Current Mirror – Mathematical Derivation
Consider a basic BJT current mirror consisting of two identical transistors Q1 and Q2.
The base and collector of transistor Q1 are connected together, forcing Q1 to operate in the active region.
Step 1 : Base-Emitter Voltage Equality
Since both transistors are identical and their bases are connected together:
VBE1 = VBE2
For identical transistors operating at the same temperature, equal base-emitter voltage results in equal collector currents.
Step 2 : Collector Current Equation
For a BJT operating in active region:
IC = IS e^(VBE/VT)
Where:
- IS = saturation current
- VT ≈ 25 mV (thermal voltage at room temperature)
Step 3 : Equal Collector Currents
Since:
VBE1 = VBE2
Therefore:
IC1 = IC2
Thus the collector current of Q2 becomes equal to the reference current.
Step 4 : Reference Current
The reference current is set by resistor R.
Iref = (VCC − VBE) / R
This current flows through transistor Q1.
Step 5 : Output Current
Since both transistors have equal collector currents:
Iout ≈ Iref
Effect of Base Current
Considering base current, the output current becomes:
Iout = Iref × β / (β + 2)
Where:
- β = transistor current gain
For large β:
Iout ≈ Iref
Important GATE Observation
- Ideal current mirror → Iout = Iref
- Practical current mirror slightly lower due to base currents
- Matching of transistors is essential
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