13 Jul 2026

Single-Stage RC-Coupled Transistor Amplifier in CE Mode

practical pg-ii transistor amplifier common-emitter

Aim

To study the voltage gain and frequency response of a single-stage RC-coupled transistor amplifier in common-emitter mode.

Apparatus

NPN transistor, DC supply, bias resistors, collector resistor, coupling capacitors, function generator, CRO, and connecting leads.

Experimental arrangement

$R_1$ and $R_2$ establish the base bias, $R_E$ stabilises the operating point, $C_E$ bypasses $R_E$ for AC, and $C_1$ and $C_2$ isolate the source and load from the DC bias.

Theory

A transistor amplifier must first be biased at a quiescent operating point in the active region. The voltage divider $R_1$-$R_2$ fixes the base voltage, while $R_E$ introduces DC negative feedback: if collector current rises, the emitter voltage rises and reduces $V_{BE}$. The collector resistor converts variations of collector current into output-voltage variations.

The coupling capacitors pass the alternating signal but prevent the generator and load from disturbing the DC bias. During the positive half-cycle of the base signal, collector current increases and the voltage drop across $R_C$ increases. The collector voltage therefore falls, giving the characteristic $180^\circ$ phase reversal of a common-emitter stage.

For a small signal, $i_c=g_mv_{be}$, where $g_m=I_C/V_T$ and $V_T\approx25$ mV at room temperature. If the emitter is effectively bypassed, the mid-band gain is approximately

\[A_v=\frac{V_o}{V_i}\approx-g_m(R_C\parallel R_L).\]

At low frequency, the reactances of $C_1$, $C_2$, and $C_E$ are appreciable and the gain falls. At high frequency, transistor junction capacitances and wiring capacitance reduce the gain. The cutoff frequencies satisfy $A_v=0.707A_{v,\text{mid}}$, and the bandwidth is $BW=f_H-f_L$.

Observations

Frequency (kHz)	Input (mV)	Output (V)	Gain $A_v$
0.05	20	0.84	42.0
0.10	20	1.12	56.0
1.0	20	1.20	60.0
10	20	1.14	57.0
100	20	0.80	40.0

Graph

The nearly flat mid-band region is bounded by the low- and high-frequency roll-off regions.

Calculation

At 1 kHz,

\[A_v=\frac{V_o}{V_i}=\frac{1.20}{20\times10^{-3}}=60.\]

The half-power gain is $0.707(60)=42.4$. The observations place the lower and upper cutoffs close to 0.05 kHz and 100 kHz respectively. Hence the approximate bandwidth is

\[BW=f_H-f_L=100-0.05=99.95\,\text{kHz}.\]

Result

The amplifier gives a mid-band voltage gain of approximately $60$. The gain decreases at low and high frequencies because of coupling/bypass capacitors and transistor stray capacitances.

Viva Questions

1. Why is the output phase reversed? An increase in collector current increases the voltage drop across the collector resistor and lowers the collector voltage.
2. What is bandwidth? The difference between upper and lower half-power frequencies.
3. Why is the emitter bypassed in a practical amplifier? To reduce AC negative feedback and increase voltage gain.

Maxima Code

Download the PG-II electronics calculation file.