In a transistor the emitter current is 6.00 mA and the collector current is 5.94 mA. Find the base current and the current gain β.
- IB = IE − IC = 6.00 − 5.94 = 0.06 mA = 60 μA.
- β = IC ÷ IB = 5.94 ÷ 0.06 = 99.
Answer: Base current = 60 μA, β = 99.
Electronics
MDCAT Physics · Physics · Physics of solids and electronics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
A p-n junction has a depletion region with no free charge carriers and a potential barrier across it. In forward bias the p-side is joined to the positive terminal: the depletion region narrows and a large current flows. In reverse bias the region widens and only a tiny current of minority carriers flows. So a diode conducts one way only, which lets it change AC into DC. This is rectification.
A transistor has an emitter, a thin base and a collector. A small base current controls a much larger collector current, so it can amplify a signal. As a switch it is either cut off (OFF, almost no collector current) or saturated (ON, largest collector current).
Logic gates work with two states only, 0 and 1.
Rules to remember
- Potential barrier: about 0.7 V for silicon, 0.3 V for germanium. Forward current = (supply voltage − barrier) ÷ series resistance
- Half-wave rectifier: one diode, uses half of each cycle. Full-wave bridge: four diodes, two conduct in each half cycle
- emitter current = base current + collector current (IE = IB + IC)
- current gain = collector current ÷ base current (β = IC/IB)
- Amplifier biasing: base-emitter junction forward biased, collector-base junction reverse biased
- AND: output 1 only if all inputs are 1. OR: output 1 if any input is 1. NOT: output is the opposite of the input
- NAND = AND followed by NOT; NOR = OR followed by NOT; XOR gives 1 when the two inputs are different; XNOR gives 1 when they are the same
Common mistake
Students join the wrong side for forward bias: it is p to positive, n to negative. In transistor sums, keep both currents in the same unit (μA or mA) before you add or divide.
Worked example
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