# 3.3 Attenuators – Pulse and Digital Circuits

##### 3.3 ATTENUATORS

An attenuator is a circuit that reduces the amplitude of the signal by a finite amount. A simple resistance attenuator is represented in Fig. 3.16. The output of the attenuator shown in Fig. 3.16 is given by the relation: From this equation, it is evident that the output is smaller than the input, which is the main purpose of an attenuator—to reduce the amplitude of the signal. Attenuators are used when the signal amplitude is very large. Let us measure a voltage, say, 5000 V, using a CRO; such a large voltage may not be handled by the amplifier in a CRO. Therefore, to be able to measure such a voltage we first attenuate the voltage by a known amount, say by a factor of 10(α = 0.1), so that the voltage that is actually connected to the CRO is only 500 V. The output of the attenuator is thus reduced depending on the choice of R1 and R2. FIGURE 3.16 A resistance attenuator FIGURE 3.17(a) The attenuator output connected to amplifier input

#### 3.3.1 Uncompensated Attenuators

If the output of an attenuator is connected as input to an amplifier with a stray capacitance C2 and input resistance Ri, as shown in Fig. 3.17(a).

Consider the parallel combination of R2 and Ri. If the amplifier input is not to load the attenuator output, then Ri should always be significantly greater than R2. The attenuator circuit is now shown in Fig. 3.17(b).

Reducing the two-loop network into a single-loop network by Thévenizing: and

Rth = R1||R2

Hence, the circuit in Fig. 3.17(b) reduces to that shown in Fig. 3.17(c).

When the input αvi is applied to this low-pass RC circuit, the output will not reach the steady-state value instantaneously. If, for the above circuit, R1 = R2 = 1 MΩ and C2 = 20 nF, the rise time is:

tr = 2.2 RthC2 = 2.2 × 0.5 × 106 × 20 × 10−9

tr = 22 ms

This means that after a time interval of approximately 22 ms after the application of the input αvi to the circuit, the output reaches the steady-state value. This is an abnormally long delay. An attenuator of this type is called an uncompensated attenuator, i.e., its output is dependent on frequency. FIGURE 3.17(b) The attenuator, considering the stray capacitance at the amplifier input FIGURE 3.17(c) An uncompensated attenuator

#### 3.3.2 Compensated Attenuators

To make the response of the attenuator independent of frequency, the capacitor C1 is connected across R1. This attenuator now is called a compensated attenuator shown in Fig. 3.18(a). This circuit in Fig. 3.18(a) is redrawn as shown in Fig. 3.18(b). FIGURE 3.18(a) A compensated attenuator FIGURE 3.18(b) Redrawn circuit of Fig 3.18(a) FIGURE 3.18(c) The compensated attenuator open-circuiting the xy branch

In Figs. 3.18(a) and (b), R1, R2, C1, C2 form the four arms of the bridge. The bridge is said to be balanced when R1C1 = R2C2, in which case no current flows in the branch xy. Hence, for the purpose of computing the output, the branch xy is omitted. The resultant circuit is shown in Fig. 3.18(c).

When a step voltage with vi = V is applied as an input, the output is calculated as follows: At t = 0+, the capacitors do not allow any sudden changes in the voltage; as the input changes, the output should also change abruptly, depending on the values of C1 and C2. Thus, the initial output voltage is determined by C1 and C2. As t → ∞, the capacitors are fully charged and they behave as open circuits for dc. Hence, the resultant output is: Perfect compensation is obtained if, vo(0+) = vo(∞)

From this using Eqs. 3.43 and 3.44 we get: and the output is αvi.  FIGURE 3.19(a) A perfectly compensated attenuator (C1 = C2) FIGURE 3.19(b) An over-compensated attenuator (C1 > C2)

Let us consider the following circuit conditions:

1. When C1 = Cp, the attenuator is a perfectly compensated attenuator.
2. When C1 > Cp, it is an over-compensated attenuator.
3. When C1 < Cp, it is an under-compensated attenuator.

The response of the attenuator to a step input under these three conditions is shown in Figs. 3.19(a), (b) and (c), respectively.

In the attenuator circuit, as at t = 0+, the capacitors C1 and C2 behave as short circuits, the current must be infinity. But impulse response is impossible as the generator, in practice, has a finite source resistance, not ideally zero. Now consider the compensated attenuator with source resistance Rs [see Fig. 3.19(d)].

If the xy loop is open for a balanced bridge, Thévenizing the circuit, the Thévenin voltage source and its internal resistance and R are calculated using Fig. 3.19(e). FIGURE 3.19(c) An under-compensated attenuator (C1 < C2) FIGURE 3.19(d) The attenuator taking the source resistance into account FIGURE 3.19(e) The circuit used to calculate the Thévenin voltage source and its internal resistance FIGURE 3.19(f) Redrawn circuit of Fig. 3.19(e)

The value of Thévenin voltage source is: and its internal resistance is: The above circuit now reduces to that shown in Fig. 3.19(f). Usually Rs (R1 + R2), hence, Rs || (R1 + R2) ≈ Rs. Thus the circuit in Fig. 3.19(f) reduces to that shown in Fig. 3.19(g).

This is a low-pass circuit with time constant τs = RsCs, where Cs is the series combination of C1 and C2; Cs = C1C2/(C1 + C2). The output of the attenuator is an exponential with time constant τs; and if τs is small, the output almost follows the input. Alternately, consider the situation when a step voltage V from a source having Rs as its internal resistance, is connected to a circuit which has C2 between its output terminals, [see Fig. 3.19(h)].

This being a low-pass circuit (can also be termed as an uncompensated attenuator), with time constant τ(= RsC2), its output will be an exponential with rise time tr, where Now consider the compensated attenuator, shown in Fig. 3.19(g), where the internal resistance of the source Rs is taken into account. The time constant of this circuit is τs(= RsCs) and the rise time is: = 2.2 RsCs where Cs = C1C2/(C1 + C2) FIGURE 3.19(g) The final reduced circuit of a compensated attenuator FIGURE 3.19(h) A low-pass circuit (uncompensated attenuator) From Eqs. (3.47) and (3.48): = [C1/(C1 + C2)] = α

where α is the attenuation constant, which tells us by what amount the signal is reduced at the output. If α = 0.5: From Eq. (3.50), it seen that the output signal from a compensated attenuator has a negligible rise time when compared to the output signal from an un-compensated attenuator. It means that the step voltage, V is more faithfully reproduced at the output of a compensated attenuator, which is its main advantage.

A perfectly compensated attenuator is sometimes used to reduce the signal amplitude when the signal is connected to a CRO to display a waveform. A typical CRO probe may be represented as in Fig. 3.20. Example 3.7 helps to further elucidate and elaborate the functioning of the attenuator circuit. FIGURE 3.20 A typical CRO probe

##### EXAMPLE

Example 3.7: Calculate the output voltages and draw the waveforms when (a) C1 = 75 pF, (b) C1 = 100 pF, (c) C1 = 50 pF for the circuit shown in Fig. 3.21(a). The input step voltage is 50 V. FIGURE 3.21(a) The given attenuator circuit

Solution: For perfect compensation, R1C1 = R2C2. Here R1 = R2.

1. When C1 = 75 pF, then the attenuator is perfectly compensated. The rise time of the output waveform is zero.

Attenuation, vo(0+) = vo(∞) = αvi = 0.5 × 50 = 25 V

2. When C1 = 100 pF, then the attenuator is over-compensated, hence vo(0+) > vo(∞).

The output at t = 0+, The output at t = ∞, From Fig. 3.21(b): and C = C1 + C2 FIGURE 3.21(b) The equivalent circuit to get the time constant for the decay of the overshoot

Time constant τ1 with which the overshoot at t = 0+ decays to the steady-state value is: Fall time tf = 2.2 τ1 = 2.2 × 87.5 × 10−6 = 192.5 μs

3. When C1 = 50 pF, then the attenuator is under-compensated. FIGURE 3.21(c) The input and output responses

The output at t = 0+: The output at t = ∞: The time constant, τ2, with which the output rises to the steady-state value is: Rise time, tr = 2.2 τ2

tr = 2.2 × 62.5 × 10−6 = 137.5 μs

The output responses are plotted in Fig. 3.21(c).

##### EXAMPLE

Example 3.8: For the circuit shown in Fig. 3.22(a), calculate and draw the output response. Determine the rise time, the magnitude of the overshoot and the time constant when the output decays to the final value.

Solution: Here R1 = R2. For perfect compensation, C1 = C2, but in the figure C1 > C2. So the attenuator is over-compensated. Also, Rs is not equal to zero, the step input of 2 V will not appear at the input of the attenuator with zero rise time. The input to the attenuator is: The rise time, tr of this input is:  FIGURE 3.22(a) The given attenuator circuit FIGURE 3.22(b) The response of the attenuator circuit

Initial response: Final response: Fall time: Overshoot = vo(0+) − vo(∞) = 1.194 − 0.995 = 0.199 V

The response is shown in Fig. 3.22(b).