Someone suggested that to ensure my MOSFET completely turns off ( switching app. about 10-40 Hz) to put a resistor between the gate and the source. I was wondering, however, will this slow down the on/off time (excuse my novice terminology)? Thanks for any help. Lucas.
What's driving the gate of your MOSFET? If it's a pin off of a microprocessor (or nearly any other IC) then your gate-source resistor won't do much.
How such a resistor would affect the switching speed would depend on what's driving the gate, but in general it would speed up turn off and slow down turn on -- the question is whether the difference would be significant or vanishingly small.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html
Thanks for the reply, My MOSFET is being indirectly driven by my printer port on my PC. I say indirectly because im actually using my printer port to engage a 2N3904 transistor to allow 10V (instead of my printer port's 3.5V) engage the MOSFET.
Im using the MOSFET to control the speed of a motor using PWM from my PC. The problem is, however, when I run my MOSFET at 10Hz it performs ok, but when I run it at, say, 40Hz it chokes because I think because my MOSFET isn't actually going on and off like it should. Anyway, thanks for you help, any more Ideas welcome.
- Lucas
You can turn a mosfet on just by touching the gate with your finger, and it will probably stay on even after take your finger off (try it). Mosfets aren't like bipolars. They don't just turn off by themselves. Mosfets need a little encouragement turning off. If you drive the mosfet with something that both sources and sinks current then you are fine. But if the drive is source-only, use the gate-source resistor.
Thanks for the quick response, but I still have a question (sorry for my inexperience). My question is, what exactly is meant by source and sink and how do I know if my supply is or isn't "sink-only"? Like i said, im using a 2N3904 to drive the MOSFET, so my intuition tells me that when my printer port is "off" I am actually driving the MOSFET gate to ground. Is this what you mean by sinking? I could be wrong. Lucas
When you reply from Google Groups it can be difficult to do so in accepted USENET format. Please read the link at the bottom of this mail for guidance on doing it right with Google.
A current source will deliver current by pulling its output high. A current sink will take it away by pulling its output low. If your '3904 is connected with the emitter to ground then it'll do a fine job of turning the MOSFET off.
If you can get your hands on an oscilloscope I'd suggest looking at the various voltages, preferably all at once. You have three to worry about directly: the voltage out of the printer port, the voltage at the buffer transistor's collector, and the voltage at the MOSFET drain. It would be hard to build a circuit too slow for 40Hz, but it can be done. More likely your PC isn't keeping up -- but you really need a scope to tell you that.
If you could post a link to a schematic, or use AACircuit to make an ASCII-art schematic, you'd get a critique.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html
Here is my circuit...
18V 8A (for motor)p.s. I hope the format is better?!? Im confused what the "right" usenet format is, sorry.
You don't have a speed problem with the MOSFET - not at 40Hz. How are you generating the timing with the PC? Normally, it is difficult at best to create timed signals on a computer with a massive opperating system do who-knows-what with your processor time.
Luhan
You're right, it is a "taboo" to use a PC to control timing aspects of a circuit. Basically, I am using code to set a pin on my prt. port high, then pausing for some time, then setting it low, pause etc... I can set my frequency by specifying the "pause" times. Anyway, I would be curious to see how bad the computer actually does at this job. Guess i would need a scope for that. oh well, thats how I am creating my PWM. Thanks, Lucas.
The gate drive circuit is fine as-is provided you don't start trying to do a 20kHz PWM on the motor. But you should have a fairly fat catch diode across the motor or you'll likely kill the MOSFET in short order. A little less likely, but so could spinning the motor backwards at higher than full unloaded speed.
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Your NPN will turn off a little slowly, but 'a little slow' in this case should be on the order of microseconds, not 10's of milliseconds (you can speed this up with a resistor from base to ground, and a lower value base resistor). You are correct that the NPN should pull the gate down sharply, unless you have a really low Hfe NPN in there -- lowering the base resistor to 10k wouldn't go amiss, but probably won't fix your problem.
The 10K pullup working against the MOSFET gate is a little slow, too, but even with some Miller effect it should be way faster than 40Hz.
Take Spehro's advise about a catch diode, by the way.
It is better. Some of the original post would have been good for context, but it's better.
The idea is that you should assume that your post is the only one available -- newsgroup servers often discard older messages, and newsreaders rarely show anything but the one message being looked at. So you should quote enough of the old thread that your message makes sense by itself, with your bit tacked on the end or interspersed.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html
IRF510s have a built in free wheel diode. Their gate capacitance is 135pF according to the datasheet, which excludes a low switching time on this part.
My guess is that you're using sleep or equivalent to generate the delays. Timers on modern operating systems usually run at 100Hz, so at
40Hz you're getting close to the limit. Try using a different technique for your delays, like for loops. You'll get 100% CPU usage with this but you can't get precise timings without sacrifice :)vic
Really? And even if you enjoy avalanching the poor thing, what if the motor inductance is more than ~600uH @8A?
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
Yes, what the other people say. You definitely need a catch diode across the load to stop you from avalanching the mosfet, like Speff says. You really need to reduce that 100K resistor, like Tim says.
Your 2N3904 has a hfe of 70min at IC=1mA. Take 0.6V Vbe of 3.5V and you have
2.9V on your 100K resistor and the Ib is 29uA. Times 70 gives you a maximum 2mA collector current. Yes that does swing 10K through 20V but it's a little bit marginal when you could gaurantee it with 10K.Your other worry is that when you switch your computer off, if the printer port output goes high impedance, then your circuit will default to being ON. Also the IRF510 is only rated for about 5A continuous and that assumes serious heatsinking.
DNA
Because Windows is a timesharing system, the computer is not always running your program. It goes bye-bye often to do many other things. As you try to generate higher frequencies, the effect is more noticeable. Basically, it just does not work for PWM.
You can use it for I2C and some other protocols because the exact timeing is not critical, only the releative times of the clock and data signals. It is often used to do very low cost PIC programmers for the same reason.
Luhan
You might try some external pwm circuit with some counter if you want to keep it simple. The better approach is proberbly to use a mcu (avr/pic) to drive this. That way you do away with any timeing issues.
There's also the dos way. Then you will be controlling on a per cycle basis. On pentium cpus you might want to look into RTC clock. That counts up for every cycle.
LOL :)
Cheers Terry
I'd also change the gate pullup from 10k to 1k. Or better yet, smack in a buffer. I doubt its too much of a problem even right up there at 40Hz, but 10k wont be winning many arguments with miller capacitance (Cgd) when the FET tries to switch.
PCs are shit for timing. My computer guy couldnt give me a guaranteed
1ms tick under windows....Hilarious when you consider my cpu will do a floating point multiply in a few ns.
Cheers Terry
Hmmm ... looks like I made a fool of myself ... I was told this was a free wheel diode, apparently it is not, so would you care to explain ?
thx, vic
Not at all.
Sure. When the MOSFET is 'on', current flows from the supply to drain to source. When the MOSFET starts to turn off, the voltage at the drain will *rise* to higher than the supply rail (due to motor inductance), most likely limited in this case by the avalanche voltage of the MOSFET. That dumps some energy (high current at high voltage, so quite high instantaneous power) into the MOSFET. The absolute maximum single pulse energy for the given part number is only 0.019J. PM motors don't tend to be as high inductance as solenoids so you might actually get away it, but it's not a good idea.
A freewheel diode across the load will limit the drain voltage to a volt or so above the positive rail, and the diode itself will dissipate very little power because the voltage is so low. In case you're wondering, the energy stored in the inductance gets used up in the motor, which is usually where you want it anyway.
The diode you see in the MOSFET datasheet (body diode) won't conduct unless the voltage at the drain drops below ground. That is actually possible under some conditions with a motor (for example if you mechanically overdrive the motor at faster than full speed or if the motor supply was suddenly pulled down with the motor spinning the back-EMF could cause the body diode to conduct). The result would be braking. But that diode is of no help dealing with the motor inductance.
Best regards, Spehro Pefhany
-- "it\'s the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
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