My book (teach yourself electricity and electronics) talks about the

static forward current transfer ratio ... ratio of collector current v.s. emitter current when we place the base at electrical ground

If the base is at ground, surely the PD between base and emitter can only be zero or the emitter is at a higher potential? (Emitter can't have a potential lower than ground right?). In which case there is a zero or reverse bias at the EB junction so surely no current should flow? I must be misunderstanding something, can anyone show me the light?

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    \$\begingroup\$ They might be referring to a common base configuration? \$\endgroup\$ – Marko Jul 14 '13 at 10:49
  • \$\begingroup\$ Thanks for the comment +1 :) Not sure I fully get it... Would the base be at ground in common base config? Isn't it normally biased at some DC offset? \$\endgroup\$ – Jimbo Jul 14 '13 at 10:52
  • \$\begingroup\$ Having read a little further I think you're right. Even though the base is at a dc bias they are calling this "grounded (for the signal)" I guess "grounded for the signal" is different from ground \$\endgroup\$ – Jimbo Jul 14 '13 at 11:13
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    \$\begingroup\$ The emitter can have a negative potential with respect to ground. To quote Wikipedia, "Voltage is a differential quantity. To measure the voltage of a single point, a reference point must be selected to measure against. This common reference point is called ground and considered to have zero voltage." \$\endgroup\$ – Wandering Logic Jul 14 '13 at 11:35

Ground is just a point of your chosing to call the 0 reference for other points. Voltages can be negative, especially since the choice of what to call 0 is arbitrary.

Read the book more carfully, and you'll probably see this section was talking about the common base configuration. In that case the base is usually held at a fixed voltage, the input is the current drawn from the emitter, and the output is the collector. Common reasons for using this configuration are:

  1. You have a current sink, but it needs to operate to a higher voltage than it can natively.

  2. You want to decouple possibly large voltage swings from a current sink. If the current sink were a bare transistor, the capacitive coupling from the widely swinging output (the collector) into the base causes negative feedback at high frequencies, which reduces bandwidth and slows transition times. With the common base transistor buffering the current source, the current source sees only the relatively constant emitter voltage of the buffer transistor. Since the base of that transistor is at a fixed voltage, it can absorb capacitively coupled current from the output without ill effects.

  3. You are using the common base transistor to make a current source. With the base tied to the 5 V supply, for example, a reasonably well controlled current sink can be made by putting a resitor in series with the emitter and either switching it to ground or not.

The current transfer ratio of a common base stage is 1 at first approximation. The emitter current (the input) is the collector current (the output) plus the base current. But the ratio of collector current to base current is the gain of the transistor. For high enough gains, the additional base current in the emitter current is small enough to ignore.

For example, let's say the gain of the transistor is 50. In this example, the output of the common base (the collector current) is 50 mA. That means the base current is 1 mA and the emitter current (the input) is 51 mA. The current transfer ratio in this example is 50 mA / 51 mA = .98. Like I said, basically 1.

In general:

current transfer ratio = gain / (gain + 1)


(Emitter can't have a potential lower than ground right?)

Hi Jimbo, consider the following ideal test circuit.

enter image description here

The base is at zero volts. The emitter voltage will be whatever negative voltage it must such that the emitter current is 1mA.

The collector current will be \$I_C = \alpha I_E \$. There will be a slight dependence on the collector-base voltage.


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