FIGURE 5.1 below shows block diagrams with different input and output states for three types of multivibrators - NSC Electrical Technology Electronics - Question 5 - 2023 - Paper 1

Block Y is identified as a monostable multivibrator. This type outputs a single pulse when triggered and returns to its stable state afterward.

Block Z is an astable multivibrator. It continuously oscillates between its high and low states without the need for an external trigger.

The variable resistor R2 adjusts the frequency of the output by changing the timing characteristics of capacitor C1. It influences the charge and discharge cycles.

Connecting the LED directly to pin 3 without resistor Rs would result in excessive current flowing through the LED, likely leading to its damage and failure due to overcurrent.

The circuit operates as a monostable multivibrator. When triggered by a switch, it causes capacitor C1 to charge through R2. The output at pin 3 (non-inverting input) transitions based on the voltage across the capacitor.

The discharge path of capacitor C1 occurs through resistor R2 back to ground once the triggering source is removed.

When the switch is open, the voltage at pin 3 is equal to the supply voltage, resulting in approximately 6 V.

The output is high (6 V) because the voltage on the non-inverting input (pin 3) is higher than the voltage at the inverting input.

When the switch is pressed, pin 3 drops to 0 V as the capacitor discharges rapidly to ground through the switch.

Upon pressing the switch, the capacitor C1 discharges rapidly, causing a momentary high output at the op-amp's output. The transition is rapid as voltage levels change between input terminals.

The circuit will trigger at a voltage level where the non-inverting input surpasses the inverting input, typically around 3 V, given standard Schmitt trigger thresholds.

A Schmitt trigger is commonly used in signal conditioning to clean up noisy signals and in generating square waves from sine wave inputs.

Using the relationship from the table, we determine that Rf = 10 kΩ at X.

The gain at point Y can be calculated as follows: $A_v = -\frac{R_f}{R_{in}} = -\frac{100 kΩ}{10 kΩ} = -10$. Hence, the gain is -10.

Using the output voltage formula for an inverting amplifier: $V_{out} = -\left(\frac{R_f}{R_{in}}\right) \sum V_{in} = -10(V_1 + V_2 + V_3)$, we find that Vout at Z is -4 V.

An increase in feedback resistance Rf increases the gain of the inverting summing amplifier, establishing a direct proportionality.