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Question 5
5.1 Name the type of multivibrator that: 5.1.1 Produces one pulse cycle of 'high' and 'low' when a trigger pulse is applied to its input. 5.1.2 Changes state when a ... show full transcript
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When the set switch is pressed, it pulls pin 2 'low' (0 V), causing the output to go 'high'. This activates the LED by providing sufficient voltage across it.
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Connecting pin 6 to ground ensures the 555 timer can reset properly by providing a reference point for the threshold voltage, allowing the circuit to function as intended.
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As the level of light increases, the resistance of the LDR decreases, which increases the voltage on the non-inverting input of the op-amp.
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The op-amp compares the voltages on its two input terminals. If the voltage on the non-inverting input is higher than that on the inverting input, the output drives to the positive saturation level. Conversely, if the inverting input is higher, the output goes to the negative saturation level.
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When the voltage on the non-inverting terminal is higher, LED1 will be ON (illuminated), while LED2 will be OFF.
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The circuit operates by comparing the input voltage to two trigger voltage levels, the upper and lower thresholds. When the input exceeds the upper threshold, the output goes high, and when it falls below the lower threshold, the output goes low, creating a stable ON/OFF state.
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An increase in the value of Rf will increase the gain of the amplifier, making it more sensitive to input signals. However, this may also lead to distortion if the output exceeds the supply limits.
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Increasing Rf beyond 78,26 kΩ is not recommended because it could push the amplifier into saturation and cause clipping of the output signal, leading to a loss of signal integrity.
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This limitation can be overcome by adjusting the supply voltage to the op-amp so that it can handle larger outputs without clipping, or by using feedback to stabilize the output.
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The output waveform will represent the charging behavior of the capacitor, showing a more gradual rise and fall time as compared to the original value.
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The output waveform will be even more gradual in charging and discharging, indicating a slower response time due to the larger capacitance.
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