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Electronics4you · 8th May 2007, 05:30 PM

Hi there,

I have an emergency. I have just builded the attached circuit, but the IGBT's went up in smoke, when I plugged in the 230V mains. It's a circuit for driving a 230V inductor motor in three phases with IGBT's controlled from a PIC.

I really need some ideas on how to drive such a motor with a simple circuit, though I need to use the motor in a few days. Have you any ideas on how to do it with no special components (like IGBT's) or how to fix my other circuit?

Any help will be appreciated
Thanks

Papabravo · 8th May 2007, 06:06 PM

What is it that prevents both upper and lower devices from being ON at the same time?
As I see it when the PIC side is unpowered the LEDs in the optos will be OFF, the output transistors in the optos will be OFF and the 1K pullup resistors will pull the inputs of the TC426 high and the outputs of the TC426 will be low and the IGBT's will all be off. This is a good thing.

When the PIC is powered on the behavior of the port pins is critical. If they are high impedance with pullups enabled then the LEDs in the optos will be ON, the transistors in the optos will be ON pulling the inputs to the TC426 LOW, allowing the outputs to go HIGH turning both IGBT's ON letting out the magic smoke. Is that what happens, or did I overlook something?

Electronics4you · 8th May 2007, 07:32 PM

It wasn't actually the IGBT's if I wrote that. The 10k resistor to the gate burned out, but the power was only 15.6V to that, which is not enough. Can the capacitor give some energy to kill the resistor?

Electronics4you · 8th May 2007, 07:58 PM

Is the 10k resistors really needed, or is the 1.5A max output from the driver IC's enough and it won't burn?

Papabravo · 8th May 2007, 08:13 PM

What was the power rating on the 10K resistors? It is hard to imagine more than 15.6 Volts across the resistors. If they were 1/10 Watt 0805 SMT resistors then

Code:

15.6V/10K = 1.56 mA
((1.56 mA)^2)*10K = 24 mW
NOT enough for smoke

The only thing the 10K resistors do is slow down the switching of the IGBT's. So when did the resistors smoke? How many of them smoked? What was the PIC doing when they smoked? Would it be possible to actually include some relevent and useful information in your post if you want some actual help instead of rank speculation?

philba · 8th May 2007, 08:35 PM

just a guess but it sounds more like 230 was across the 10K resistor - that's 5W. poof! Probably a wiring error.

Electronics4you · 8th May 2007, 08:38 PM

Only five of the resistors smoke. They started to smoke, when a just pluged in the main supply. The PIC wasn't connected then. I therefore didn't had the time to measure the voltage to the resistors.

I tested the circuit with a 20V DC supply (soldered to the cap) before adding the 230V. It worked fine and only 15.6V to the resistors. The motor (a smaller one) started running with the supply of 5V. Is it possible that the 230V mains is interrupting the driver IC's?

Papabravo · 8th May 2007, 09:27 PM

Could you clarify what you mean by:
"Is it possible that the 230V mains is interrupting the driver IC's?"

The gate of an IGBT like the gate of a FET does NOT draw appreciable DC or quiescent current. The only way I see for five resistors out of six to be destroyed is a wiring mistake. Without more information I don't see any possibility of helping you further.

The driver chips are connected to 15.6VDC and the output side of some optos also connected to 15.6VDC. Now you still, after two requests, have not told us what the power rating or size of the 10K resistors was. If they were 1/10 watt you might need at least 150 mW for some amount of time to make them go up in smoke. Working backwards

Code:

150 mW / 10K = 15e-6
SQRT(15e-6) = 3.87e-3
3.87e-3 * 10K = 38.7V

So back of the envelope you would need almost 39 Volts across the 10K to produce enough current to cause them to smoke.

This suggests a wiring error since you have no source of 39V on your board.

Once again you have failed to answer the original question which was: What prevents BOTH IGBT's from being on at the same time when the PIC is not in control?

Electronics4you · 8th May 2007, 09:32 PM

The power rating for the resistor is/was 250mW so that should be more than enough, and nothing is currently protecting the IGBT's from being ON. They might have to be pulled up in order for the driver IC to pull them down.

Papabravo · 8th May 2007, 09:40 PM

But..But..if you pull up the gates won't they be on ALL the time?!
Oh...I see you mean pull up the inputs to the driver IC.
But they are already pulled up by the resistor on the output of the optos.

You need to look for a current path that took out 5 of 6 resistors. What is different about the one that was saved?

I guess you're just going to have to buy me a plane ticket to Denmark. Lovely country, wouldn't mind another visit?

LOL

Electronics4you · 8th May 2007, 09:56 PM

Yeah, I'll just go through the entire circuit again and again until the error is fixed.

Thanks

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8th May 2007, 06:06 PM	(permalink)
Papabravo	What is it that prevents both upper and lower devices from being ON at the same time? As I see it when the PIC side is unpowered the LEDs in the optos will be OFF, the output transistors in the optos will be OFF and the 1K pullup resistors will pull the inputs of the TC426 high and the outputs of the TC426 will be low and the IGBT's will all be off. This is a good thing. When the PIC is powered on the behavior of the port pins is critical. If they are high impedance with pullups enabled then the LEDs in the optos will be ON, the transistors in the optos will be ON pulling the inputs to the TC426 LOW, allowing the outputs to go HIGH turning both IGBT's ON letting out the magic smoke. Is that what happens, or did I overlook something? Last edited by Papabravo; 8th May 2007 at 06:32 PM.

8th May 2007, 07:32 PM	(permalink)
Electronics4you	It wasn't actually the IGBT's if I wrote that. The 10k resistor to the gate burned out, but the power was only 15.6V to that, which is not enough. Can the capacitor give some energy to kill the resistor?

8th May 2007, 07:58 PM	(permalink)
Electronics4you	Is the 10k resistors really needed, or is the 1.5A max output from the driver IC's enough and it won't burn?

8th May 2007, 08:13 PM	(permalink)
Papabravo	What was the power rating on the 10K resistors? It is hard to imagine more than 15.6 Volts across the resistors. If they were 1/10 Watt 0805 SMT resistors then Code: 15.6V/10K = 1.56 mA ((1.56 mA)^2)10K = 24 mW NOT enough for smoke The only thing the 10K resistors do is slow down the switching of the IGBT's. So when did the resistors smoke? How many of them smoked? What was the PIC doing when they smoked? Would it be possible to actually include some relevent and useful information in your post if you want some actual help instead of rank speculation? Last edited by Papabravo; 8th May 2007 at 08:15 PM.*

8th May 2007, 08:35 PM	(permalink)
philba	just a guess but it sounds more like 230 was across the 10K resistor - that's 5W. poof! Probably a wiring error.

8th May 2007, 08:38 PM	(permalink)
Electronics4you	Only five of the resistors smoke. They started to smoke, when a just pluged in the main supply. The PIC wasn't connected then. I therefore didn't had the time to measure the voltage to the resistors. I tested the circuit with a 20V DC supply (soldered to the cap) before adding the 230V. It worked fine and only 15.6V to the resistors. The motor (a smaller one) started running with the supply of 5V. Is it possible that the 230V mains is interrupting the driver IC's?

8th May 2007, 09:27 PM	(permalink)
Papabravo	Could you clarify what you mean by: "Is it possible that the 230V mains is interrupting the driver IC's?" The gate of an IGBT like the gate of a FET does NOT draw appreciable DC or quiescent current. The only way I see for five resistors out of six to be destroyed is a wiring mistake. Without more information I don't see any possibility of helping you further. The driver chips are connected to 15.6VDC and the output side of some optos also connected to 15.6VDC. Now you still, after two requests, have not told us what the power rating or size of the 10K resistors was. If they were 1/10 watt you might need at least 150 mW for some amount of time to make them go up in smoke. Working backwards Code: 150 mW / 10K = 15e-6 SQRT(15e-6) = 3.87e-3 3.87e-3 * 10K = 38.7V So back of the envelope you would need almost 39 Volts across the 10K to produce enough current to cause them to smoke. This suggests a wiring error since you have no source of 39V on your board. Once again you have failed to answer the original question which was: What prevents BOTH IGBT's from being on at the same time when the PIC is not in control?

8th May 2007, 09:32 PM	(permalink)
Electronics4you	The power rating for the resistor is/was 250mW so that should be more than enough, and nothing is currently protecting the IGBT's from being ON. They might have to be pulled up in order for the driver IC to pull them down.

8th May 2007, 09:40 PM	(permalink)
Papabravo	But..But..if you pull up the gates won't they be on ALL the time?! Oh...I see you mean pull up the inputs to the driver IC. But they are already pulled up by the resistor on the output of the optos. You need to look for a current path that took out 5 of 6 resistors. What is different about the one that was saved? I guess you're just going to have to buy me a plane ticket to Denmark. Lovely country, wouldn't mind another visit? LOL Last edited by Papabravo; 8th May 2007 at 09:44 PM.

8th May 2007, 09:56 PM	(permalink)
Electronics4you	Yeah, I'll just go through the entire circuit again and again until the error is fixed. Thanks