Such a clean explanation of the low-pass filter, and smoothening capacitance....Really appreciate the efforts of this professor in explaining the topics
Now i know where did my college professor learnt these concepts from .......... 😍😍😍😍😍 These lectures are treasure for the humankind. Love you prof. Thank you so much.
is there someone can explain me that in the half-wave rectifier case, i dont konw why the diode turn off when the output voltage of capacitor reach the peak，why dose it stay constant ?. just from the Mathematical equations(V_o=V_in-V_D), V_o should decrease with V_in decreasing
cause the capacitor is either absorbing the voltage or releasing it, once Vin is not powerful enough to push the voltage towards the capacitor, the capacitor is releasing voltage. the current goes backwards, thus turn off the diode.
Last equation does not make sense for me. Ripple Amplitude = (V0 - VD,on) / (R1 * C1 * fin). If we have a small device, the R1 is going to be small so the ripple is going to be big. That does not make sense for me.
R1 is an approximation for the power draw of the device, not its size (small or large device). A small R1 indicates a device with high power requirements ( P = V^2/R, small R means high power). The more power it needs, the more current it will draw hence the smaller the value for R1.
It depends on the input impedance of the load device. So the charger is designed for a specific load. For an example, if we are talking about mobile phone chargers then all the mobile phones will have more or less same input impedance.
so basically to decrese ripple we need to increase capacitance. can anyone tell me what if we put a buffer in between cap and load then i guess load won't affect the voltage across cap so i guess there won't be any discharging?
Awesome lecture razavi sir Thanks a lot A very silly mistake over the lecture after 25:00 Output voltage = Vo - Vd, on ( wrong) Correct is = Vin - Vd, on
21:42 the rate of change of capacitor voltage should be lower than the slope of input voltage i.e the capacitor will charge slowly and than the input voltage change Plzz correct me if i am wrong
Sir, why can't we plot the transfer (I/O) characteristics for the half-rectifier diode with smoothening capacitor at 33:14 ? Since, by applying KVL, V(out) = 0, for V(in) V(D, on).....So, shouldn't it be a straight line on the input axis till V(D, on) and then an inclined line with slope = 1 thereafter?
uptill V(in) < V(d, on) the diode is off so in that case you are correct that the V(out) will be just zero. But as soon as V(in) > V(d, on) the rate at which the V(in) is provided starts affecting the V(out) as in a capacitor the current voltage relationship is I = C (dV/dt) so you can notice that the output current or even the output Voltage will depend on how "fast" the V(in) is provided and the slope is not just one.
Think about the voltage difference between the anode and cathode of the diode just after the pick. The cathode (the capacitor actually keeps that fixed as it can't discharge through the diode in the reverse direction) is charged up to (V0 - Vd,on) and the anode starts to go below V0, hence the voltage across the diode right after the pick is less than Vd,on which is required to keep the diode on.
@@Rafi8712-h1j Hi, I liked your explanation ..i have one question regarding negative cycle ..if the vin is becoming negative then the terminal of the capacitor that was previously negative what would happen to it..would the one end of wire be positive because of negative vin and the other end be negative because of negative terminal of capacitor not discharging ..won't that voilate some law of circuit theory of wire not having potential differences ? Thanks.
Consider the input voltage as 10.8 volts as the peak. So the Voltage above the capacitor is 10 volts. For current to push into capacitor voltage above capacitor should be less than the input voltage. But here input voltage is going back to 0. But the voltage above capacitor remains constant as capacitor stores voltage. It does not track the input voltage. In case of resistor, voltage follows or tracks the input voltage. Hope it's clear! If not anyone can reply back!
