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.
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!
Most of us come to know the beauty of any subject, only when taught by a legend like you, sir.
Such selfless catering of knowledge! I wish your good health and long life so that you can keep spreading the light longer.
Such a clean explanation of the low-pass filter, and smoothening capacitance....Really appreciate the efforts of this professor in explaining the topics
These lectures are so addictive ❤️
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.
I was today years old when I realized I could scroll down while the video was full screen.
you are best electronics teacher i have ever seen, thank you very much for ur great effort sir....
Superb! Salute to you Sir!!!
legendary lecture
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
只有当二极管的正向偏置电压达到Vd,on 二极管才会开启，也即 Vx-Vcap >= Vd,on，当交流电压源达到第一个波峰后，Vcap = Vx-max-Vd,on,此时二极管的开启变为了 Vx >= Vx-max,显然，交流电源无法维持二极管的开启，因而，过了第一个波峰后二极管将一直处于关闭状态，电容无法放电，自然保持电压不变
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.
Thank Y MR.Razavi.
Such a wonderful explanation!!!
What a amazing prof!
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.
awesome stuff
explained greatly!
This is amazing!
Thank sir you are great professor
40:39 Vout not equal 2*V0 - VD,on (but corrected to Vin - VD,on = Vo - VD,on at 40:55)
I think you're right, since 2*V_0-V_D,on is a maximum reverse bias, not the voltage across capacitor
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?
40:33 *Voltage across capacitor is V0-VDon
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
Vin =vo
Vo = max( | Vin | )
شكرا يدوكتور
hi doctor razavi
where we can find the note that u save it in smartboard in this course???
please help us
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.
Thanks
Where did he mention about a low pass filter before?please Let me know...
nope
Lecture 9.
For point 1 R is pull down resistor
17:10 stay constant
20:05
Better tha the useless Profs at UTEP, way better
I didn't get the point when he excluded the option of output voltage going up 18:37
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!
@@mytech3833 Thanks, that helped
Birba des edw re loko poso wraia ta leei
Can we download the notes anywhere ?
28:32
check