Timeline for Should gain of this differential circuit be absolute value?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Feb 4 at 16:16 | vote | accept | kile | ||
| Dec 18, 2023 at 19:42 | history | edited | JYelton | CC BY-SA 4.0 |
deleted 5 characters in body; edited title
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| Dec 17, 2023 at 23:07 | comment | added | kile | @G36 Could you please have a look at my answer. What's wrong here? | |
| Dec 17, 2023 at 23:06 | comment | added | kile | @internet Could you please have a look at my answer. What's wrong here? | |
| Dec 17, 2023 at 15:53 | answer | added | kile | timeline score: 0 | |
| Dec 16, 2023 at 17:50 | comment | added | Hearth | That's a bizarre manufacturing error. I know it's just a textbook example but I'm struggling to imagine how any problems at the manufacturing stage could alter the width of one transistor and not another! | |
| Dec 16, 2023 at 15:43 | comment | added | G36 | But this is not the case here. We do not have virtual GND because gm1 is not equal to gm2. So the middle point will not be at 0V. And this is why your method does not give the correct result. Using your method we have the M1 gain RD/(1/gm1) and M2 gain is RD/(2 * gm1) and the overall differential gain is 3*gm1*RD | |
| Dec 16, 2023 at 15:32 | comment | added | kile | @G36 the source of M1 and M2 are virtual grounded if it's symmetrical. | |
| Dec 16, 2023 at 15:25 | comment | added | G36 | Do you have in mind your method (grounding the M1 and M2 source)? | |
| Dec 16, 2023 at 15:23 | comment | added | kile | @G36 Is symmetrical valid in this case? | |
| Dec 16, 2023 at 15:21 | comment | added | G36 | I also prefer the T-model. electronics.stackexchange.com/questions/536586/… Using it we can see by inspection that the asymmetrical voltage gain is RD/(1/gm1 + 1/gm2), and if gm2 = 2*gm2 we have 2/3*gm1*RD Thus, for differential output (Vout1 - Vout2) the gain will be twice this value the value 4/3*gm1*RD | |
| Dec 16, 2023 at 15:18 | comment | added | internet | check this: yourimageshare.com/ib/fqwo6Yy3wO | |
| Dec 16, 2023 at 15:04 | comment | added | kile | @internet Could you please show me the small signal circuit diagram to make it easier to understand you equation? | |
| Dec 16, 2023 at 14:00 | comment | added | internet |
You can always draw the small signal model to calculate gain but for this circuit I can do it in my head. Assume that you use T model for the mosfets and ignore the output impedance. id1 = (Vin1 - Vin2)/(1/gm1 + 1/gm2) Vout = Vout1 - Vout2 = id1*RD - (-id1)*RD = 2id1*RD = 2RD(Vin1 - Vin2)/(1/gm1 + 1/gm2) So Vout/(Vin1-Vin2) = 2RD/(1/gm1 + 1/gm2) = 4/3*gm1*RD with gm2 = 2*gm1
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| Dec 16, 2023 at 13:48 | comment | added | kile | @internet What should you use to analyse this circuit? | |
| Dec 16, 2023 at 12:19 | comment | added | internet | Your calculation seems correct but the result is wrong because you cannot split the circuit like that as it's not symetric. The voltage vgs of M1, M2 in this case are not same magnitude as you drew. | |
| Dec 16, 2023 at 10:20 | history | edited | kile | CC BY-SA 4.0 |
edited body
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| Dec 16, 2023 at 9:58 | history | asked | kile | CC BY-SA 4.0 |