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1. Homework Statement
Question attatched in image file.
2. Homework Equations
3. The Attempt at a Solution
A1.b) Input will be a sine wave 6Vpp . When the input goes positive, D1 conducts and short circuits the ‘first’ resistor. D2 doesn’t conduct (it acts as an open circuit) and so Vout = voltage across R in parallel = 3V.
When the input goes negative D1 doesn’t conduct and D2 does. Therefore, the voltage goes through the ‘first’ resistor but the second resistor in parallel is short circuited. Therefore Vout = 0v since the diode is ideal.
A1.C) The output from the opamp will be = to vin . 1 (negative feedback) .When vin goes positive, D conducts and Vout = 0.7V
When Vin goes negative D acts as an open circuit, so the Vout = 1.5V

Obviously the answers above somewhat incomplete since a graph of v_in/v_out is also needed, but i just wanted someone to check if i have got the right idea.
regards,
Mo.
Question attatched in image file.
2. Homework Equations
3. The Attempt at a Solution
A1.b) Input will be a sine wave 6Vpp . When the input goes positive, D1 conducts and short circuits the ‘first’ resistor. D2 doesn’t conduct (it acts as an open circuit) and so Vout = voltage across R in parallel = 3V.
When the input goes negative D1 doesn’t conduct and D2 does. Therefore, the voltage goes through the ‘first’ resistor but the second resistor in parallel is short circuited. Therefore Vout = 0v since the diode is ideal.
A1.C) The output from the opamp will be = to vin . 1 (negative feedback) .When vin goes positive, D conducts and Vout = 0.7V
When Vin goes negative D acts as an open circuit, so the Vout = 1.5V

Obviously the answers above somewhat incomplete since a graph of v_in/v_out is also needed, but i just wanted someone to check if i have got the right idea.
regards,
Mo.
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