Circuit & Component Check

[Dominic-Luc Webb]

If you run the 555 from a divider, it will have a hard time providing
any significant base drive current without collapsing its supply. I
would replace the divider with a 5 volt regulator.

I actually did have a plan to use an LM317 to regulate the 555 Vcc at 5 V.
Another way I have run this circuit is with separate 9 volt (to 555 Vcc)
and 20 volt or more (to coil) supplies. Point of exercise was to give a
higher voltage to the primary winding of a step up transformer than the
555 is capable of. The idea was to drive a higher voltage from the 555 via an
NPN. In this case I am implementing what you call the "PUMP & DUMP"
method (volts * time product is increased).

At this point, I can get my transformers to work OK in the low kHz (1-2
kHz). I have also been thinking to try driving them via solid state or
induction type relays. For higher frequency transformers (more than 30 kHz),
a transistor would be needed because it would be difficult or impossible to
use a relay. But at 1-2 kHz, most relay types look like another option.

That said, if you have a 100 ohm resister between a 5 volt powered 555
and the base of a grounded emitter NPN transistor, that leaves you
with 5 volts minus pull up saturation drop of the 555 minus the base
emitter drop of the transistor, across the 100 ohm resistor. Lets say
that the pull up saturation voltage is about (based on the middle left
graph on page 5 of http://cache.national.com/ds/LM/LM555.pdf )
1.5 volts and the base to emitter voltage of about .8 (saturated
switches have more base to emitter drop than ones operating in the
linear mode), so that leaves you with about 5 - 1.5 - 0.8= 2.7 volts
across the 100 ohm base resistor for a base current of about 27
milliamps.

If the transistor has a saturated gain (remember, you get a lot less
gain when the collector voltage gets near or below the base voltage)
of something like 20 to 50 that allows a collector current of
somewhere between .54 and 1.35 amperes. But the 555 supply has to be
able to deliver the 555 consumption and the 27 milliamperes of output
current while holding the supply steady at 5 volts.

Is all this making sense?

--
John Popelish

I need some time on the rest. I think I have not been taking into account
the pull up saturation voltage of the 555. Until now, I have not really
understood this detail and have relied on a potentiometer at the NPN base
and using this to tune the circuit for best output. I did note about 2.7
volts as you mention at the resistor and wondered how it came about. I
need to read up on this.

Dominic

 

[Dominic-Luc Webb]

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?

Dominic

 

[John Popelish]

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