" Jag ghoomeya thaare jaisa na koi" Translation- I roamed around the world and found no one like you... This differential amplifier topic was so confusing to me.... but you explained it so smoothly... thank you so much sir... Love from INDIA...
43:05 No. Even ignoring that the CMRR is ideally infinite, the CMRR as calculated does not remains the same. The A_vcm remains the same (b/c 200 kΩ +1 kΩ ~ 200 kΩ ), but the differential gain A_vid with the introduction of Re dropped by a factor of 5x (1+ gm*Re) as intended & calculated a few minutes earlier in the video. Therefore the CMRR as calculated becomes 160 (and not 800). PS: this comes to show that even great Professors can have slips of the pen...
Nope, you almost got me, but think about the definition of CMRR @22:10. It's the ratio between Av_diff and Av_cm. The numerator of Av_cm always has the expression of Av_diff. So the CMRR is simplified as the denominator of Av_cm, which is CMRR=1+2*gm*REE which is still approx. of 800.
@@nurahmedomar No. When we introduce Re the Av_diff is no longer gm*Rc, but rather ~ -Rc/Re . However in the Av_cm expression your numerator is still gm*Rc (no longer Av_diff therefore). Look, with Re the Av_cm = -Rc/(rm + Re + 2Ree) ~ -Rc/(rm + 2Ree) because 2Ree >> Re (200 Kohm >> 1 Kohm). So basically with or without Re, Av_cm remains approximately the same. Av_diff however drops by a factor of 5, and so the CMRR must drop by 5x from 800 to 160.
@@justpaulo I think you are right. After adding the source degeneration resistor Re, the Av_diff changes, but Av_cm numerator does not change, so it cannot cancels out simply.
Hello sir, I've got some doubts, 1. Why is the common-mode gain just the gain of the half circuit? Shouldn't we take the difference of both half circuits, so AvCm = 0 ideally? 2. @35:35 200μΑ flowing through RC is the max case, so that's the lowest dip of the collector node. In this case, the lowest dip is 4V so the Base input must be 4.5 V minimum. So shouldn't the Common-mode input's minimum be 4.5V to something higher than that, why is it 1V - 4.5V? (I'm speaking under the assumption that the common-mode input is the base voltage, I think I'm wrong somewhere but I'm not able to put my finger on it) Thank you very much
1. It is the same gain that you find because you apply the same voltage in both input and then you obtain the same output voltage variation in both the citcuit parts 2. When the collector voltage is 4 V then the BJT is approaching the saturation, and can't go further below that value. If the BJT has to remain in forward active mode, VBC has to be lower than 0,5V so that the BC junction is not directly biased. So, using KVL we obtain that the maximum value for Vi- (the voltage of the base of right BJT) is 4V+0.5V = 4.5V. The minimum value for Vi- has discussed previous in the video: the voltage across the current source, because this is realized with a transistor, is minimum 0.2V-0.3V, the minimum voltage, of the other (left side) BJT in forward active region is approximately 0,7V, then the left BJT base voltage is at least 0,7V+0,3V= 1V.
1) Yes, you are right. Differentially & ideally the AvCm =0 and so CMRR is ideally infinite (even if Ree is finite). 2) The base can't go above 4.5V b/c otherwise the Base/Collector p/n junction would start to be forward biased and the NPN transistor would leave the active region and enter the saturation region.
43:36 Sir, you said that when we add a resistor to the emitter, shouldn't the CMRR goes down a lot? because it was 1+2g_mR_EE. But now, it is 1+2g_mR_EE/(1+g_mR_E).Where R_E is the emitter resistance
Yes, you are right. The differential gain dropped by a factor of 5x (1+g_mR_E) while the common mode gain remained the same. Therefore the CMRR should drop by 5x (the 1+g_mR_E factor you mention). However there's another issue that makes irrelevant that 5x gain drop. Common mode wise both branches have the same input and the same gain ideally. When you take the output differentially they cancel out and therefore the common mode gain is actually 0 (zero) even when Ree is finite. So, ideally the CMRR is infinite.
No, see the definition of CMRR @22:10. It's the ratio between Av_diff and Av_cm. The numerator of Av_cm always has the expression of Av_diff. So the CMRR is simplified as the denominator of Av_cm, which is CMRR=1+2*gm*REE which is still approx. of 800.
Hello Professor ! Can I have access to handouts of these lectures and to assignments by any chance ? That will be great help in addition. Thank you in advance.
22:22 However, I think that differentially the common mode gain is *ideally* 0 (zero) b/c both branches have the same input and the same gain. Therefore I believe the CMRR is *ideally* infinite even if Ree is finite.
No, if Ree is finite, you would amplify the common mode voltage. This is very common in differential amplifiers where you amply the common noise seen by both inputs, which is not desirable. That's why in may datasheets this CMRR parameter is extremely important. Typically you see 70dB~90dB of CMRR range in datasheet, not infinite.
@@nurahmedomar No. Note that I used the words "differentially " and "ideally". By ideally I mean no mismatches (in Rc or the BJT's). And by differentially I mean taking out the output differentially (vop - von). If there are no mismatches, each half circuit is exactly equal with finite or infinite Ree. The gain will be equal for both half circuits and equal to -Rc/(rm + 2Ree). Therefore, when you take the output differentially (vop - von) they cancel out. Of course that in the real world it's inevitable to have mismatches and therefore the CMRR will never be infinite even is the output is taken differentially. In that case the larger Ree the better.
@@justpaulo If you say Ree is ideal, and infinite. I agree with you, the CMRR is also infinite. The mismatch you are talking about is more profound for differential gain, it does not affect the common mode gain much.
As differential voltage grows it will steer the bias current onto one side, woudn't that change the gm as it is determent by Ic/Vt? So we wouldn't have linear amplification of Vid, or is Vid considered small signal and thus always having gm = Iee/2Vt?
If you look at the large signal curve you start to get some clipping once most of the current is on one side or the other. The gm is only accurate in the approximately linear region. If you really want to you can do small signal model everywhere via a piece-wise linear model. Which is where you cut up your large signal into a bunch of small signals and do the small signal analysis for each region.
You're right. Gm is changing with the input, and that's why we're assuming a very small perturbation so that we can assume gm is constant and equals to quasi-point collector current (biasing current) over the thermal voltage. You just have mixed large signal analysis with small signal, and that is not correct. You have a gm around biasing point, and that gm is changing for large swing input, thus changing gain with input voltage and finally distortion in the output. While solving in small signal model, you assume that all of these things are not there, since this model is linear.
" Jag ghoomeya thaare jaisa na koi"
Translation- I roamed around the world and found no one like you...
This differential amplifier topic was so confusing to me.... but you explained it so smoothly... thank you so much sir... Love from INDIA...
Hello Prof. Ali Hajimiri,
I just want to THANK YOU for your lectures. Stay blessed ❤️
46:38 With the 2Rs & split current source topology you also gain some headroom (200mV in the given example)
43:05 No. Even ignoring that the CMRR is ideally infinite, the CMRR as calculated does not remains the same.
The A_vcm remains the same (b/c 200 kΩ +1 kΩ ~ 200 kΩ ), but the differential gain A_vid with the introduction of Re dropped by a factor of 5x (1+ gm*Re) as intended & calculated a few minutes earlier in the video. Therefore the CMRR as calculated becomes 160 (and not 800).
PS: this comes to show that even great Professors can have slips of the pen...
Nope, you almost got me, but think about the definition of CMRR @22:10. It's the ratio between Av_diff and Av_cm. The numerator of Av_cm always has the expression of Av_diff. So the CMRR is simplified as the denominator of Av_cm, which is CMRR=1+2*gm*REE which is still approx. of 800.
@@nurahmedomar No. When we introduce Re the Av_diff is no longer gm*Rc, but rather ~ -Rc/Re . However in the Av_cm expression your numerator is still gm*Rc (no longer Av_diff therefore).
Look, with Re the Av_cm = -Rc/(rm + Re + 2Ree) ~ -Rc/(rm + 2Ree) because 2Ree >> Re (200 Kohm >> 1 Kohm). So basically with or without Re, Av_cm remains approximately the same.
Av_diff however drops by a factor of 5, and so the CMRR must drop by 5x from 800 to 160.
@@justpaulo I think you are right. After adding the source degeneration resistor Re, the Av_diff changes, but Av_cm numerator does not change, so it cannot cancels out simply.
Hello sir, I've got some doubts,
1. Why is the common-mode gain just the gain of the half circuit? Shouldn't we take the difference of both half circuits, so AvCm = 0 ideally?
2. @35:35 200μΑ flowing through RC is the max case, so that's the lowest dip of the collector node. In this case, the lowest dip is 4V so the Base input must be 4.5 V minimum. So shouldn't the Common-mode input's minimum be 4.5V to something higher than that, why is it 1V - 4.5V? (I'm speaking under the assumption that the common-mode input is the base voltage, I think I'm wrong somewhere but I'm not able to put my finger on it)
Thank you very much
1. It is the same gain that you find because you apply the same voltage in both input and then you obtain the same output voltage variation in both the citcuit parts
2. When the collector voltage is 4 V then the BJT is approaching the saturation, and can't go further below that value. If the BJT has to remain in forward active mode, VBC has to be lower than 0,5V so that the BC junction is not directly biased. So, using KVL we obtain that the maximum value for Vi- (the voltage of the base of right BJT) is 4V+0.5V = 4.5V. The minimum value for Vi- has discussed previous in the video: the voltage across the current source, because this is realized with a transistor, is minimum 0.2V-0.3V, the minimum voltage, of the other (left side) BJT in forward active region is approximately 0,7V, then the left BJT base voltage is at least 0,7V+0,3V= 1V.
1) Yes, you are right. Differentially & ideally the AvCm =0 and so CMRR is ideally infinite (even if Ree is finite).
2) The base can't go above 4.5V b/c otherwise the Base/Collector p/n junction would start to be forward biased and the NPN transistor would leave the active region and enter the saturation region.
@@stefano.a You are absolutely correct in answering both questions. I totally agree with you!
43:36 Sir, you said that when we add a resistor to the emitter, shouldn't the CMRR goes down a lot? because it was 1+2g_mR_EE. But now, it is 1+2g_mR_EE/(1+g_mR_E).Where R_E is the emitter resistance
Yes, you are right. The differential gain dropped by a factor of 5x (1+g_mR_E) while the common mode gain remained the same. Therefore the CMRR should drop by 5x (the 1+g_mR_E factor you mention).
However there's another issue that makes irrelevant that 5x gain drop.
Common mode wise both branches have the same input and the same gain ideally. When you take the output differentially they cancel out and therefore the common mode gain is actually 0 (zero) even when Ree is finite.
So, ideally the CMRR is infinite.
No, see the definition of CMRR @22:10. It's the ratio between Av_diff and Av_cm. The numerator of Av_cm always has the expression of Av_diff. So the CMRR is simplified as the denominator of Av_cm, which is CMRR=1+2*gm*REE which is still approx. of 800.
Hello Professor ! Can I have access to handouts of these lectures and to assignments by any chance ? That will be great help in addition. Thank you in advance.
22:22 However, I think that differentially the common mode gain is *ideally* 0 (zero) b/c both branches have the same input and the same gain. Therefore I believe the CMRR is *ideally* infinite even if Ree is finite.
No, if Ree is finite, you would amplify the common mode voltage. This is very common in differential amplifiers where you amply the common noise seen by both inputs, which is not desirable. That's why in may datasheets this CMRR parameter is extremely important. Typically you see 70dB~90dB of CMRR range in datasheet, not infinite.
@@nurahmedomar No. Note that I used the words "differentially " and "ideally". By ideally I mean no mismatches (in Rc or the BJT's). And by differentially I mean taking out the output differentially (vop - von).
If there are no mismatches, each half circuit is exactly equal with finite or infinite Ree. The gain will be equal for both half circuits and equal to -Rc/(rm + 2Ree). Therefore, when you take the output differentially (vop - von) they cancel out.
Of course that in the real world it's inevitable to have mismatches and therefore the CMRR will never be infinite even is the output is taken differentially. In that case the larger Ree the better.
@@justpaulo If you say Ree is ideal, and infinite. I agree with you, the CMRR is also infinite. The mismatch you are talking about is more profound for differential gain, it does not affect the common mode gain much.
nice video!!! informative
As differential voltage grows it will steer the bias current onto one side, woudn't that change the gm as it is determent by Ic/Vt? So we wouldn't have linear amplification of Vid, or is Vid considered small signal and thus always having gm = Iee/2Vt?
If you look at the large signal curve you start to get some clipping once most of the current is on one side or the other. The gm is only accurate in the approximately linear region. If you really want to you can do small signal model everywhere via a piece-wise linear model. Which is where you cut up your large signal into a bunch of small signals and do the small signal analysis for each region.
You're right. Gm is changing with the input, and that's why we're assuming a very small perturbation so that we can assume gm is constant and equals to quasi-point collector current (biasing current) over the thermal voltage.
You just have mixed large signal analysis with small signal, and that is not correct. You have a gm around biasing point, and that gm is changing for large swing input, thus changing gain with input voltage and finally distortion in the output. While solving in small signal model, you assume that all of these things are not there, since this model is linear.