Rather, the diodes are used as switches.
Remember that the dynamic resistance (rd) of a
diode is approximated by rd = 26 mV/Id.
With no current (Id = 0), rd is very high and
the diode switch is open. If Id becomes, say, 10
mA then rd becomes 2.6 Ω and the diode
switch is closed. The carrier input supplies the
Referring to Figure 2, if point X is positive
with respect to point Y, diodes D1 and D2
conduct and connect points A to A’ and B to
B’, thus allowing the input to flow to the
output. But if point Y is positive with respect
to point X, diodes D3 and D4 conduct and
connect points A to B’ and B to A’. Now the
input flows to the output but with a 180° phase
shift. In effect, the input is multiplied by -1 and is inverted.
The audio is multiplied by the carrier input. Note that the
carrier current is a bipolar square wave (switches between
+I and -I); you want those diodes to open and close
quickly and cleanly.
Assuming all the diodes are closely matched and
assuming the transformer center-taps are exactly in the
middle, then the current from the carrier input will flow
equally both ways from the middle of the transformers.
That means there is no net flux in the transformers from
the carrier, so it doesn’t couple to the output; the carrier
is suppressed. (In real life, some carrier will leak through;
but with good design, you could get 60 dB of carrier
I didn’t want to just build a circuit with three
audio transformers. What else is there? Well, another
diode switch type of circuit is a balanced bridge
modulatoras shown in Figure 3. Resistors RS and RL form
a voltage divider with RL » RS. With the
diode switches open, the output is
Vout = [ RL / (RL + RS) ] x Vin ≈ Vin.
With the diode switches closed, a
very low resistance (RD) is in parallel
with RL. Since RD « RL, we have
Vout ≈ (RD / RS) x Vin ≈ zero. This
circuit effectively multiplies the voice
signal by a unipolar square wave
(switches between +I and 0): half the
time Vout ≈ Vin; and half the time
Vout ≈ 0.
Figure 3 is easier to build than
Figure 2, but it still requires a
transformer. However, if multiplying
by a square wave will do the job, why
not use an analog switch IC? Such ICs
are inexpensive, and no transformers
■ FIGURE 3
voice changer. A DC supply is connected to tie-block TB1.
Diode D1 protects against reverse polarity. (I’ve learned
the hard way that such protection is a good idea.) D2 is a
green LED power indicator. Capacitor C11 decouples
audio from the power rail. IC3 is a 78L05 to supply a
regulated + 5 volts to an electret microphone which plugs
into 3. 5 mm stereo jack J1. R3 connects the five volt bias
to the microphone’s built-in amplifier (see symbol in
Figure 5) and also sets its gain.
IC4 needs both positive and negative supply rails.
Since we are using a single voltage to power the board,
we need to create a separate audio ground at half the
supply voltage. That’s done with 1% resistors R4 and
R5. With respect to audio ground, actual ground looks
like a negative voltage. Capacitor C6 decouples the
audio ground by creating a low impedance AC path to
The power supply you use for this project should be
■ FIGURE 4
Figure 4 is a schematic of the
May 2008 41