-low-s...

Hope this helps!
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Not applicable
sorry i forget to mention that but i have connected connected vs to the source already still not working help me with the bootstep circuitry with value of cap.
thank you for your reply indeed
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Not applicable
thanks for your reply i will tell you when i done what u have told me
thanks
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Not applicable
2684.attach

still my circuit is not working please help with capacitors value
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Not applicable
2684.attach

still my circuit is not working please help with capacitors value
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User270
Level 5
5 solutions authored First solution authored
Level 5
To find the right value of the bootstrap capacitor and for other hints I think this design tip document will help you.
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Not applicable
does my circuit is correct.?
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srhim1971
Employee
10 replies posted 5 replies posted First question asked
Employee
Hello Zuhaib,

Please keep in mind that a high side driver requires the bootstrap capacitor to be charged in order to turn on the switch. This means that driver needs to be used in such a way that it is switching at higher frequencies (kHz) and should have a way to refresh the bootstrap capacitor each cycle.

Check to confirm whether or not the voltage across the bootstrap capacitor (VB to VS) > than the UVLO for the IR2110. This voltage should be in the vicinity of the ~VCC voltage. Also, as asked before, it is not clear what is the value of your bootstrap capacitor.

My recommendation to you is to take scope measurements of the signals and timings and compare these against the links that are mentioned in the thread within the reply above.
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Not applicable
HELLO
ITS CLEARLY MENTIONED ON TOP OF MY CIRCUIT . 22UF (63 v) IS THE VALUE OF MY BOOTSTRAP CAPACITOR THAT IS BETWEEN VB AND VS..
I JUST WANT TO KNOW THAT MY VS IS CONNECTED TO SOURCE WHICH IN TURN IS CONNECTED WITHY THE POSITIVE TERMINAL OF BATTERY SO HOW WILL I CHARGE MY BOOTSTRAP CAPACITOR
THANKS IN ADVANCE
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tengD
Moderator
Moderator 5 sign-ins 10 replies posted 5 replies posted
Moderator
Zuhaib,

As I already mentioned, "Only when you solved the charging issue for Cbs, then the circuit can work properly. " Actually there's no charging loop for Cbs as in your circuit.

Normally the charging loop is from Vcc --> bootstrap diode --> Cbs-->Vs--> low side mosfet--> COM; but in your application, Vcc=12V, Vs is connected to the battery, it is 14.5V, and low side Mosfet is OFF. So the charging loop doesn't work. So there's no enough voltage on Cbs, then HO will keep low, the circuit doesn't work as you expected.

For your application, I would suggest you to use our smart high side switch, for example, BTS441, here is the link for the datasheet:
https://www.infineon.com/dgdl/Infineon-BTS441TG-DS-v01_21-EN.pdf?fileId=5546d4625a888733015aa9afd699...
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