The twin-T complementary amplifier circuit with filter selector

The filter circuit of this main circuit will sent to through to the signal that middle frequency. At the same time is attenuated the others frequency into input signal to R1 through the base of emitter circuit as complementary form (T1/T2). And the feedback from the emitter lead of T1 and T2 through twin-T to input of the complementary amplifier (T3/T4) at others frequency from feedback.

Then, out to the collector lead of T3 and T4. By there are the phase opposite the input signal. Thus, at a moderate frequency of the twin-T has been reduced significantly.

The twin-T complementary amplifier circuit
The twin-T complementary amplifier circuit


Therefore, The signal level at the middle frequency has a few only. Which will be sent out of the collector lead of T3 and T4 from both T1 and T2. To get it to expand and cause reduced again. This final signal can connect out of the emitter lead of T1 or T2

The Q level of the filter is approximately A/4. The “A” is the amplifier rate of the T3 / T4. Which is equal 2R1 (R2 and R4 are equal), the Q of the circuit, it is about 500.

Since we use the complementary Amplifiers, So the signal has low distortion. This feature can be used as a modifier harmonic distortion sine before the distortion measure.

The middle frequency of the filter is set to f = 1 / (2?RC). From the items are in the makes the circuit, it is the middle frequency of 1 KHz. Both P1 and P2 to adjust the filter to get the highest output frequency.

What is more? Is it hard? Look Basic circuit.

Complementary common-emitter circuit

This complementary common-emitter circuit has special feature that can be applied in 2 model.
– Low distortion buffer
– The low power output circuit

How it works
In Figure 1 quiescent current flow through T1 and T2 as voltage (U) and R1, R2 sequence. Look at the circuit will see that difference normal circuit at both base lead is connected together. But usually will connected to the diode, which current that flow through diode will cause input resistances changes, result to quiescent current also changes.

Complementary common-emitter circuit

In this circuit the quiescent current of T1 has value equal (U – 0.6) / R1 , and T2 = (U – 0.6)/R2
Assume the current gain of T1 and T2 high and same voltage across R3 will lower effect to the circuit. Normally R1 will use to equal R2 and relationship of the C2, C3 and C4 depend on the lowest frequency of the circuit.

If T1 and T2 have the current gain same together. And R1 = R2 will no DC voltage show at R3 cause not need to put C1. And if this circuit get signal from Op-amp then can remove both R3 and C1 away.

This circuit is designed as the class A buffer circuit. By Power loss in R4 is I^2*R4 ; I = (U-0.6)/ R
R4 is lower; R= R1 – R2

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This Post Has 3 Comments

  1. Thnaks.

  2. Thanks.

  3. Hi Salim Khan,
    Thanks for your feedback.

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