On Fri, Sep 27, 2002 at 07:47:58PM -0700, Tom Messenger wrote:
> Wagner wrote:
>
> >10 kOhms between base and emitter should cut the NPN when contol is
> >floatting.
> >When driving, the voltage divided will still be able to drive the NPN.
>
> Of course, Wagner meant to say that the current is divided, not the voltage
> since the base emitter junction will clamp the 10K to .6 volts.  The
> difference is that now you supply the transistor base current plus .6V/10k
> or 60uA extra.

Right.

>
> What you will need to verify here is whether the 10k turns the device off
> fast enough. You didn't mention operating frequency, I think.  At "low"
> freqs, 10K is fine. At "high" freqs, 10K is an open. You are probably
> working around, lemme guess, 38-40khz? this will be right where these
> effects begin to take place so check it out with a scope. If it doesn't
> turn off fast enough, change the 10k to something a bit lower. If your base
> voltage is .6V, then a 1k will only need 600uA.

Of course it's 40khz as that's the standard freqency.

Lowering the resistance seems like a good idea and will suck down very little
current over that 0.6V drop. For starters I think I'll do a 470 series with
a 2.2k pulldown. That'll feed the transistor nearly 10ma of current which
will drive it into hard saturation and give the 2.2k the 272uA of current
with no problem.

>
> The deal here is that for quick operation, both your drive resistor and the
> turn off resistance need to be checked to verify that they are doing their
> jobs.
>
> Hope this helps rather than confuses you. Good luck.

It helps a lot. No confusion whatsoever.

Thanks,

BAJ

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