Driver circuit for IGBT

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HI EVERONE
could you help me?
I want to design circuit FOR driving Gate_IGBT (GT60N321)
to control the speed of dc motor (180 volt _ 2 HP)
I want to control the Igbt-Gate by pic microcontroller (pwm_pin)
and this is the circuit but it isnot work
 

Once you change the motor side, you can get rid of R3 since the gate is already pulled low by Q3.
 

CataM said:
Once you change the motor side, you can get rid of R3 since the gate is already pulled low by Q3.
Click to expand...

I agree with you, except if the 12v at the isolated side was not obtained from the same power source of the 180v, otherwise in the power failure of the push-pull driver rail, I guess that `weak` condition could generate an indeterminate state at the IGBT gate.
 

I guess that `weak` condition
Click to expand...
First of all the pull down used here, topology-wise (regardless of values used) is not weak, is actually a very strong pull down (hard to beat with other methods). Besides, this kind of pull down improves dv/dt immunity as well.

What you have described could happen (and hence an explicit pull down resistor is needed) if the U1 opto-transistor is swapped with R6, but not in the present configuration.
The default state of this configuration is OFF, as one can easily see when the opto-transistor is OFF. Even without power supply on the push-pull stage and opto-transistor working or not, the IGBT's gate is still being pulled low by the already described components.
If you look sharp, Q3+R6+R5 are effectively a pull down resistor, so the use of another one is redundant.

In the case when the opto-transistor is swaped with R6, the default state (when opto OFF) is the one given by the PSU of the driving stage. So in that case, if the PSU has an unknown state (or not even connected physically to the circuit), then yes, the gate without an explicit pull down has an unknown state (obviously assuming opto not working and hence not pulling down the gate).
 

First of all the pull down used here, topology-wise (regardless of values used) is not weak, is actually a very strong pull down (hard to beat with other methods). Besides, this kind of pull down improves dv/dt immunity as well.
Click to expand...

Perhaps I did not explain it clearly, but I referred to the weak condition as the lack of the pull-down resistor.

you can get rid of R3 since the gate is already pulled low by Q3
Click to expand...

I was not sure how much Q3 could ensure assertion for zero volt in the absence of polarization in the Q2-Q3 arm, but I may actually be wrong in my assumptions.
 

Thank you my friends
but ultimeately I' d like to know the intact circuit :lol::lol::lol::lol::lol::lol::lol::lol:

The only mistake is The motor should be in the transistor collector side, not in the emitter.
 

It's not clearly in which regard the circuit isn't working, even the load connection is no fault, but probably not optimal. I can imagine that the circuit suffers from interferences picked up by the high impedance opto coupler and base node, but that's just a guess.
 
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    CataM

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FvM said:
I can imagine that the circuit suffers from interferences picked up by the high impedance opto coupler and base node, but that's just a guess.
Click to expand...

I never leave the base floating. Usually connect something like a 22 k between the base-emitter. It reduces noise pickup and speeds up the turn-off of the opto.
 
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    CataM

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Hi all
I HAVE put snubber circuit for IGBT (C1-R9) C1(100nf-400volt) - R9(25ohm-5watt)
What are the correct values c1- R9 ?
and i have put R8 (100k) as a volt divider to get about 13 volt on the collecter of Q2
and this is The final electrical diagram
 

I think you should make R8 lower than 100k. You can see below that the DC level to the driver only stabilize near the zener voltage when R8 (R4 in my diagram) is under under 30k with 10 kHz switching.
 

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It depends on the final construction. If the connection to the IGBT is made via some plug/connector on the pcb, I usually prefer to have a permanent fixed resistor soldered over the G-E terminals to prevent someone from unplugging the gate drive, and you have an open floating gate. If there is no way of accidentally disconnecting the gate, then it can be left out.
 
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    CataM

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hello
I have noticed some things
the value of Main power supply is 265 volt and the motor is 180 volt - 2HP
I have chenged R3 TO 10K OHM
the ground in BAT1(12 VOLT) is the same ground in Main power supply (265VDC)?
I have tried this circuit , but I noticed that node (A) heated up a lot and it give a little somke
I need to know what is the wrong and why did this happen to my circuit ?
and how could i solve this problem?
and there is alot of noise
Please help me
 

I have tied resistance(1 ohm - 10 watt) in series with motor to divided
The voltage (the main power supply)(265 v dc)

the frequency is 500hz
the Min/Max Duty (30-180)
 

Hello
I have tried the circuit in The image below
It have worked but I have noticed that there is alot of noise and
I noticed tHE IGBT was Destructive
I need to know what is the wrong and why did this happen to my circuit ?
and how could i solve this problem?

 

Hi,

?? What´s the IRG4PC50U good for?
I assume it does nothing. It doesn´t even prevent high voltage.

Klaus
 

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