The timestamps for the different topics covered in the video: (For more information check the description ) 0:23 What is a precision rectifier? How is it different from a regular rectifier circuit? 4:00 How precision rectifier circuit works (Circuit analysis of Precision rectifier) 8:51 Modified Precision Rectifier 16:02 Design of full wave rectifier using the precision half wave rectifier
to anyone who is confused at why dividing by AOL then have patience for 2 to 3 minutes, actually this act of dividing is being done to see equivalently how it affects the voltage difference between positive and negative terminal of op amp. rest is in the video itself , thankyou from Bihar, India
So glad you made this video. My textbook fucking sucks. It literally makes all of these claims without proving or showing how it arrived at these statements. I need to see why something is as stated, not just take it for granted. Your videos are the background info so many of us really need, thank you good sir.
Your analysis methods are sometimes get way too vague leaving so many doubts.Please while making assumptions provide a little more information like in this video you assumed the voltage (V_) to be 0V(while analysing the circuit for negative input ) now if one goes by the virtual ground concept, one gets confused how can it happen when V+ has to be V- ( and V+ is Vin).
First of all, Thank you for a great video. If I understood correctly, the purpose of explaining the super diode circuit is to demonstrate how the assumption that the diode in the negative feedback loop is approximately ideal can be justified, with its voltage drop being negligible. Thanks to this assumption, the problem of the simple half-wave rectifier circuit is resolved. However, in the explanation of the super diode circuit, it is not clear to me how you derived the expression for V_F' . Is the V_F' actually a representation of the V_{in} voltage, where when you showed that V_F' is approximately zero, it means that a slightly greater input voltage than 0V is required for the diode to conduct? And this is derived from the general expression for any operational amplifier, which is V_{IN} = V_{out}/{A_{OL} ? Additionally, if you made such an assumption, why did you then use the value of V_F and not V_F' in the expression for V_{out}' ? Thus, the assumption that the diode is approximately ideal is not reflected here. I appreciate if you could explain that.
12:23 diode D1 is conducting some forward current flows through it. Then the current will pass through D2 also isn't it? Then how D2 won't conduct? Which path does current from D1 will follow?
Sir, at 1st case, we assume the diode is conducting, but after the explanation, you said, only when Vin>0, then the diode is conducting. But, why? First we assume, and the we prove vin=vout. But, how can we say that??
Let me try to explain. Stop the video at 3:09 and look at the circuit. As you asked, why should not diode conduct only when Vin is negative? Well if Vin is negative, output from the op amp will be negative too. This will reverse bias the diode and current cannot sink in through output. Since Vout is connected to ground through resistor, output voltage will also be zero.
@@curiosity551 When vin>0 diode is forward biased and we have negative feeback and Vout is fed back to negative input. Op amp action makes the inputs the same so Vin=Vout . Difference between inputs is 0 volts. When Vin less than ground the op-amps output goes to negative supply voltage and diode is reversed bias and Vout = 0 because there is no negative feeback since diode is reversed biased (open circuited) . The output resistor pulls vout to ground since there is no current.
Great, but filtering the precision full wave rectifier is going to have more ripple than full wave rectification by conventional means. Frank Reiser M.S.
I don't understand how the open loop gain can be used to prove the diode's forward voltage doesn't effect the output voltage when the op amp is closed loop. Can anyone explain?
Why we have taken open loop gain when you have connected the op amp in feedback configuration. Shouldn't you have divided the drop with closed loop gain
Sir, during positive half cycle output voltage is positive. So, diode should be reverse biased, because positive terminal connected to n side of diode?
Thank you for creating the material with easy to understand explanations. I have a question with respect to the modified precision rectifier. Why is the input provided to the inverting input of the op-amp? is this change a part of the modification?? Please explain. Thanks in advance.
I have one doubt why we use opamp although we made analysis on opamp circuits carried out using resistors what happen when we made the experiments without opamp
I don't understand in the case 12:21 when V_in > 0 there is the current passing from the point of 0V through D1 and to the point of -0.7V while this circuit isn't closed ?
Although it seems, the circuit is not complete through an external resistors, but the current flowing through D1 will go through an internal circuits of the op-amp. And the path will get completed through an internal circuit components of the op-amp. I hope, it will clear your doubt.
11:58 Why is Vo equal to 0? Is it because there's a negative voltage at the anode of the second diode? I don't see how it would be 0 if Vo is connected to ground through 2 different resistors. Wouldn't the voltage drop across those resistors?
Think it as if two passive resistors are connected at one end, and the other end of both resistor is connected to the ground. In that case, the voltage at the common node (where both resistors are connected ) would be 0 V. I hope, it will clear your doubt.
Because the concept of virtual ground is applicable whenever the op-amp is used in the linear region. When the positive feedback is used then op-amp is not operating in the linear region.
The diode is connected at the output of the op-amp in the feedback configuration. So, the effect of this diode drop on the input will be equal to Vf / Av. So, from the input side perspective, it behaves like an ideal diode. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Dear Sir, I really appreciate your effort on making such good teaching videos. Regarding your explanation, I understood (hopefully) the assumptions you made, and seems normally that the voltage across the diode to be as minimum as possible in order that the circuit to behave as a voltage follower (when Vin > 0). However, can you point me a further reading resource on how this is achieved knowing that the silicon diode (in this case) has to consume at least 0.7V in order to close the circuit (to be seen as a voltage follower)? Should I think on some very fast time changing states of the internal opamp circuitry, that somehow balances this overall steady state? Or in fact, the diode really stays around 0V, as in practice there will still be a very small current passing through it and in this fashion closing the circuit? Thank you for your attention! Best regards,
Here we are assuming that the diode is in the reverse biased condition and hence acts as an open circuit. For the diode to be in a reverse-biased condition the voltage at the anode should be less than the cathode. Here, in that condition, the voltage at the cathode is zero. That means the diode will remain in the reverse bias condition when the voltage at the anode is negative. Or in other words, the output of the op-amp is negative. And it will happen when the input Vin is negative. (Because in that condition, the output of the op-amp will be negative saturation voltage as it is operating as a comparator in the open loop condition). I hope this will clear your doubt.
ALL ABOUT ELECTRONICS sir you just said Vin< 0 in the beginning then how come the output will be zero, instead it should be (-R2/R1)Vin for an open loop circuit where D2 is conducting .Kindly explain I’m confused.
I have explained that in the video during the circuit analysis. Please watch it 5:00 onwards. If you still have any doubt after that then do let me know here.
It is in the closed loop configuration. The concept which was used to establish the concept of virtual ground is used here. Just after 4:25 in next 2-3 mins,I have explained that.
@@ALLABOUTELECTRONICS sir at 11:30 you said that opamp is operated in linear region and we can apply the concept of virtual ground so my doubt is that if it was not in linear region then could we apply or not the concept of virtual short plz reply sir
for purposes of explanation at 11:37 : at the junction of R1, R2 is at "virtual ground" ; be careful with this idea ! this voltage is very, very small, it can NOT be 0v. for positive Vin]pk inverting input is +0.6v/A]open_loop or about 0.6uv for A]open_loop = x1,000,000 0.6v is presumed to be the fwd voltage of D1
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The timestamps for the different topics covered in the video: (For more information check the description )
0:23 What is a precision rectifier? How is it different from a regular rectifier circuit?
4:00 How precision rectifier circuit works (Circuit analysis of Precision rectifier)
8:51 Modified Precision Rectifier
16:02 Design of full wave rectifier using the precision half wave rectifier
to anyone who is confused at why dividing by AOL then have patience for 2 to 3 minutes, actually this act of dividing is being done to see equivalently how it affects the voltage difference between positive and negative terminal of op amp. rest is in the video itself ,
thankyou from Bihar, India
You are a life saver when it comes to my electronics courses....
Are Bhai
Are you alive?🙄
I have been trying to understand this since 2 days...finally a good explanation of the negative feedback equations! Thank you very much!
So glad you made this video. My textbook fucking sucks. It literally makes all of these claims without proving or showing how it arrived at these statements. I need to see why something is as stated, not just take it for granted. Your videos are the background info so many of us really need, thank you good sir.
thank you for everything.. best lecturer for my degree..
yr video on PRECISION RECTIFIER Is outstanding And Is very WELL DESIGNED.
thanks you
s.vatsa
The best explanation as usual...👍👍👍
You are a life saver when it comes to my electronics courses....
Your analysis methods are sometimes get way too vague leaving so many doubts.Please while making assumptions provide a little more information like in this video you assumed the voltage (V_) to be 0V(while analysing the circuit for negative input ) now if one goes by the virtual ground concept, one gets confused how can it happen when V+ has to be V- ( and V+ is Vin).
Beautifully explained Hats off !!!!!!!!
Neat explanation....much needed before VIVA...thanx
I'm still a little bit confused, why Vo is equal to zero when Vin < 0, for ideal op amp, V+ = V-, so the Vo should be always equal to Vin 🤔
The best explanation as usual...👍👍👍
First of all, Thank you for a great video.
If I understood correctly, the purpose of explaining the super diode circuit is to demonstrate how the assumption that the diode in the negative feedback loop is approximately ideal can be justified, with its voltage drop being negligible. Thanks to this assumption, the problem of the simple half-wave rectifier circuit is resolved. However, in the explanation of the super diode circuit, it is not clear to me how you derived the expression for V_F' . Is the V_F' actually a representation of the V_{in} voltage, where when you showed that V_F' is approximately zero, it means that a slightly greater input voltage than 0V is required for the diode to conduct?
And this is derived from the general expression for any operational amplifier, which is V_{IN} = V_{out}/{A_{OL} ?
Additionally, if you made such an assumption, why did you then use the value of V_F and not V_F' in the expression for V_{out}' ? Thus, the assumption that the diode is approximately ideal is not reflected here.
I appreciate if you could explain that.
your my mentor thank you very much and Im sorry to correct mentor..God bless.
Very nice explanation
Thanks a lot 🙏
Your explanation is so simple to understand. Thank you so much for uploading the videos.
12:23 diode D1 is conducting some forward current flows through it. Then the current will pass through D2 also isn't it? Then how D2 won't conduct? Which path does current from D1 will follow?
Really a variety approach is taken to explain.... Very nice 👌👌👌
Well explained... Thank you very much.
Your videos are very very useful to me in Analog elecronics subject, but at 7:48 I think.. " Matlab kuch bhi ho raha hai"😅
Thank you sir good explanation
Thanku sir,ur exlaplanation is excellent ..👏👍
What do you study?
Sir, at 1st case, we assume the diode is conducting, but after the explanation, you said, only when Vin>0, then the diode is conducting. But, why?
First we assume, and the we prove vin=vout. But, how can we say that??
To prove that case, you need to take some assumption. With that assumption if you are getting correct answer that means your assumption is correct.
Let me try to explain. Stop the video at 3:09 and look at the circuit.
As you asked, why should not diode conduct only when Vin is negative? Well if Vin is negative, output from the op amp will be negative too. This will reverse bias the diode and current cannot sink in through output. Since Vout is connected to ground through resistor, output voltage will also be zero.
@@curiosity551 When vin>0 diode is forward biased and we have negative feeback and Vout is fed back to negative input. Op amp action makes the inputs the same so Vin=Vout . Difference between inputs is 0 volts. When Vin less than ground the op-amps output goes to negative supply voltage and diode is reversed bias and Vout = 0 because there is no negative feeback since diode is reversed biased (open circuited) . The output resistor pulls vout to ground since there is no current.
Nice explanation..
Crystal clear explanation 🙏
Great video! Can you explain in simple terms how the adding part of creating the full rectifier works? Thanks!
Really good video. Thank you!
Great, but filtering the precision full wave rectifier is going to have more ripple than full wave rectification by conventional means.
Frank Reiser M.S.
SO helpful. Thank you!!!
I don't understand how the open loop gain can be used to prove the diode's forward voltage doesn't effect the output voltage when the op amp is closed loop. Can anyone explain?
EXCELLENT as usual
Excellent explaination !
Why we have taken open loop gain when you have connected the op amp in feedback configuration. Shouldn't you have divided the drop with closed loop gain
Sir, during positive half cycle output voltage is positive. So, diode should be reverse biased, because positive terminal connected to n side of diode?
F
Exactly my doubt
Same doubt... Hey Raza did u get it cleared?
Good one sir....
I m also a teacher of electronic gate net jam
loving it!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Why we are using two diodes? Can we use any other no of diodes there?
Awesome Explanation.
Nicely explained.
Thank you for creating the material with easy to understand explanations.
I have a question with respect to the modified precision rectifier.
Why is the input provided to the inverting input of the op-amp? is this change a part of the modification??
Please explain.
Thanks in advance.
Yes, it is a part of the modification. (To restrict the output of the op-amp from going into the saturation)
ALL ABOUT ELECTRONICS Thank you very much for for the quick reply.
Plz make a video on op amps internal circuit
I have one doubt why we use opamp although we made analysis on opamp circuits carried out using resistors what happen when we made the experiments without opamp
I don't understand in the case 12:21 when V_in > 0 there is the current passing from the point of 0V through D1 and to the point of -0.7V while this circuit isn't closed ?
Although it seems, the circuit is not complete through an external resistors, but the current flowing through D1 will go through an internal circuits of the op-amp. And the path will get completed through an internal circuit components of the op-amp. I hope, it will clear your doubt.
awesome explanaition!!!!
how did you write the effective voltage drop across the diode in closedloop configuration as Vf/ openloop gain .. tiem 4:10 sec in the video
scroll down he has answered to this question
When diode D1 IS CONDUCTING, ULTIMATELY WHERE THE CURRENT IS GOING? PLEASE EXPLAIN
We need a video on Precision clipper.
Also with some reference voltage.
Sir ,Please make video on absolute value output circuit 😊
sir nice , to tell about design
thanks sir
11:58 Why is Vo equal to 0? Is it because there's a negative voltage at the anode of the second diode?
I don't see how it would be 0 if Vo is connected to ground through 2 different resistors. Wouldn't the voltage drop across those resistors?
Think it as if two passive resistors are connected at one end, and the other end of both resistor is connected to the ground. In that case, the voltage at the common node (where both resistors are connected ) would be 0 V. I hope, it will clear your doubt.
you confuse me alnost everytime but I watch your videos because there are no other options available
i fuckin love this guy
what if we consider the offset voltages?
Sir Why Virtual Ground only for -ve feedback ?
Because the concept of virtual ground is applicable whenever the op-amp is used in the linear region. When the positive feedback is used then op-amp is not operating in the linear region.
Thanks to bolna re harami
@@abhijithanilkumar4959 dude i was feeling the same thing,ungrateful people all around
When V>0, I don't understand why D2 does not conduct.
thank u sir, can i use opamp based power supply fr radio receiver
I want to ask that.....is it same as inverting linear half wave rectifier with positive output voltage
Yes, you can also see it from the transfer characteristic at 14:58
If the question is explain half wave precision rectifier only means from where we should write
Can anyone help me please
Nice one
Amazinggg!!!
Sir , how does the effective voltage drop get divided by open loop gain 4:22
The diode is connected at the output of the op-amp in the feedback configuration. So, the effect of this diode drop on the input will be equal to Vf / Av. So, from the input side perspective, it behaves like an ideal diode.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Dear Sir,
I really appreciate your effort on making such good teaching videos.
Regarding your explanation, I understood (hopefully) the assumptions you made, and seems normally that the voltage across the diode to be as minimum as possible in order that the circuit to behave as a voltage follower (when Vin > 0).
However, can you point me a further reading resource on how this is achieved knowing that the silicon diode (in this case) has to consume at least 0.7V in order to close the circuit (to be seen as a voltage follower)? Should I think on some very fast time changing states of the internal opamp circuitry, that somehow balances this overall steady state? Or in fact, the diode really stays around 0V, as in practice there will still be a very small current passing through it and in this fashion closing the circuit? Thank you for your attention!
Best regards,
Why do we use load resistance in precision rectifier
Can anyone explain how at 8.16time .. The voltage should be negative
Here we are assuming that the diode is in the reverse biased condition and hence acts as an open circuit. For the diode to be in a reverse-biased condition the voltage at the anode should be less than the cathode. Here, in that condition, the voltage at the cathode is zero. That means the diode will remain in the reverse bias condition when the voltage at the anode is negative. Or in other words, the output of the op-amp is negative. And it will happen when the input Vin is negative. (Because in that condition, the output of the op-amp will be negative saturation voltage as it is operating as a comparator in the open loop condition). I hope this will clear your doubt.
Please help Build an opamp based current to voltage converter that will give an output of 0v at 4 mA and 5v at 20mA
sir ge u r great
Why isnt the concept virtual short is applied when vin < 0? Applying virtual short concept for vin< 0 yields Vo = Vi, im confused 🤔
It won't be. Let me explain. Let's say, Vin
Harami sale
Tereko sir itna time leke explain kiyela aur tereko thanks bolne ko nhi hota
Thank you sir. I also had the same doubt.
ALL ABOUT ELECTRONICS sir you just said Vin< 0 in the beginning then how come the output will be zero, instead it should be (-R2/R1)Vin for an open loop circuit where D2 is conducting .Kindly explain I’m confused.
How do I size the component values?
Why effective voltage drop = VF / open loop gain?????
I have explained that in the video during the circuit analysis. Please watch it 5:00 onwards.
If you still have any doubt after that then do let me know here.
12:34 can't we use inverting opamp Vout formula? : Vout=-(Rf/R1) Vin?
The diodes are also in the feedback loop.
please make a video on clamper using opamp.......there are no satisfactory resources available online
Soon I will make a video on the clamper circuit using op-amp.
Great video! Where you from? I don't understand much your accent (I'm not english native speaker).
India
@@ALLABOUTELECTRONICS 😂😂
at 4:25 why effective voltage drop found out by dividing by open loop gain ?
It is in the closed loop configuration. The concept which was used to establish the concept of virtual ground is used here. Just after 4:25 in next 2-3 mins,I have explained that.
@@ALLABOUTELECTRONICS thanks for replying sir
@@ALLABOUTELECTRONICS sir at 11:30 you said that opamp is operated in linear region and we can apply the concept of virtual ground
so my doubt is that if it was not in linear region then could we apply or not the concept of virtual short
plz reply sir
please, can you explain the varactor??
Damn this question was asked in jam 2021.
Tq
wt is the minimum and maximum input voltages applied to the precision rectifier
That depends on the opamp and supply voltages you choose/have.
what is meant by linear region and virtual ground ?
Please check this video. You will get it.
th-cam.com/video/AuZ00cQ0UrE/w-d-xo.html
for purposes of explanation at 11:37 : at the junction of R1, R2 is at "virtual ground" ; be careful with this idea !
this voltage is very, very small, it can NOT be 0v.
for positive Vin]pk inverting input is +0.6v/A]open_loop or about 0.6uv for A]open_loop = x1,000,000
0.6v is presumed to be the fwd voltage of D1
why (V(0) + 0.7V ) / A(OL) = 0 at 6:52
Because the open loop gain of the op-amp is very high. Typically 10^5 to 10^6.
So, (Vo + 0.7V)/ Aol will approximately equal to zero.
@@ALLABOUTELECTRONICS thanks for reply
Mighty Mighty MIT
At 10:36 how come D2 is off and vo=0
Please watch the video 11:00 onwards, I have already explained it.
4:18
How this formula came sir???
assuming diode voltage as output voltage. then input voltage* gain is output voltage nah... so, Vf = GAIN * vf`
vo=R2/R1 Vin and not Vo=- R2/R1 Vin...
👍👍👍👍👍
0:36 😂😂😂
?
😂😂😂🤣🤣🤣🤣
Preferisco che spiegazione di elettronica soprattutto in italiano spagnolo è vanno bene anche in inglese le altre lingue straniere non ci capisco niente
are you a Gujju ?