Problem about IR2110

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I can't quite understand what you mean by " high side HO & LO the two inputs should become opposite input? "

Can you please clarify?

As for the 0.1uF capacitor - I had read that a ceramic capacitor must be used in parallel with the bulk capacitor if the bulk capacitor is electrolytic. If it's tantalum, then it's not necessary.
 
Tahmid said:
I can't quite understand what you mean by " high side HO & LO the two inputs should become opposite input? "

Can you please clarify?
Click to expand...

I means that the High side input(Hin) & Low side input(Lin) should be what for your given circuit?, while i want to use the MOSFET as a high-side switch.
 

If you want to use MOSFET as high-side switch and use both MOSFETs in the circuit, provide the signal input to HIN and the inverted signal to LIN.

If the load is a battery, you will need to "isolate" it from the output. An easy (but possibly lossy) solution would be a diode in the path. Use a schottky to minimize loss.

Hope this helps.
Tahmid.
 
Tahmid said:
If the load is a battery, you will need to "isolate" it from the output. An easy (but possibly lossy) solution would be a diode in the path. Use a schottky to minimize loss.

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

By 'isolate' did you mean when pwm signal is low the battery voltage can't back to the IR?
please clear me


Also, is there need any 'dead time' to set?
Thanks?
 

The battery voltage can't come back to the IR2110. Placing a diode in the path will prevent that.

If you're using both MOSFETs, some deadtime must be provided. Otherwise the MOSFETs will turn on simultaneously for a very small time during which time, they'll cause a short-circuit.

Hope this helps.
Tahmid.
 
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Also, more asking,
Tahmid said:
The battery voltage can't come back to the IR2110. Placing a diode in the path will prevent that.
Tahmid.
Click to expand...
Yes, you already talked to use shotky diode. Is there any problem if i use UF4007?

Tahmid said:
If you're using both MOSFETs, some deadtime must be provided. Otherwise the MOSFETs will turn on simultaneously for a very small time during which time, they'll contribute a short-circuit.

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

Brother, I don't know how to set 'dead time'. Please give me a hinks how can i set up the 'Dead time'
Please help me.
 

Mizan.Bangladesh said:
Also, more asking,

Yes, you already talked to use shotky diode. Is there any problem if i use UF4007?
Click to expand...

If current is low, then you can use UF4007. The current must be within bounds of the capacity of the selected diode. Keep some margin for safety. Since the UF4007 is rated for 1A, you can use it if the current in your circuit will be less than about 700mA.

Using a schottky means lower losses. However, you can use an ultrafast diode instead of the schottky if the power loss isn't too significant.


Brother, I don't know how to set 'dead time'. Please give me a hinks how can i set up the 'Dead time'
Please help me.
Click to expand...

This should come from your driving circuitry.

You could use only the high-side MOSFET and use a diode instead of the low-side MOSFET. Or keep the low-side MOSFET there, but don't use it as a MOSFET - just utilize the body diode, although this is a lossy solution. If you do this, you won't need to drive this MOSFET and so won't have to worry about deadtime.

Hope this helps.
Tahmid.
 


"just utilize the body diode"? I am not clear the second fact you mentioned here.
Would you please explain?
 

If you don't drive the MOSFET on, what you have left is essentially a diode - due to the MOSFET internal diode, which we often refer to as the body diode. This will serve the purpose of the diode required a buck converter - the one from high-side MOSFET source to ground (anode to ground, cathode to source).

Hope this helps.
Tahmid.
 
Tahmid said:
If you don't drive the MOSFET on, what you have left is essentially a diode - due to the MOSFET internal diode, which we often refer to as the body diode.

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

Brother this is new to me where i know it is like switch when Vgs=+V we get output else zero. Please give explain with example.
 

This is an N-channel enhancement MOSFET:



Due to the construction of the MOSFET, there is a "parasitic" anti-parallel diode (body diode) in the MOSFET - from drain to source. So, when you're not driving the MOSFET, the drain and soruce act like a diode - drain being the cathode and source the anode.

If you want to know more, you can Google and get loads of information on this. :smile:

Hope this helps.
Tahmid.
 
Brother,

Following is the your suggested circuit diagram. Am i right?



Secondly,
Tahmid said:
some deadtime must be provided. Otherwise the MOSFETs will turn on simultaneously for a very small time during which time, they'll cause a short-circuit.

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

if i know you about my circuit/ code. would you help me about "Dead time"?

So many thanks
 

Brother, Special thanks to you.

Tahmid said:
Your IR2112 configuration is wrong. Moreover, you have not shown the output LC filter. I thought you had planned to construct a buck converter.
Click to expand...

Yes proteus don't contain IR2110. Is now right? please give me LC value.
 

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No, the configuration is still not right.

D3 should be between VC and VB. Remove the short from VS to VC. You need to use capacitor(s) between VB and VS. You need to connect VS to the high-side IGBT source.

At such low voltages, don't use IGBTs. Use MOSFETs.

Connect gate-to-source resistors for each MOSFET/IGBT.

Use D2 after the LC filter.

Hope this helps.
Tahmid.
 
Hi Brother,

Since, You suggested me i can use shotky diode in stead of LO side MOSFET. i do that & my complete circuit diagram is given below:


But i want to share my practical experience which i done today.
when i implemented the circuit diagram, it gave me strange output!
1. when i gave Hin with 50% duty cycle it gave me output 6.32V for 5.2V given to drain!

2. without power off, When i removed Hin input then it give me also output of 4.5V for 5.2V drain voltage!

So, finally i couldn't apply the circuit where my expectation is: to get different output for different PWM cycle.
What is my wrong please explain me.


I also tried with the following circuit,


which gave me output 2.8V for input 6V with 50% pwm duty cycle.
This output is similar to expected.
But what is the limitations of this circuit.
Please clear me. I am eagerly waiting for your kind response.
Thanks
 

What load did you use to test the output?

Instead of 5V on the drain, use 12V or higher and check again. In that case, the 1st circuit should work and the 2nd one won't.

Ensure that for the IR2110, you used VDD = +5V.
 

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