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An ideal voltage transfer curve illustrates the relationship between the input and output voltages of a device, such as an amplifier or a comparator. In an ideal scenario, the curve shows a sharp transition from low output to high output as the input crosses a certain threshold, indicating perfect switching behavior without distortion or delay. The ideal curve typically resembles a steep "S" shape, where the output voltage rapidly changes from 0 to maximum as the input voltage exceeds the threshold. In practice, real devices exhibit non-ideal characteristics, leading to gradual transitions and potential saturation effects.

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What is the Clipping circuit transfer function?

Transfer function is the relationship between output and input of a circuit. In the clipping circuit, the general transfer function is such that the transfer curve Av is less than 1 for passive limiter and greater than 0 to limit the maximum and minimum voltage value.


Draw and explain the UJT characteristics?

The Unijunction emitter current vs voltage characteristic curve figure(a) shows that as VE increases, current IE increases up IP at the peak point. Beyond the peak point, current increases as voltage decreases in the negative resistance region. The voltage reaches a minimum at the valley point. The resistance of RB1, the saturation resistance is lowest at the valley point.Ip and Iv, are datasheet parameters; For a 2n2647, IPpand IVvare 2µA and 4mA, respectively. Vp is the voltage drop across RB1 plus a 0.7V diode drop; see Figure (b). Vv is estimated to be approximately 10% of VBB.


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basically a diode flows an exponential curve Vs current if you try to double the voltage drop by increasing the voltage it should self destruct


Are diode forward and zener voltages the same?

A: Both diodes have the same curve in the forward direction however if the zener voltage is reverse it will breakdown at a particular voltage and remain conducting at the voltage. A regular diode will not do that the to voltage will fold back after breakdown to any voltage


Explain quantitatively why the maximum power transfer curve is not symmetrical about the point where the source resistance equals load resistance?

when we look at the curve ,, we can see that before the peak point curve has greater slope as compared to the slope after the peak point .. the reason is PL is given as I^2RL ,,, current is a squared term here . before peak point current is greater so overall change in power is much greater but after peak point RL is greater and current is less now the load resistance is not a squared term... so slope will be less. therefore the curve is not symetrical

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The Unijunction emitter current vs voltage characteristic curve figure(a) shows that as VE increases, current IE increases up IP at the peak point. Beyond the peak point, current increases as voltage decreases in the negative resistance region. The voltage reaches a minimum at the valley point. The resistance of RB1, the saturation resistance is lowest at the valley point.Ip and Iv, are datasheet parameters; For a 2n2647, IPpand IVvare 2µA and 4mA, respectively. Vp is the voltage drop across RB1 plus a 0.7V diode drop; see Figure (b). Vv is estimated to be approximately 10% of VBB.


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