emitter followers / Darlington - need help

Michael, did you replace the 2nd transistor (the one with its base connected to the other's emitter) in that schematic with a mosfet?

Jon

Sure did. Is that a bad idea? They are both TO-220s.

Michael

I think so. I don't know what you are doing wrong that is keeping the mosfet on all the time, but an emitter follower provides no voltage gain, but actually loses a diode drop off the input voltage, just to forward bias the transistor.

If I were you, I would start with a logic level mosfet (one with a low turn on voltage that could be connected directly to the parallel port output). Otherwise, you need a 10 to

Such an amplifier can certainly be built with one to three transistors, but you can also get integrated gate drivers that accept the parallel port signals as inputs and provide full supply swing output with a low enough resistance to charge the gate capacitance very quickly, to provide low loss switching.

For instance:

Ok, thanks for the info. So they make logic level mosfets, eh... less than a buck each at Mouser.

He's got the emitter of the first transistor floating on the mosfet gate, if I'm reading him right. When the BJT turns off, the emitter just hangs there with little more than leakage to allow the mosfet gate to fall. So it stays on, I think.

Also, it may be a good idea to look for isolation here. I gather this is to drive some motor ("I'd like to be able to control a small motor via my parallel port.") I don't know what's providing that power, but it may be nice to galvanically isolate these two devices; the PC port and the motor circuit.

Jon

Jonathan Kirwan wrote: ...

So, swap them? Have the mosfet drive my TIP31?

...

I've got a pak of 8 NiMHs providing the power (~9.6V or so). Are you saying there's a chance the parallel port could see this power (hence the need for isolation)?

At least the gate wouldn't float.

You have an NMOS and an NPN device. Since your parallel output is going to be less than 5V max (unstated, but let's assume 3V max for talking purposes to cover laptops and other evils) or close to ground reference, and that is about all you have to work with, the mosfet won't turn on well. It's Vto is about 3.2V, I think.

Keeping the two devices, you might try something only a little different:

This is just off the cuff.

Well, batteries are good. There is still some potential when the motor feeds back energy -- inductive kickback. But try the above and see if that helps any.

Jon

I think you've got it kind of upside down. The parallel port is nominally a "TTL Output", which meand a weak pullup to something guaranteed to be more than 2.7V, if you're sinking less than some ridiculously low amount of current, like 4 mA or something (you could probably look it up), but the point is, you're not getting enough drive to your darlington.

I wouldn't use a darlington. I'd use a PNP, like a 2N3906 or 2N4401or so, driving a reasonable NPN; actually your TIP31 could work here, since the PNP can provide arbitrary base drive:

Good Luck! Rich

That looks interesting... I think I'll try.

I actually am using a laptop. Really low voltage, especially downstream of my 2k resistor (which I use for R2).

Why the 47k R1 though?

Thanks,

Michael

It pulls the gate of the IRF530 up when the TIP31A is off. Otherwise, you'd have a similar problem as before. The value of 4.7k is a guess. You could try the 2k, as well. Should be about the same behavior, I was just trying to lighten the load on the parallel port a little bit. R1 will need to be smaller values for faster operation, I suppose. I didn't look at the IRF530 capacitance; and for now, I don't care to find out. But it will limit the turn-on rate of the IRF530, which could be a small problem. Would need to know a lot more about what you are wanting to try, though. And by then, there would probably be a much better topology to use. So why bother, for now? Mostly, I'm just wondering how that seems to work using the parts you have on hand.

Jon

...

Gotcha. Thanks.

Basically just trying to learn. I'm using a lamp for now... once it works properly, I'll use my motor, with capacitors to smooth out the power.

I already built a 555-based PWM thingie off a schematic off google; that works great. I just thought with a parallel port's 8 pins, I could potentially control 8 motors simultaneously, from a DOS program on my laptop. It's a learning exercise (and I have so much to learn!)

Michael

Okay. I looked. 1 nanofarad territory. That means on-times in the small 10s of microseconds, roughly. If you aren't toggling around faster than 1000 Hz or so, you are probably okay.

Jon

Oh. Uh... thanks. I'm running at 100 Hz.

:-o

Michael

Emitter followers don't amplify the voltage. In fact the voltage to the FET gate will be reduced by one Vbe.

A 'logic level' gate drive FET would do the job simply.

Graham

Yup. Unfortunately, he has a TIP31A and a IRF530, right now. ;)

Jon

gate

This will likely be rectified after my next order to Mouser. Dad needs some parts, too... we will submit one large order to save on shipping.

Michael

This is actually what I'd had in mind:

(with the parallel port and my DOS software taking the part of the 555 and variable resistor.)

Yes, I was planning on sticking a diode across the motor.

Are 100nF and 100uF caps between +12VDC and ground still a bad idea...?

Thanks,

Michael

snipped-for-privacy@gmail.com wrote: (snip)

Not a good idea to smooth the output from a mosfet switch with a capacitor. The capacitor charges up to almost the same voltage, regardless of the pulse width. Better to put a diode across the motor, and let the motor's inductance average the pulses. This may require a higher pulse frequency to get a smooth rotation. If you want to try a capacitor across the motor, you will need some additional inductor between the mosfet and the parallel motor and capacitor, with the catch diode moved to the drain, instead of being across the motor.

Not at all. They should connect as closely as possible between the motor diode node and the source of the mosfet (or emitter if you use a junction transistor). This provides a very close source of charge that keeps the pulses out of the supply lines.