■ WITH RUSSELL KINCAID
In this column, I answer questions about all
aspects of electronics, including computer
hardware, software, circuits, electronic theory,
troubleshooting, and anything else of interest
to the hobbyist. Feel free to participate with
your questions, comments, or suggestions.
Send all questions and comments to:
Join us as we delve into the
basics of electronics as applied
CLASS A/B AMP
QCan you tell me how the AC
and DC analysis are done
on the class A/B transistor
amplifier (Figure 1) and
answer these questions?
1) Why is the signal input between
2) Does the signal voltage affect the
3) How can I use loop equations on
■ FIGURE 1
this circuit configuration?
— Sal Bivona
AThis is a non-linear circuit, so
you can’t do an accurate
analysis without knowing the
resistance values and voltages. However, you can linearize the
circuit by assuming a constant 0.6 volt
for the diode drop and Vbe. Then, Ve1
= Vin and the solution is trivial except
that at some point, as Vin increases, the
current through D1 will become zero
and the output will be constant (the AC
output will be clipped).
The signal input is between the
diodes because it is a symmetrical
circuit and you want a symmetrical
output signal. The presence of input
signal will increase the current output
of the circuit but since DC bias is
defined as the condition in the absence
of signal, the question is a non-sequitur.
Since the circuit is symmetrical,
analysis of the NPN circuit will also
apply to the PNP circuit. The only question of interest is: Where will the output be clipped? Let us find that point.
• Since Id1 = 0, then
Ib1 = (Vcc - Ve1 + . 6)/R1
• Also, Ve1 = (Ib1 + Iq1)*(R3+R5)
• And, Iq1 = Beta*Ib1
• This leads to: Ve1 = [(Vcc - Ve1 +
. 6)/R1]*(Beta+1)*(R3 + R5)
• To simplify, let K = (Beta+1)*
• Then: Ve1 = [(Vcc+. 6)*K] – Ve1*K
• Therefore the clipping point is:
Ve1 = (Vcc+. 6)*K/(K+1)
QWould you happen to know
where or how I can obtain
Howard Sam’s Photo Fact?
AThis site will give you many
and there is always: www.
samswebsite.com/ or you can write to:
Sam’s Technical Publishing, 9850 E.
30th St., Indianapolis, IN 46229.
60 HZ PICKUP CIRCUIT
QI have a need for a circuit
which will pick up the 60 Hz
power line signal as radiated
and generate a zero-cross
The circuit is battery powered
so it is not connected to the AC
mains. The circuit’s main source is a
solar-panel array. I would expect an
AM-ferrite type antenna and op-amp
or FET device could be used.
— Paul KJ4UO
AYou don’t say how far from
the power line you are. At any
rate, the AM radio antenna
that is designed for MHz will
not be efficient at 60 Hz. While it is true
that the US Navy transmits 10 kHz signals long distances, it requires a massive
antenna. The 60 Hz transmission lines
are designed not to transmit, so any sig-