Re 1k 10uF
between the collector and emitter
called the base that controls the
current flow through the transistor.
A gap so small that it took nearly
six years between 1948 and 1953
to perfect the first reliable commercial transistor: the CK722.
logic. In the NPN configuration, a logic HIGH turns
on the transistor. In the
PNP version, a logic LOW
turns on the transistor.
Make sure you adjust
A history note. When I
was a tyke and new to
transistors — having cut
my teeth on voltage-controlled vacuum tubes — I heard about this back-to-back diode transistor analogy — and
tried to build one myself using 1N34A
diodes. Guess what? Didn’t work.
The secret to the transistor’s
transconductance is the tiny gap
If your application is for logic
switching, all you need to do is
exchange the emitter and collector in
your design so that the forward/reverse
bias rule is maintained, as shown in the
two bottom circuits. The catch is, when
you change sexes, you also change
Q. I wish to convert the 6. 3
VAC heater voltage in my
EICO HF- 32 tube amplifier from
AC to DC. I started with a full-wave bridge rectifier (no center-tap) of the power transformer,
then connected two 1,000 uF caps in
parallel for filtering. I stopped at this
point to take some measurements,
which was approximately 9. 5 volts
DC with 2 mV AC ripple with no load.
This is fine and good and agrees
with theory. But when I connected the
heaters, the voltage dropped to 5. 3
volts DC and the ripple increased to 2
volts AC! I can understand the DC
drop of an unregulated supply, but
I’m baffled as to the ripple increase. If
anything, it should decrease! Any
ideas on the ripple increase?
— John Agugliaro, CET
NUTS & VOLTS
Circle #88 on the Reader Service Card.
A. Sorry, but you have it backwards.
The more current you draw, the
greater the ripple. How come? When
AC voltage is full-wave rectified, you get
a bunch of peaks with very deep valleys — think the Grand Tetons (Figure
3). When you place a capacitor across
this ripple, it charges to the peak voltage, then discharges in the valley. The
discharge rate of the capacitor is proportional to the load resistance; i.e., the
output current. The more current you
draw, the faster the filter cap discharges
— and the greater the ripple.
For one volt of ripple at one amp
you need 8,300 uF. For example, one
volt of ripple at 500 mA is 8,300/2 or
4,150 uF. To reduce the ripple to 10
mV at 3. 5 amps, you need 2.9F (yes,
farads!). As you can see, your 2,000
uF cap combo is a pathetic attempt
at filtering the DC. One solution is to
use a supercap. But they are rated at
just 2.5 volts, which means you’ll