Section Index | Owatonna Steele County Amateur Radio |
20-Jul-2007 |

Emitter-Follower Amplifier The Emitter-Follower (EF) amplifier is also known as the common-emitter amplifier. It is commonly used where buffering or isolation is necessary, combined with power to drive a heavy load. It provides: - Modest power gain.
- High input impedance (few KΩ).
- Low output impedance.
- No voltage gain.
- No phase shift between the input and output.
**Power Gain**: Ratio of the input power to the output power.
**Cutoff**: Collector current reduced to zero.
**Q-Point**: Quiescent or resting values of collector current (Icq) and voltage (Vceq) with no input signal applied.
**Ic**: Collector current.
**Ib**: Base current.
**Ie**: Emitter current.
**Vbe**: Voltage between the base and emitter.
**Vce**: Voltage between the collector and emitter.
**Zin**: Input impedence, the equivalent AC impedence looking into the amplifier input.
**Zout**: Output impedence, the equivalent AC impedence looking into the amplifier output.
**re**: Internal emitter impedence = 25 mV / Ieq. (Note small 'r')
**Rs**: Signal source impedence.
**//**: Symbol used to indicate components in parallel.
Most amplifiers need to work with input signals having both positive and negative voltage swings. To accomplish this, the amplifier is designed in a Class A configuration, the collector current is offset from zero without an input signal present. This is easily accomplished by biasing the base current (Ib), using R1 and R2. The emitter resistor (Re) ensures the collector voltage can make large swings without trying to exceed (Vcc) or drop to ground (0VDC).
The input capacitor (Cin) and output capacitor (Cout) provide an
- Choose the basic operating parameters:
- Vcc = 12 VDC, the power supply voltage.
- Icq = 5mA, keeping power dissipation low.
- Vceq = 6 VDC, about one-half of Vcc.
- Beta = 150, based on the specifications of the transistor.
- Vbe = 0.7 VDC, typical for a silicon transistor.
- Calculate the value of Re:
- Vcc ~ Vce + Ie * Re -- therefore
- Re = ( Vcc - Vceq ) / Icq -- so
- Re = ( 12 V - 6 V ) / 5 mA = 6 V / 0.005 A = 1,200 Ω = 1.2 kΩ
- Calculate the base current:
- Assuming Ic ~ Ie, Ic = Ib * Beta -- therefore
- Ib = Icq / Beta -- so
- Ib = 5 mA / 150 = 0.005 / 150 = 0.000033 A = 33 µA
- Assuming Ic ~ Ie, Ic = Ib * Beta -- therefore
- Establish current through R1 and R2 equal to 10 times Ib as a rule of thumb:
- IR1R2 = Ib * 10 -- so
- IR1R2 = 33 µA * 10 = 330 µA
- Calculate the voltage across R2:
- Vb ~ Vbe + Ic * Re -- so
- Vb ~ 0.7 V + 5 mA * 1.2 kΩ = 0.7 V + 0.005 A * 1,200 Ω = 6.7 V
- Calculate R2 with Ohm's Law:
- R2 = VR2 / IR2 -- so
- R2 = 6.7 V / 330 µ = 20,300 Ω = 20.3 kΩ
-- selecting 22 kΩ as a standard value
- The voltage across R1 is calculated with a voltage divider:
- VR1 = Vcc - VR2 -- so
- VR1 = 12 V - 6.7 V = 5.3 V
- Again using Ohm's Law, calculate R1:
- R1 = VR1 / IR1 -- so
- R1 = 5.3 V / 330 µ = 16,060 Ω = 16.06 kΩ
-- selecting 15 kΩ as a standard value
- Zin is determined:
- Zin = R1 // R2 // Zb = 1 / { 1 / R1 + 1 / R2 + 1/ [ Re * ( Beta + 1) ] } -- so
- Zin = 1 / [ 1 / 15 kΩ + 1 / 22 kΩ + 1 / 1.2 kΩ * ( 150 + 1 ) ] = 8,510 Ω = 8.5 kΩ
- Assuming Rs = 50Ω:
- Ze = ( Rs // R1 // R2 ) / ( Beta + 1 ) + re -- therefore
- Ze = ( 50Ω // 15 kΩ // 22 kΩ ) / ( Beta + 1 ) + ( 25 mv / 5 mA ) = 49.72 / 151 + 5 = 5.31 Ω
- Zout is calculated as:
- Zout = Ze // Re -- so
- Zout = 5.31 Ω // 1.2 kΩ = 5.3 Ω
Based on |

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