Differential Amplifier

Article: Andy Collinson

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Some of TINA's simulation abilities are performed on the circuit below. Q1 and Q2 form a single ended differential amplifier. The input is Q1 base and the output taken from Q1 collector. The output is further amplified by Q3, the load resistor RL has a feedback fraction, R10. This feedback is fed back to differential input Q2 and increases bandwidth, but with controlled gain. The circuit is shown below:

Manual Analysis

Before any ac analysis can be performed a dc bias solution must be calculated. R1 and R2 bias Q1 base at half supply voltage or 6V. The r

In a differential amplifier the base voltages of Q1 and Q2 must be equal so that current is evenly distributed in the emitter resistor, R5. Therefore the base of Q1 and Q2 is 6 V. Q1 base emitter voltage is 0.7 V so the voltage on the emitter will be 6 - 0.7 or 5.3 V. The current through R5 is therefore of 5.3 / 4.7k or 1.13 mA. As this is the combined emitter current of Q1 and Q2 the emitter current is halved so Ie Q1=Q2 or 1.13 / 2 = 0.565 mA. Ignoring the base current, collector current is approximately same as emitter current, i.e Ic = 0.565mA. The intrinsic emitter resistance r

Open Loop Gain

Overall feedback is sampled by the 10k load resistor RL and R10 470 ohms. This feedback fraction (B) is 0.47k / 10k = 0.047 and is applied at Q2 base via C4 and R9. To measure open loop gain the feedback must be broken so if R9 (base end) is grounded the open loop gain can be calculated.

The gain of the differential stage Q1 and Q2 is Rc / 2 Re. As the collector load is R4 in parallel with the input impedance of Q3. Assuming a β of 250 for a BC557 the input impedance of Q3 is βr

The gain of the differential amp is :

Q3 has a gain of Rc / r

The overall open loop gain (A) is the product of both stages or:

8.5 * 141 = 1198.5 =61.5dB

Shorting R9 to ground and running an ac frequency analysis in Tina yields the following result. The midband gain is indeed 60dB and calculated results were very close to this value.

Open Loop Gain

Closed Loop Gain

As the feeback by RL and R10 is series voltage feedback then the closed loop gain is :

A/(1+AB) = 1198.5 / (1 + (1198.5 * 0.047)) = 20.91 or 26.4dB

The calculated values again relate very closely to the actual bode plot produced by Tina, as shown below:

Closed Loop Gain

Input Impedance

To measure open loop input impedance the feedback loop must be opened. If R9 (base end) is grounded the open loop input impedance can be calculated. The input signal sees R1 in parallel with R2 and R3 (as C2 is short circuit at audio frequencies) plus the input impedance of the differential stage Q1. This is β x 2 r

R1 || R2 || R3 || [250 x 2 x 46 + 15k/250] = 2.78k || 23.06k = 2.48k

As series feedback is used, the closed loop input impedance is raised by (1 + loop gain). As loop gain is AB or 53.5 the input impedance is now 2.48k * 56.3 = 139.62k. However this is still in parallel with R1,R2 and R3 so closed loop input impedance is:

6.2k || 6.2k || 27k || 139.62k = 2.73k

The actual input impedance calculated by Tina is shown below. It is 2.78k for mid band.

Input Impedance

Output Impedance

The open loop output impedance sees the impedance into Q3's collector (high) in parallel with R8, the (load + R10) and R7 15k (as C3 is a short at audio frequencies. The output impedance is therefore: 2.2k || (10k + 0.47k) || 15k = 1.62k

However as series feedback is used the output impedance is reduced by (1 + loop gain). It is:

1.62k / (1 + 53.5) = 29.7 ohms

The output impedance calculated by Tina is shown below. The actual result is 25.23 ohms.

Output Impedance

Maximim Input Level

Maximum input level is about 210mV pk-pk (150mV RMS). Above this input level the output clips and distortion increases significantly.