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[SOLVED] TLP 250 or IR 2110 , Which one to use?

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Hi Everybody,

I am currently trying to build a 500W, 100kHz buck boost converter, with Vin = 25to 35 Vout = 25 / 50 V. Switch MOSFET : IRF 3710
I am a bit confused about a number of things.

1. Should i be using a high side + low side driver like IR 2110 or a optocoupler based TLP 250 would suffice?
i feel ir 2110 is redundant since i need to drive only one mosfet. But i have heard TLP works satisfactorily only upto 25 kHz or so (?????).

2. Do i need to use a bootstrap circuit at all or is it required since the output / source is not at ground zero potential ( i am clearly a bit confused about when and where the bootstrap circuit is needed )

3. I have tried using ir 2110 , following the design equations given in the application note ( C> .1 uF) , i had used bootstrap capacitor ( electrolytic) C = 1uF , lower C = uF and bootstrap diode 1N4148. The resistor in series with the gate was 100 ohm ( random selection ). But with these values the test circuit didnt workout . I would like to know , on what basis do we fix the values of the R and the lower C?

Thanks in advance..... :)
 

Depending on the specific topology, a bootstrapping driver like the IR2110 may not work, since some topologies (like a basic buck boost) require the transistor drive to swing both above and below ground. In such cases you need a truly isolated gate driver. So you would need something like the opto coupler driver, and an isolated power supply for it.
 
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lxmn said:
Hi Everybody,

I am currently trying to build a 500W, 100kHz buck boost converter, with Vin = 25to 35 Vout = 25 / 50 V. Switch MOSFET : IRF 3710
I am a bit confused about a number of things.

1. Should i be using a high side + low side driver like IR 2110 or a optocoupler based TLP 250 would suffice?
i feel ir 2110 is redundant since i need to drive only one mosfet. But i have heard TLP works satisfactorily only upto 25 kHz or so (?????).

2. Do i need to use a bootstrap circuit at all or is it required since the output / source is not at ground zero potential ( i am clearly a bit confused about when and where the bootstrap circuit is needed )

3. I have tried using ir 2110 , following the design equations given in the application note ( C> .1 uF) , i had used bootstrap capacitor ( electrolytic) C = 1uF , lower C = uF and bootstrap diode 1N4148. The resistor in series with the gate was 100 ohm ( random selection ). But with these values the test circuit didnt workout . I would like to know , on what basis do we fix the values of the R and the lower C?

Thanks in advance..... :)
Click to expand...

1. What kind of a buck boost converter are you using? Is it a single switch inverting buck-boost converter or 2/3 switch non-inverting converter?

2. When source is not at ground potential, and the load is taken through source to ground, the MOSFET is configured as a high-side switch and then, high-side driving is required. One such way to drive the high-side MOSFET is through a diode-capacitor bootstrap method. So, if you're driving a high-side MOSFET this way, you need to use bootstrap.

3. What, in the test circuit, did not work out? Did the circuit simply not work? Did it get burnt? Did you verify that input is correct? You should upload your schematic. There may be errors there.

The TLP250 on its own is a low-side driver. You can design high-side driving circuitry employing a TLP250 using bootstrapping or an isolated power supply, but it's not designed as a high-side driver.

Hope this helps.
Tahmid.
 
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hi mtwieg,

Thnx for the quick response
i am using the basic inverting buck boost configuration so i guess i need a isolated driver like tlp 250.
But as hamid had mentioned the source is not connected to the ground in the basic buck boost configuration, so then i would need a bootstrap circuit right?
so finally which one to go for??

---------- Post added at 14:37 ---------- Previous post was at 14:33 ----------
Tahmid said:
1. What kind of a buck boost converter are you using? Is it a single switch inverting buck-boost converter or 2/3 switch non-inverting converter?

2. When source is not at ground potential, and the load is taken through source to ground, the MOSFET is configured as a high-side switch and then, high-side driving is required. One such way to drive the high-side MOSFET is through a diode-capacitor bootstrap method. So, if you're driving a high-side MOSFET this way, you need to use bootstrap.

3. What, in the test circuit, did not work out? Did the circuit simply not work? Did it get burnt? Did you verify that input is correct? You should upload your schematic. There may be errors there.

The TLP250 on its own is a low-side driver. You can design high-side driving circuitry employing a TLP250 using bootstrapping or an isolated power supply, but it's not designed as a high-side driver.

Hope this helps.
Tahmid.
Click to expand...

hi Tahmid,

thanx for the reply

i am trying to buld the single switch basic inverting buck boost configuration which doesnt have the source at ground potential. So as you have put forward i would need a bootstrap circuit to drive the high side switch like an ir 2110 but this inference just contradicts what mtwieg had suggested ( ???? ). Please clarify me on this.

The test circuit just didnt work. Nothing was burnt . I will upload the schematics that i used for ir2110 asap.
and sorry i splet your name wrong in mtwieg' reply.... :)
 

I have attached the schematics that i used to test the ir 2110 driver.
 

The IR2110 will work in the above due to the R load which will allow charge of the bootstrap cap before start up and when the mosfet is off.
 
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lxmn said:
I have attached the schematics that i used to test the ir 2110 driver.
Click to expand...
This isn't a buck boost converter (it's not a DC-DC converter of any kind, just a DC chopper). But it should still be able to turn on the FET and power the load. Are you sure the SD pin is pulled low? Does the LO output work if you provide PWM to the LIN pin?
 
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mtwieg said:
This isn't a buck boost converter (it's not a DC-DC converter of any kind, just a DC chopper). But it should still be able to turn on the FET and power the load. Are you sure the SD pin is pulled low? Does the LO output work if you provide PWM to the LIN pin?
Click to expand...

This isnt my actual circuit. I just thought of testing the ir2110 since it was the first time i was working with one..

The SD pin wasnt pulled low, it was kept open. Guess i should try that out. I didnt try the Lin pin input either . Will try both.
Thanks anyways.
 

From where are you giving the signal? Microcontroller or PWM IC?

You need to pull pin 11 low. You should use a smaller gate resistor although that won't stop the MOSFET from turning on.

Connect pin 11 to ground. Provide the input and observe the output across the load.
 

Tahmid said:
From where are you giving the signal? Microcontroller or PWM IC?

You need to pull pin 11 low. You should use a smaller gate resistor although that won't stop the MOSFET from turning on.

Connect pin 11 to ground. Provide the input and observe the output across the load.
Click to expand...
The gating signal is from a microcontroller.
Ya, i should have pulled the pin 11 low. Will try that out.
But what other resistance value to use 10 ohm?
 

What is the frequency of the signal? 10 ohm is okay for most frequencies unless it's very high. But, seeing as the input is from a microcontroller, the frequency won't be too high (assuming that you're using a PIC/AVR/ other 8-bit microcontroller) and you can use 10 ohm resistor.
 

Tahmid said:
What is the frequency of the signal? 10 ohm is okay for most frequencies unless it's very high. But, seeing as the input is from a microcontroller, the frequency won't be too high (assuming that you're using a PIC/AVR/ other 8-bit microcontroller) and you can use 10 ohm resistor.
Click to expand...
The frequency is 100 kHz, with 65 % duty cycle.
In fact i have tried out with 22 ohm and i got the output .
The problem with the circuit now is that ir2110 is getting heated up. what could be wrong now?
 

the IR2110 doesn't like operating at high frequencies - there are extensive notes on the data sheet relating operating conditions to power diss / temp rise in the chip
 

100KHz should be doable with a IR2110, so long as the fet is not very huge. And if you use an external gate resistor then that should dissipate most of the gate drive power, rather than the IR2110.
 

mtwieg said:
100KHz should be doable with a IR2110, so long as the fet is not very huge. And if you use an external gate resistor then that should dissipate most of the gate drive power, rather than the IR2110.
Click to expand...

It is the internal level shifting pulses that cause the heating - and this goes up with high side voltage so at 500V the freq is quite limited, the external mosfet makes little difference for Rg > 10 ohm. Regards, Orson Cart
 

lxmn said:
The frequency is 100 kHz, with 65 % duty cycle.
In fact i have tried out with 22 ohm and i got the output .
The problem with the circuit now is that ir2110 is getting heated up. what could be wrong now?
Click to expand...

How hot does it get?
Also, remember to place a 1k resistor between the gate and source of the MOSFET. Otherwise you'll face problems, especially at higher voltages.
 
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sorry , about the late reply.
It doesnt get too hot but when compared to normal IC it is working at a bit higher temperature.
Now if i leave out the heating issue i have a problem with the mosfet it seems to miss out some pulses. I have attached the output across the load resistor.
You can see that only during the half of the time the mosfet is turned on (???)
 

Hello there, there is a limit to the max duty cycle you can have as the bootstrap cap needs to charge thru the load in your test circuit, so 100% is not possible, if the recharge in the off time is not sufficient (even at 50% duty) then after a while the high side driver wil stop due to the UVLO on the IC, then the cap will charge thru the load and the process repeat...
 
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Orson Cart said:
Hello there, there is a limit to the max duty cycle you can have as the bootstrap cap needs to charge thru the load in your test circuit, so 100% is not possible, if the recharge in the off time is not sufficient (even at 50% duty) then after a while the high side driver wil stop due to the UVLO on the IC, then the cap will charge thru the load and the process repeat...
Click to expand...

Hi Orson cart ,how do we get to know the maximum duty cycle at which ir 2110 can work satisfactorily. I need it to be driving the mosfet at 100kHz with 65% duty cycle. Is it possible to extend the maximum duty cycle limit by changing the bootstrap elements or something or is at possible to run the ir 2110 at a higher duty ratio at all?
 

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