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For the given circuits and input waveform determine the output waveform
Diode is off when Vi < 5
When Vi > 5, Vo = 5V
For the given circuits and input waveform determine the output waveform
Diode is off when Vi + 2 > 0.
For the given circuits and input waveform determine the output waveform
For V1 < 4 the diode is on otherwise its off.
For the given circuits and input waveform determine the output waveform
During +ve half cycle when Vi < 8 both the diodes are off. For V1 > 8, D1 is on. Similarly for the negative half cycle.
At regulated poer supply, Is = 30-9/15k or 1.4 mA. The load current will remain less than this current.
For the circuit shown diode cutting voltage is Vin = 0. The ripple voltage is to be no more than vrip = 4 V. The minimum load resistance, that can be connected to the output is (in kilo ohm)
The circuit inside the box in fig. P3.1.31. contains only resistor and diodes. The terminal voltage vo isconnected to some point in the circuit inside the box. The largest and smallest possible value of vo most nearly to is respectively
The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage.
The Q-point of the zener diode in the circuit shown below is
A 2-port network is driven by a source Vs = 100 V in series with 5 ohm, and terminated in a 25 ohm resistor. The impedance parameters are
The Thevenin equivalent circuit presented to the 25 ohm resistor is
100 = 25I1 + 2I2, V2 = 40I1 + 10I2
Also V2 = 160I1 + 6.8I2
using above equations we get, Vth = 160 V and Rth = 6.8 ohm.
800 = 10V1 – V2 and 3V2 = 5V1 = 0.
11 videos|57 docs|108 tests
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11 videos|57 docs|108 tests
|