Styx said:Well When you turn of a power device there will be a voltage overshoot due to the stray inductance (reminds me I need to write that Power electronics cookbook for this forum...)
Now since you have 100V FET's you are not avalaching them at turn-off.
YAN-1 said:Styx said:Well When you turn of a power device there will be a voltage overshoot due to the stray inductance (reminds me I need to write that Power electronics cookbook for this forum...)
Now since you have 100V FET's you are not avalaching them at turn-off.
Well I didn't understand that. Can you please elaborate on what I should do?
As for the gate drive, I am driving the gates through opto-couplers that are switched by the PIC pins and are used to pass the 5 Volts to the MOSFET gates. Sometimes I drive the gates from the PIC directly. In either case, there are of course pull-down resistors (10 k ohm). I will add the capacitor to the input of the system. Do you recommend changing the MOSFET? As for the gates, I didn't know I was supposed to provide them with that much current.
YAN-1 said:So since I have a 100 V FET, I shouldn't worry about getting a larger one as long as I keep my wires short? I mean the problem is not in the FET itself in terms of its specs?
YAN-1 said:Well thanks a lot. I've downloaded many articles related to gate drive and I'm reading. But I just have one question. If the MOSFETs were never actually fully turned on, how come the power supply always indicated a draw of full current and the motor turns (for a while ofcourse before the FETs are damaged)? Is it because it never reaches the full ON state but it is acting as an amplifier now (linear region) and it passes the current but yet heats up a lot?
YAN-1 said:Well, I read some articles and I'm getting the TC4427 MOSFET driver to drive the gates instead of using a transistor pair. This IC can provide a peak current of 1.5 Amps at 18 Volts to the gate. I did some calculations and if I'm right, my MOSFET needs around 0.43 amps of peak current for the gates to switch on so this IC should do the job at a VGS of 12 volts. Should I place a limiting resistor between the IC and the gate since it can provide me with more than I need or will the MOSFET charge faster and stop drawing current and it'll be ok? As for the inputs of the driver, it will be switched on and off through the opto-couplers. The opto-coupler will take the signal from the PIC and will pass on the PWM to the driver. That should work, right? Thanks a lot.
Styx said:I am a bit confused you say the chip is 18V but then talk abt a Vgs of 12V
Have you set it to provide 12V of something (via its supply rail).
Apart from that looking good.
YAN-1 said:Styx said:I am a bit confused you say the chip is 18V but then talk abt a Vgs of 12V
Have you set it to provide 12V of something (via its supply rail).
Apart from that looking good.
Well the IC takes a voltage range between 4.5 to 18 Volts. I'll be using 12 Volts since I already have that at the input of my voltage regulator which outputs 5 volts to my logic circuit. The peak current of the IC is 1.5A at 18V so it'll be around 1 Amp at 12 Volts assuming linearity! Do you think that I should limit that current through a series resistor? Anyways, the frequency of the motor control signals is such that each pulse is on for 1 ms. During that 1 ms, the ON pulse is really a PWM signal with a frequency of 4 KHz (so that 4 pulses are given to it withen 1 ms) with a duty cycle that varies according to closed-loop control through current sensing. Does that make any sense?
To be honest, I'm a bit scaredI'm gonna get the driver tomorrow and will (once again) build the cct. If the MOSFETs get damaged this time, I don't know what I'll do! Is there any way by which I can test the gate drive for, say, one phase only or do I have to be running the whole thing for me to see the real thing because of the autotransformer effect of the unipolar stepper motor? Thanks again.
Hi! Well thanks a lot for your help. I went over the figures and they're correct. However, turns out that I might have made a mistake in my previous calculations and I ended up getting an 18-ohm resistor for the gates and I somehow now get around 350 mA. I also changed the operating frequency because I just couldn't get to the PWM frequency that will not draw high current and will not cause the motor to stall or vibrate at the same time. Anyways, I still got current control through measuring the voltage drop across a resistor that is in series with all the phases and whenever the current goes over the ratings, I switch it off for 8 microsec and when it falls below that, I turn it on for 30 microsec. So I'm not sure about the frequency at which the MOSFETs are being switched on or off but I'm sure it can be calculated from the RL circuit of a certain phase. The bottom line is, it's working so far! I just hope it keeps up like that and no MOSFET’s get cooked in the near future because I just can't handle it now! I just never knew there was so much behind MOSFET drive to take into consideration. Thanks again.
Nichola Victor Abdo
For simplicity assume charging/discharging curve is standard exponential curve of a capacitor (in practice it isn't due to the Millar effect
I = V/R* ( exp(-t/RC) )
Now the time it takes for that curve to reach 5 times constant is going to be insignificant to your switching time so:
The instantious power in the gate resistor is IR
P = (V*V/R)*( exp(-t/RC) )
now if we integrate that over 5 time constants worth we will get the energy needed to switch the gate on/off
E = -CVV(-1 + exp(-5) )
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?