## 3 Transistor Op-Amp ```

The 3 transistor op-amp shown was designed for use as a 20 milliwatt
but lacks efficiency since the output is single ended and draws around
60mA from AA batteries, or 120mA for a stereo application.

Parts Selection:

For a power output of 20mW (RMS), the peak output power will be 40mW
and the peak output current will be I^2*R =.04 and I(peak) = 32 mA.
The negative peak of the output signal occurs when transistor Q3 is
completely turned off, so the output voltage is just the result of
a voltage divider 40 ohms and R4. If the peak current is 32 mA, then
the peak output voltage will be E=IR = .032*40 = 1.28 volts. Since the
negative supply voltage is 3 and the peak load voltage is 1.28, there
will be 3-1.28 = 1.72 volts across R4. So, we need a resistor that drops
1.72 at 32 mA or R4 = E/I = 1.72/.032 = 54 ohms, or 56 standard value.

Next consideration was the base and collector currents for Q3. Since the
output voltage swing is symmetrical, the peak voltage across R4 will be
3 + 1.28 or 4.28 and the current in R4 will be I= E/R = 4.28/56 = 76mA.
The peak collector current of Q3 will then be 76mA plus 32 milliamps in
the load or, 108mA total. This seems a little high for a 2N3906, maybe
some other transistor would be better.

Next consideration was the steady state DC current for the differential
pair of transistors (Q1,Q2). The DC currents of Q1,Q2 should be somewhat
balanced (with no signal) so that both transistors carry near equal DC
currents. And this current should be somewhat greater than what is needed
by Q3. The measured hFE gain of the 2N3906 at 100mA was about 100, so the
peak base current of Q3 should be around 1 mA. The DC current for Q1,Q2
was then chosen to be about 3 times greater, or 3mA for each transistor
(6 mA total). It could be larger, but would increase battery drain.

Next step was to work out the common emitter resistor R3. Since the base
of Q2 is at ground, the emitters will be about -700mV and the voltage
across R3 will be 3-0.7 = 2.3, and the value of R3 will be
E/I = 2.3/0.006 = 390 ohms. The total current in R3 will then be 6mA,
or 3mA for each transistor Q1 and Q2.

Next step was to assign a value to the Q3 e/b resistor (R2). In the
balanced (no signal) condition, Q3's base current will be about 0.5mA and
the remaining current for R2 will be 3mA - 0.5mA = 2.5mA, so the resistor
value would be R2=Vbe/.0015 = 0.7/.0025 = 280 ohms (270 standard value).
Resistor R1 will have the same value and is used to measure the DC current
in Q1 which should be about 3mA, so the voltage across R1 should read
around 810mV if things are working right.

The amplifier gain was selected to be ten or (20dB) which defines the ratio
of R5 to R6, or 1K and 10K. This makes the input impedance about 1K since
the inverting configuration is used. For a higher input resistance, larger
resistor values could be used (10K 100K) etc. Or you could ground the 1K
imput resistor R6 and use the non-inverting input point (base of Q2) as a
high impedance input.

Notes:

A small 1000pF cap was added to limit the bandwidth to around 100kHz and
suppress oscillations. The value was found by experiment.

The circuit seems fairly linear (low distortion) as indicated in the
scope picture below. The top waveform is the ramp input from a function
generator, and the lower waveform is the inverted output of the op-amp
with a 40 ohm load. So, it's actually a plot of the rising input ramp
and falling output ramp (2.25v p-p). Notice there is a slight bend in
the output line indicating some non-linearity (distortion) but hard
to see.

The DC offset at the output will move negative as the battery voltage falls.
It measured about +50mV at full supply voltage and drifted negative to
-150mV with the supply at +/- 2.5 volts. This can be corrected with a
small adjustment to R3. Lowering R3 will increase the current in Q1/Q2
and move the DC offset positive, and visa versa. Another approach is to
replace R3 with a constant current source (one extra transistor) which
should maintain the DC offset near zero as the supply voltage falls.
But then it wouldn't be a 3 transistor opamp anymore. It would be a
4 transistor opamp.
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