After going through the concept of sampling gates, let us now try to understand the types of sampling gates. Unidirectional sampling gates can pass either positive or negative going pulses through them. They are constructed using diodes.

A unidirectional sampling gate circuit consists of a capacitor C, a diode D and two resistors R₁ and R_L. The signal input is given to the capacitor and the control input is given to the resistor R₁. The output is taken across the load resistor R_L. The circuit is as shown below.

According to the functioning of a diode, it conducts only when the anode of the diode is more positive than the cathode of the diode. If the diode has positive signal at its input, it conducts. The time period in which the gate signal is ON, is the transmission period. Hence it is during that period in which the input signal is transmitted. Otherwise the transmission is not possible.

The following figure shows the time periods of the input signal and the gate signal.

The input signal is transmitted only for the time period during which gate is ON as shown in the figure.

From the circuit we have,

The anode of the diode is applied with the two signals (V_S and V_C). If the voltage at the anode is indicated as V_P and the voltage at the cathode is indicated as V_N then the output voltage is obtained as

$$V_o = V_P = (V_S + V_C) > V_N$$

So the diode is in forward biased condition.

$$V_O = V_S + V_1 > V_N$$

Then

$$V_O = V_S$$

When V₁ = 0,

Then

$$V_O = V_S + V_1 : Which : means : V_O = V_S$$

Ideal value of V₁ = 0.

So, if V₁ = 0, the entire input signal appears at output. If the value of V₁ is negative, then some of the input is lost and if V₁ is positive, additional signal along with input appears at the output.

This whole thing happens during transmission period.

During non-transmission period,

$$V_O = 0$$

As diode is in reverse biased condition

When voltage on anode is less than voltage on cathode,

$$V_S + V_C < 0 : Volts$$

During non-transmission period,

$$V_C = V_2$$

$$V_S + V_2 < 0$$

Magnitude of V₂ should be very high than V_s.

$$|V_2| &Gt; V_S$$

Because for the diode to be in reverse bias, the sum of the voltages V_S and V_C should be negative. V_C (which is V₂ now) should be as negative as possible so that though V_S is positive, the sum of both the voltages should yield a negative result.

Special Cases

Now, let’s see a few cases for different values of input voltages where the control voltage is of some negative value.

Case 1

Let us take an example where V_S = 10V and V_C = -10v (V₁) to -20v (V₂)

Now, when these two signals are applied, (V_S and V_C) then the voltage at anode will be

$$V_P = V_S + V_C$$

As this is about transmission period, only V₁ is considered for V_C.

$$V_O = (10V) + (-10V) = 0V$$

Hence the output will be a zero, though some amount of input voltage is being applied. The following figure explains this point.

Case 2

Let us take an example where V_S = 10V and V_C = -5v (V₁) to -20v (V₂)

Now, when these two signals are applied, (V_S and V_C) then the voltage at anode will be

$$V_P = V_S + V_C$$

As this is about transmission period, only V₁ is considered for V_C.

$$V_O = (10V) + (-5V) = 5V$$

Hence the output will be 5 V. The following figure explains this point.

Case 3

Let us take an example where V_S = 10V and V_C = 0v (V₁) to -20v (V₂)

Now, when these two signals are applied, (V_S and V_C) then the voltage at anode will be

$$V_P = V_S + V_C$$

As this is about transmission period, only V₁ is considered for V_C.

$$V_O = (10V) + (0V) = 10V$$

Hence the output will be a 10 V. The following figure explains this point.

Case 4

Let us take an example where V_S = 10V and V_C = 5v (V₁) to -20v (V₂)

Now, when these two signals are applied, (V_S and V_C) then the voltage at anode will be

$$V_P = V_S + V_C$$

As this is about transmission period, only V₁ is considered for V_C.

$$V_O = (10V) + (5V) = 15V$$

Hence the output will be a 15 V.

The output voltage gets affected by the control voltage applied. This voltage adds up to the input to produce the output. Hence it affects the output.

The following figure shows the superimposition of both the signals.

We can observe that during the time when only gate voltage is applied, the output will be 5v. When both the signals are applied, V_P appears as V_O. During the non-transmission period, output is 0v.

As it is observed from the above figure, the difference in the output signals during transmission period and non-transmission period, though (with V_S = 0) input signal is not applied, is called as Pedestal. This pedestal can be positive or negative. In this example, we get a positive pedestal in the output.

Effect of RC on Control voltage

If the input signal is applied before the control voltage reaches the steady state, there occurs some distortion in the output.

We get the correct output only when the input signal is given when control signal is 0v. This 0v is the stable value. If the input signal is given before that, distortion occurs.

The slow rise in the control voltage at A is due to the RC circuit present. The time constant which is the result of RC affects the shape of this waveform.

Pros and Cons of Unidirectional Sampling Gates

Let us have a look at the advantages and disadvantages of unidirectional sampling gate.

Advantages

• The circuitry is simple.

• Time delay between input and output is too low.

• It can be extended to more number of inputs.

• No current is drawn during non-transmission period. Hence under quiescent condition, no power dissipation is present.

Disadvantages

• There’s interaction between control and input signals (V_C and V_S)

• As the number of inputs increase, the loading on control input increases.

• Output is sensitive to control input voltage V₁ (upper level of V_C)

• Only one input should be applied at one instant of time.

• Because of slow rise time of the control signal, the output may get distorted, if the input signal is applied before reaching the steady state.