##### 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 *R*_{1} and *R*_{2}.

**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 *C*_{2} and input resistance *R _{i}*, as shown in Fig. 3.17(a).

Consider the parallel combination of *R*_{2} and *R _{i}*. If the amplifier input is not to load the attenuator output, then

*R*should always be significantly greater than

_{i}*R*

_{2}. The attenuator circuit is now shown in Fig. 3.17(b).

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

and

*R _{th}* =

*R*

_{1}||

*R*

_{2}

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

When the input *αv _{i}* is applied to this low-pass

*RC*circuit, the output will not reach the steady-state value instantaneously. If, for the above circuit,

*R*

_{1}=

*R*

_{2}= 1 MΩ and

*C*

_{2}= 20 nF, the rise time is:

*t _{r}* = 2.2

*R*

_{th}C_{2}= 2.2 × 0.5 × 10

^{6}× 20 × 10

^{−9}

*t _{r}* = 22 ms

This means that after a time interval of approximately 22 ms after the application of the input *αv _{i}* 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 *C*_{1} is connected across *R*_{1}. 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), *R*_{1}, *R*_{2}, *C*_{1}, *C*_{2} form the four arms of the bridge. The bridge is said to be balanced when *R*_{1}*C*_{1} = *R*_{2}*C*_{2}, 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 *v _{i}* =

*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

*C*

_{1}and

*C*

_{2}.

Thus, the initial output voltage is determined by *C*_{1} and *C*_{2}. 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, *v _{o}*(0

^{+}) =

*v*(∞)

_{o}From this using Eqs. 3.43 and 3.44 we get:

and the output is *αv _{i}*.

**FIGURE 3.19(a)** A perfectly compensated attenuator (*C*_{1} = *C*_{2})

**FIGURE 3.19(b)** An over-compensated attenuator (*C*_{1} *> C*_{2})

Let us consider the following circuit conditions:

- When
*C*_{1}=*C*, the attenuator is a perfectly compensated attenuator._{p} - When
*C*_{1}*> C*, it is an over-compensated attenuator._{p} - When
*C*_{1}*< C*, it is an under-compensated attenuator._{p}

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 *C*_{1} and *C*_{2} 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 *R _{s}* [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 (*C*_{1} *< C*_{2})

**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 *R _{s}* (

*R*

_{1}+

*R*

_{2}), hence,

*R*|| (

_{s}*R*

_{1}+

*R*

_{2}) ≈

*R*. Thus the circuit in Fig. 3.19(f) reduces to that shown in Fig. 3.19(g).

_{s}This is a low-pass circuit with time constant *τ _{s}* =

*R*, where

_{s}C_{s}*C*is the series combination of

_{s}*C*

_{1}and

*C*

_{2};

*C*=

_{s}*C*

_{1}

*C*

_{2}/(

*C*

_{1}+

*C*

_{2}). The output of the attenuator is an exponential with time constant

*τ*; and if

_{s}*τ*is small, the output almost follows the input. Alternately, consider the situation when a step voltage

_{s}*V*from a source having

*R*as its internal resistance, is connected to a circuit which has

_{s}*C*

_{2}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*(= *R _{s}C*

_{2}), its output will be an exponential with rise time

*t*, where

_{r}Now consider the compensated attenuator, shown in Fig. 3.19(g), where the internal resistance of the source *R _{s}* is taken into account. The time constant of this circuit is

*τ*(=

_{s}*R*C

_{s}_{s}) and the rise time is:

= 2.2 *R _{s}C_{s}* where

*C*=

_{s}*C*

_{1}

*C*

_{2}/(

*C*

_{1}+

*C*

_{2})

**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):

= [*C*_{1}/(*C*_{1} + *C*_{2})] = *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) *C*_{1} = 75 pF, (b) *C*_{1} = 100 pF, (c) *C*_{1} = 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, *R*_{1}*C*_{1} = *R*_{2}*C*_{2}. Here *R*_{1} = *R*_{2}.

- When
*C*_{1}= 75 pF, then the attenuator is perfectly compensated. The rise time of the output waveform is zero.Attenuation,

*v*(0+) =_{o}*v*(∞) =_{o}*αv*= 0.5 × 50 = 25 V_{i} - When
*C*_{1}= 100 pF, then the attenuator is over-compensated, hence*v*(0_{o}^{+}) >*v*(∞)._{o}The output at

*t*= 0+,The output at

*t*= ∞,From Fig. 3.21(b):

and

*C*=*C*_{1}+*C*_{2}**FIGURE 3.21(b)**The equivalent circuit to get the time constant for the decay of the overshootTime constant

*τ*_{1}with which the overshoot at*t*= 0^{+}decays to the steady-state value is:Fall time

*t*= 2.2_{f}*τ*_{1}= 2.2 × 87.5 × 10^{−6}= 192.5*μ*s - When
*C*_{1}= 50 pF, then the attenuator is under-compensated.**FIGURE 3.21(c)**The input and output responsesThe output at

*t*= 0+:The output at

*t*= ∞:The time constant,

*τ*_{2}, with which the output rises to the steady-state value is:Rise time,

*t*= 2.2_{r}*τ*_{2}*t*= 2.2 × 62.5 × 10_{r}^{−6}= 137.5*μ*sThe 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 *R*_{1} = *R*_{2}. For perfect compensation, *C*_{1} = *C*_{2}, but in the figure *C*_{1} *> C*_{2}. So the attenuator is over-compensated. Also, *R _{s}* 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, *t _{r}* 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 = *v _{o}*(0

^{+}) −

*v*(∞) = 1.194 − 0.995 = 0.199 V

_{o}The response is shown in Fig. 3.22(b).