I read in sci.electronics.design that Paul Burridge
[email protected]>) about 'Are small signal npn transistors really so
different from one another?', on Sun, 21 Dec 2003:
Incidentally, is there a single identifiable parameter in the Spice
model of a transistor that would show up its susceptibility of going
into saturation and the criticality of correct biasing? Sorry if that
sounds like a load of bollocks but I'm still recovering from a
hangover.
You need to keep Vce above 1 V to prevent the capacitance's getting too
big, even if Vce(sat) is only 200 mV.
Forget your Spice model and look at the device parameters that *humans*
can understand. Beta, rbb', Icmax, Vceo, Vce(sat), Rtheta, Tjmax ....
(;-)
RC amplifier stages are quite easy, if you are not pushing for really
maximum performance. I think you have a 9 V supply. OK, I'm not even
going to specify the device, beyond 'small-signal silicon, RF rather
than audio', and the operating frequency is well below fbeta. If it
isn't, common-emitter is the wrong circuit configuration anyway.
[This is an example of the 'all transistors are the same' rule of
circuit design. It takes experience to know when to apply it, and when
you must apply the 'all transistors are different' rule.]
We'll assume that you want to get 1 V peak-to-peak out; you may want
much less, but if so you could eliminate a normally important design
step. vce (the instantaneous value) mustn't go below 1 V, so we set Vce
= 1.7 V to give a bit of margin for device variations. We set Ve = 1 V
for biasing. Vc is then 2.7 V. We want as much gain as we can
reasonably get, so we'll set the Ic at 5 mA (check if possible with the
data sheet that gm or yfe hasn't dropped off at this current; it should
be approximately 39 x 5 mA = 195 mS (millisiemens).
To drop (9 - 2.7) V at 5 mA we need Rc = 1.25 kohms (nearly). If this is
the only load on the stage, the gain is gmRc = 244 times. It might not
be stable at this gain, due to feedback effects in the device and
circuit; if so, reduce Ic until it is stable. If the stage feeds a 50
ohm load, the gain is 195 x 0.05 = 9.75.
To make Ve = 1 V at Ic = 5 mA, we need Re = 200 ohms. Take beta = 50
nominal (it's an RF device; an audio device would have much higher
beta). We set the base divider current at 0.2 times Ic, i.e. 1 mA, so
that it's a good bit larger than Ib = 5mA/50. If current drain and input
impedance are not critical, we could make the current higher, so that Ic
varies less with beta.
With Ve = 1 V, Vb needs to be 1.7 V. The total base divider resistance
is 9/1 = 9 kohms. So the resistors are 1.7 kohms and 6.3 kohms.
I've left the resistor values as the non-preferred numbers that come off
the calculator. If you want to use 220 ohms, 1.8 k and 6.8 k, just
calculate back what that does to Vc. It probably won't make much
difference, but if it does, just tweak the three values until satisfied.
The input resistance is 26 x beta/Ic = 260 ohms. Maybe add 5 ohms for
rbb'; I hope that's OK for your application. AC-couple across the input
a resistor of value (265 x 50)/(265 - 50) = 62 ohms to get a 50 ohm
input resistance.
You see what happens if you try to cascade two such stages? The low
input impedance of stage 2 reduces the gain of stage 1 to 195 x 0.26 =
51. You'd probably win a bit more if you ran stage 2 at a lower current.
There is a formula for optimising the collector currents in such a
2-stage amplifier for maximum gain. Maybe you can do the mathematics.
Yes, this is a crude analysis, but it's easy and it puts you in the
right ball-park. There are lots of ways of refining it without getting
into difficult calculations. Personally, I'd build it and measure it at
this stage! But you could SPICE it first, if the lab is too cold at this
time of year.
I just hope I didn't make any stupid mistakes in the arithmetic, which I
often do.(8-(
