Electronic – Why is the common emitter input insensitive to base width modulation

In a common base circuit, the signal input source is applied to the emitter terminal. The "lump" of semiconductor that seperates base and emitter is a forward biased diode. The base is held at a constant voltage and therefore the emitter signal voltage causes a signal current to flow thru the base terminal.

Irrespective of whether the circuit is common emitter or base, the signal voltage across the forward biased B-E diode always produces a signal current through the base terminal. That signal current is determined by the diode biasing point i.e. how forward biased that junction is AND the magnitude of the signal.

That base signal current (even if the real input is the emitter) is amplified by the current gain and this results in a much larger signal current in the collector. The collector signal current (superimposed on the DC quiescent current of course) also flows thru the emitter terminal (plus of course the much smaller base signal current irrespective of where the signal source is connected).

So, an input connected to the emitter (common base) has to be able to cope with collector current AND the little bit of current that ends up flowing thru the base terminal.

With an amplifier like this, you want to bias the collector to more or less mid-rail, this maximises the +/- swing it can make from the quiescent state before hitting the top rail, or running out of voltage drop across the transistor. Choosing 5v, as half the rail voltage, is about right, though it could be a tad higher.

Once we have chosen the output voltage, we choose a collector resistor. The range of sensible values is quite wide at this stage. Unless we are asked to make a certain output impedance, then 1k to 10k sort of range is reasonable. They have chosen 10k. With 5v on the collector, this means we need a collector current of 500uA.

Now we need to choose, again fairly arbitrarily, the voltage across the emitter resistor. It needs to be enough to swamp variations in transistor Vbe with batch variation and temperature variation. It needs to be small enough to not use up all of the available rail volts. I tend to go for about 10% of rail, as I'm quite conservative. This circuit author has gone for 5% of rail, which is still OK. This is 0.5v, so now we can compute the value of the emitter resistor as 1k.

Now we have the emitter voltage, we need to add Vbe to find the base voltage. I tend to call Vbe 0.7v, ending up at 1.2v. But the beauty of this type of biassing is, it can be a bit off, and still work well. Once you've built it, and see what Vbe you get with your transistor at your current, then you can adjust slightly.

Finally, choose R1 and R2 as a potential divider to give your base voltage. The base voltage gives you their ratio, but you have a further choice, over a relatively wide range, of what the actual values are. Too small, and they load the input excessively. Too big, and variations in the base current will alter the base voltage. You can design for a nominal base current, but a good design will tolerate variations in beta over at least a 2:1 range. As the emitter current is fixed by Re, beta variations will alter the base current the transistor draws.

With an output impedance of 10k, the choice of 12k and 100k for R1/R2 seems a bit low, but then I guess this is just an exercise question. A higher base voltage would allow a larger R2, and therefore higher input impedance.