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I’m just thinking. How come that PIC output was damaged? I found out that PIC output can safely sink and source 25ma. So, with the last circuit diagram. Steve suggests a 1k base resistor of BC547. This means, even base of BC547 was shorted to +12V. current flow in 1K resistor was 12MA. So, the last circuit diagram was not the cause of partially stuffed PIC.
I think 5 pic’s was damaged on when 300 ohms was used as base resistor of BC547 which have a 40ma current flow when BC547 base and collector was shorted. . Exceeding source and sink current PIC output of 25ma..
I think 5 pic’s was damaged on when 300 ohms was used as base resistor of BC547 which have a 40ma current flow when BC547 base and collector was shorted. . Exceeding source and sink current PIC output of 25ma..
No, what is more likely is that the rapid switching of current to the coil which happened after the gate driver was added combined with the poor wiring resulted in large voltage spikes appearing in the circuit.
These could have entered the PIC via the driver (the OP mentions that one of these transistors died).
In pretty much all cases, you can short a PIC output to either supply rail without causing any damage as the outputs are effectively current limited. Do it to too many of them and you can exceed the device power dissipation though. (Don't take this as a recommendation to short outputs to the supply rail.
These could have entered the PIC via the driver (the OP mentions that one of these transistors died).
In pretty much all cases, you can short a PIC output to either supply rail without causing any damage as the outputs are effectively current limited. Do it to too many of them and you can exceed the device power dissipation though. (Don't take this as a recommendation to short outputs to the supply rail.
Lets see now......you have a PIC on a small board, long wires hanging out everywhere,
and a spark plug sitting about a foot away......
That spark plug is nothing but an EMP device, and every damn wire on that table is going to see a spike when that plug fires!
PIC's are not designed to withstand that kind of EMI.....no way, no how!
and a spark plug sitting about a foot away......
That spark plug is nothing but an EMP device, and every damn wire on that table is going to see a spike when that plug fires!
PIC's are not designed to withstand that kind of EMI.....no way, no how!
GonzoEngineer, I've never had reason to use an IGBT in this manner, and I'm simply considering it to be a mosfet with different characteristics )).
How do you calculate the turn on time for an IGBT based on the available gate current? I note that the gate voltage tents to follow a similar curve to a mosfet, but the datasheet doesn't mention the gate charge as such (I presume "On State Gate charge" is the magic number).
Looking at figure 16 in that datasheet is seems that most of the switching occurs with the gate voltage between about 6 and 7 volts and over a charge of about 10nC. Would it be reasonable to (as a ballpark figure) to assume switching occurs in the time it takes to transfer this 10nC of charge to the gate? I'm assuming that the earlier portion of the charge effectively gets you up to the Vgs(th) and that further charge us essentially superfluous as the device is beyond it's linear region at that point.
If so, a 10k resistor from a 12V source would transfer 10nC across (say) 6.5 volts in about 15uS? -- and that's somewhat faster than I expected.
)).How do you calculate the turn on time for an IGBT based on the available gate current? I note that the gate voltage tents to follow a similar curve to a mosfet, but the datasheet doesn't mention the gate charge as such (I presume "On State Gate charge" is the magic number).
Looking at figure 16 in that datasheet is seems that most of the switching occurs with the gate voltage between about 6 and 7 volts and over a charge of about 10nC. Would it be reasonable to (as a ballpark figure) to assume switching occurs in the time it takes to transfer this 10nC of charge to the gate? I'm assuming that the earlier portion of the charge effectively gets you up to the Vgs(th) and that further charge us essentially superfluous as the device is beyond it's linear region at that point.
If so, a 10k resistor from a 12V source would transfer 10nC across (say) 6.5 volts in about 15uS? -- and that's somewhat faster than I expected.
I'm planning to modify my multispark cdi project from IR2155 self oscillating H-driver into PIC16F84 for spark pulse triggering. I found out that RC waveform on RC input of IR2155 was distorted causing missing pulse or misfiring due to noise on ground from high voltage spark.
I don't have EMP problem with my multispark cdi board even it was 1 foot away from spark plug.. IR2155 ic was only damaged when I forgot to discharge filter capacitor before removing the IC or re-insert it again on ic socket. It's also Mosfet with IGBT.
How did you figure 6v? 12v/10.5k=1.14ma, 500x1.14ma=.57v
Don't know much about pics but if he would use the transistor that I recommended it should work without a pull up resistor. With the irfbe30 that has a threshold of 2.55 and full on at 4.33v it should work without the pull up resistor. Why reinvent the wheel? I built a couple of circuits using bjts and a driver,like you did, but they only put out 3000v at the best.
This is very simular to the tester that I use.http://www.ebay.com/itm/Zareba-RSVT...922?pt=LH_DefaultDomain_0&hash=item416282f572
That sounds like an interesting project. I hope that you start a new thread. I built a capacitor discharge ignition, but it only puts out one spark if I remember correctly. But it was over 9000v! I don't know how much over because my tester only goes to 9000v. I wish I had scoped it to know for sure. What are you using the circuit for?
Hey another new word for me . "partially stuffed " means ruined, shorted or defective.
A new word for me too. At least we're learning new words! ha
The ESD protection diodes will not allow the output to rise more than 0.7V above the supply voltage, so 5.7V.
With your 10K pullup, and 500R resistor in series with the output, you have a voltage divider with 12V at one end, 5.7V at the other, and a total resistance of 10500 ohms.
This gives a current of 0.6mA.
0.6mA across the 500R resistor drops 0.3V, therefore the gate has a voltage of 6 volts.
edit:
I looked briefly, but I didn't see your recommendation.
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I have tried bjts before, but they never worked well.
You realise he's not using a BJT, don't you?
Ok, now I understand. I guess the pic would never drive that igbjt transistor. That's what I didn't know about pics. But if the pic puts out that much voltage it should easily drive the irfbe30 that I recommended. And if the pic doesn't put out that much voltage the pull up resistor would still work with the irfbe30, because it is full on at 4.33 volts according to my experiments with it. There might even be better transistors than that one, because it has a high on resistance of .3 ohms. A lower voltage transistor might work, but I chose that one because I didn't know how much flyback voltage there would be. It is rated at 500 volts if I remember correctly.
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I thought igbjt stood for insulated gate bi junction transistor.
GonzoEngineer, I've never had reason to use an IGBT in this manner, and I'm simply considering it to be a mosfet with different characteristics
How do you calculate the turn on time for an IGBT based on the available gate current? I note that the gate voltage tents to follow a similar curve to a mosfet, but the datasheet doesn't mention the gate charge as such (I presume "On State Gate charge" is the magic number).
Looking at figure 16 in that datasheet is seems that most of the switching occurs with the gate voltage between about 6 and 7 volts and over a charge of about 10nC. Would it be reasonable to (as a ballpark figure) to assume switching occurs in the time it takes to transfer this 10nC of charge to the gate? I'm assuming that the earlier portion of the charge effectively gets you up to the Vgs(th) and that further charge us essentially superfluous as the device is beyond it's linear region at that point.
If so, a 10k resistor from a 12V source would transfer 10nC across (say) 6.5 volts in about 15uS? -- and that's somewhat faster than I expected.
15us is a long time. What many people don't pay attention to is the inductance of the drive circuitry.
One thing I have learned in the Pulsed Power business is that R and C are your best friends, but L is like your freaking ex wife!
To turn on an IGBT, you are essentially charging the Miller Capacitance of the gate.
You need a high current driver (TC426 or TC427) and you have to make sure the supply to the driver has a very low ESR Capaitor with the stored energy to hit that bastard gate as hard and as fast as you can.
You need to charge that gate as fast as possible. And you also need to turn off that gate just as fast! That is why a lot of IGBT drivers, (unlike MOSFET drivers) swing from +18V to -18V.
Don't bother reading those silly "Threshold Voltage" numbers.....they are talking about the linear range.....which is where you want to stay out of when switching. That's where the heat builds up in the die.....and the smoke comes out if you stay there too long.
Hope this answers your questions.....
It's IGBT
Yeah, it's rated as having Rds of 3 (not 0.3) ohms at Vgs = 10V
According to the specs, at 4.33V Vgs and 25C, you'll get an Igs of about 200mA.
I don't think it's the device I'd choose.
Note that the tolerances for mosfets tend to vary somewhat more than for some other types of transistors, but I'm not certain that your testing (even so) was correct.
Or maybe you have a different device?
Ha, ha,
I just looked at the data again. The current goes down when temp. goes up, just like a mosfet.Similar threads
