Re: Circuit & Component Check
From: John Popelish (jpopelish_at_rica.net)
Date: 01/13/05
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Date: Wed, 12 Jan 2005 19:14:21 -0500
Dominic-Luc Webb wrote:
>
> > > > Driving any switching transistor with a large voltage and large series
> > > > resistor approximates a current source, which aggravates the turn on
> > > > and turn off time problems. Slightly more complicated drive schemes
> > > > (adding a base to emitter resistor to drain the stored charge out
> > > > faster at turn off and paralleling the series resistor with a small
> > > > capacitor to drive the transitions harder) can result in significant
> > > > efficiency improvements.
>
> I have actually seen a couple circuits in which this was done, long ago.
> No hint was given about why these components were in this config or how
> to estimate values. Any idea where this is documented, and in particular,
> with some idea how to calculate the resistance and capacitance needed?
I usually divert at least 1/10th of the base drive from the series
resistor to the emitter, assuming about .6 volts Vbe. For example if
you have 2.7 volts across your base resistor for a drive of 27 ma
drive current, I would put a resistor no higher than .6V/.0027A=220
ohms base to emitter. THis will drain charge out of the base, even if
the 555 output does not go all the way to zero volts (and it won't
while current is passing back to it from the base).
The optimum capacitor value depends on several factors, including the
base stored charge (smaller transistors generally have smaller stored
charge, so using the biggest one you can find does not improve
everything), the driving signal swing and rise and fall time, and the
pulse width. You don't want this cap to drive the base so far into
reverse bias that it breaks down the emitter junction, or it will
slowly degrade the current gain if the transistor. If this happens,
you can clamp the reverse voltage with a signal diode base to emitter.
And the series resistor and the paralleled speedup capacitor have to
have a time constant less than the minimum on or off time so things
get back near center before the next transition has to be handled. I
usually look at the base voltage waveform with a scope and select a
capacitor that causes Vbe to go at least to zero on turn off (all the
base charge sucked out before the resistors take over to hold zero
volts), after I have gotten the two resistors selected to provide the
required saturation voltage, Vce, during the largest load current. If
the driver swing is very large, compared to Vbe on, I sometimes put a
low value resistor in series with the speed up cap to limit the peak
driver current to a safe value during the transitions.
The whole idea of this network is a recognition that there are two
current requirements for base drive. One is the DC requirement to
force the transistor to maintain a low Vce during the on time, at the
highest load current and the other current is required only at the
transitions to drive stored charge into and back out of the base
emitter junction to produce fast turn on and turn off for high
efficiency switching.
-- John Popelish
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