if you mean the 2 microamperes it's because you are interested in signal to be amplified! You want the microphone signal to be 50mv not the microfone and operation point voltage. The operation point helps you to get in the region you want for the transistor to do what you want. But the important part you want to make 50mV is the 5mV from the microphone
i have a doubt that when the device is at operating point i.e, greater than threshold voltage then when we are not applying any signal then output should be zero and when we apply some signal then it should be amplified but when there is no signal we are able to get some output at load then how to identify at output that input is not applied?
In gm vs Vgs graph what does professor mean when he says "remember we are still in saturation region" does that mean Vds is greater than Vgs-Vth. But then how will amplifier do more amplification when you are bounding the current bcs of saturation
Thank you sir for ur lectures but i have a small doubt in the attempts you made, you didnot give any vds how can the MOSFET act like current source then
I have a question about the circuit you draw in 23:33. where does the power P = Id^2 * R actually come from? Does it come from the Battery? If so, there is no current flowing into the gate, right?
I am confused as to which expression of gm to use when the values of Id, Vgs, W/L, mu_n*C_ox are all available. Because the three expressions give three different answers. For example, if W/L = 10/0.25, Vgs = 1 V, Id = 1 mA, mu_n*C_ox = 100*10^-6 If I try to find gm using the three different expressions I am getting different values with each expression. One expression gives 2.4mS, another gives 3.33mS , the other one gives 2.828 mS. Can somebody please point out what I am missing here? Thank you in advance
vds just creates potential difference in n-channel and if u increase vds that potential difference will increase current flow will be more in the n channel .but for all this to happen n channel should be there which will be formed only when vgs>=vth .hence if no gate voltage then no current irrespective of the vds.
At 11:00 if pinch-off voltage moves away from vds, they won't be any channel after pinch-off voltage, so how the current flows? anyone please explain this?
there's no charge carrier inversion, but there's still an electric field across the channel length that sweeps the electrons across the channel that were sourced by the Vds. It's not like we created a vacuum in that area.
let's say we biased the circuit and a sinusoidal signal comes at Vgs. Are there any chances that the transistor go out of saturation region in the negative half of the signal, if it becomes Vds
@@jyothiyanapu3524 change in current taken because amplification is of input ac signal .if you will not take the current difference then output voltage will contain dc biasing voltage of the input which isn't needed ,we just want output voltage wrt to input ac signal (5mv).you can think of small signal model then you will understand why taken the difference.
Greetings !!! , background music , concepts ....loved it !!!!
Thanks for all these vidoes. Yo're the lord of the electronics sir.Thanks to you ı can understand all concepts of BJT and FET.
That cliffhanger :)
this would outsell game of thrones anyday!
Thanks professor Razavi!
Loved it !! Graphs tells everything....
very straight and understandable lecture. Thanks sir
Why we should take change in current regarding calculation of resistance in third attempt of amplifier design at 34:38 ??
if you mean the 2 microamperes it's because you are interested in signal to be amplified! You want the microphone signal to be 50mv not the microfone and operation point voltage. The operation point helps you to get in the region you want for the transistor to do what you want. But the important part you want to make 50mV is the 5mV from the microphone
i have a doubt that when the device is at operating point i.e, greater than threshold voltage then when we are not applying any signal then output should be zero and when we apply some signal then it should be amplified but when there is no signal we are able to get some output at load then how to identify at output that input is not applied?
through capacitve coupling, it is possble to isolate the AC signal from the DC offset
Thank you so much sir. You have made MOSFET concepts very easy to understand. Thanks a lot for creating this lecture series.
Prof. Behzad Razavi:Electronics :: Prof. Walter Lewin:Physics
If only we had professors like you and Walter Lewin in India!
Do you know other good proffessors that offer free lecture related to engineering/ electrical engineering?
@@gamar1226MIT opencourseware anant Agarwal, Patrick Mercier vlsi , chembiyan T rc circuits.
@@manideepp2229 Wow you have saved me buddy. As an electrical student i am disappointed that i am not aware of the existence of such a helpful website
Supub 😊😊 , guruji 🙏🙏🙏🙏🙏🙏
In gm vs Vgs graph what does professor mean when he says "remember we are still in saturation region" does that mean Vds is greater than Vgs-Vth. But then how will amplifier do more amplification when you are bounding the current bcs of saturation
Mr. Mike :P
Absolutely amazing!
@24:49 second attempt seems to be wrong as there is no V_DS.
great lecture sir
Thank you sir for ur lectures but i have a small doubt in the attempts you made, you didnot give any vds how can the MOSFET act like current source then
he said that in the last minute :)
I have a question about the circuit you draw in 23:33. where does the power P = Id^2 * R actually come from? Does it come from the Battery? If so, there is no current flowing into the gate, right?
okay you answered the question :D There is no current Id, because there is no Drain-Source voltage
why does Vout have a 180 degree phase difference from Vin?
Increasing voltage causes increase in conduction or decrease in resistance and it causes voltage drop to decrease across the terminal of the device.
I am confused as to which expression of gm to use when the values of Id, Vgs, W/L, mu_n*C_ox are all available.
Because the three expressions give three different answers.
For example, if W/L = 10/0.25, Vgs = 1 V, Id = 1 mA, mu_n*C_ox = 100*10^-6
If I try to find gm using the three different expressions I am getting different values with each expression. One expression gives 2.4mS, another gives 3.33mS , the other one gives 2.828 mS.
Can somebody please point out what I am missing here?
Thank you in advance
What is the vth ? I think these values will not evaluate the Id mentioned
Vth is 0.4V
Check if the MOS is in saturation or not.
You need to calculate Id, not declare it, since it's dependent on Vgs. I get 720 uA using your numbers. Then g=2.4 mS in all cases.
What happens if we do not apply the gate voltage and go on increasing the drain voltage? what will be the effect?
Simple the n-channel will not be formed unless Vgs > = vth. So no current will flow.
vds just creates potential difference in n-channel and if u increase vds that potential difference will increase current flow will be more in the n channel .but for all this to happen n channel should be there which will be formed only when vgs>=vth .hence if no gate voltage then no current irrespective of the vds.
This is the best
perfect Lecture ❤❤
Awesome sir !!!!
Awesome!
16:04 is it large-signal model?
yes
Great lecture !
Indeed!
At 11:00 if pinch-off voltage moves away from vds, they won't be any channel after pinch-off voltage, so how the current flows? anyone please explain this?
there's no charge carrier inversion, but there's still an electric field across the channel length that sweeps the electrons across the channel that were sourced by the Vds. It's not like we created a vacuum in that area.
lets build an amplifier :)
let's say we biased the circuit and a sinusoidal signal comes at Vgs. Are there any chances that the transistor go out of saturation region in the negative half of the signal, if it becomes Vds
yes it can go out of saturation thats why its impotant to bias the circuit such that this doesn't happen.
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Pppfade+8
Why we should take change in current regarding calculation of resistance in third attempt of amplifier design at 34:38 ??
coz Vout=-IdRl(when KVL is applied we get Vout +IdRl=0)
even i have the same doubt? why they took change in current
@@jyothiyanapu3524 change in current taken because amplification is of input ac signal .if you will not take the current difference then output voltage will contain dc biasing voltage of the input which isn't needed ,we just want output voltage wrt to input ac signal (5mv).you can think of small signal model then you will understand why taken the difference.