большое спасибо за ваш труд. всегда очень интересно смотреть ваши лекции. Потому что кроме функционала хотелось бы увидеть математику обеспечивающую функционал, но при этом не утонуть вы буквах.
OTA are also stable with capacitive loads! That's been useful on a few of my projects, and a convenient way to tune feedback loops in converters with few external passives
@@sambenyaakov I thought most VFB opamps need a series output resistor to add phase margin when driving capacitive loads as an integrator? I think TI has a good app note on VFB vs OTA and their respective integrator topologies.
@@benm4784 To stabilize an OP Amp in closed loop with a large capacitive load, you add a series resistor and take the feedback at the output of the Op Amp not from the cap. A transconductance is a current source, that is, has a large series resistance at the output. So if you try to operate it in CLOSED LOOP you can not not sample the signal before the Cap.
AS far as I know real professional audio, and stuff I work on, has everything digital. The units I mentioned have about 1% distortion for 100 mVpp input, not very professional to me. And BTW are the OTA you talk about one of the units I have pointed out? If not which?
@@sambenyaakov Behringer PMP6000 uses V13700 as compressors on the input to the class-D power amp stage. Doubt 1% distortion would matter here because it's just going to speakers that add their own distortion anyway. Soft 2nd and 3rd harmonic distortion is much more tolerable than hard high frequency effects such as class B crossover distortion.
Dear professor, thank you for an excellent video (as always). I am wondering: Being an open loop topology, it should be easier to make this kind of amplifier working up to a very high frequency, in comparison to trying to do so with a classical global feedback of an op-amp. That would make it a great application for a fast ADC preamplifier. Unfortunately, those parts that you've shown have GBP only of 2 MHz. This suggests there might be some caveat there. Maybe the transistors are not fast enough or some parasitic capacitances (or inductances -- as it is all current driven inside) are impeding high speed operation.
Dear Professor, thank you for the great video! I think there is a slight mistake at 19:15. When I worked out the equation, I found that Io = Iin * (IB/ID) (e.g. no factor of 2).
Thanks for checking the calculation. I think though that there might be a confusion as what is Io, at the collectors or after the current mirrors. The ratio between these two is 2.
Hello Professor. Could you please axplain why did we multiply (I_C1+2I_B) by the alpha factor to calculate the I_out? Also, what is the rationale to assume the base currents of the Q1 and Q2 equal? Is it an approximation or a fact?
Thank you @@sambenyaakov , for your response. I was referring the calculations in Slide #7 (around 5'47"), I could not understnad why we multiply the I_C1+2I_B by ALPHA (line 2 in the slide.) And also in the same slide, what is the reason behind the assumption of identical base currents in the analysis. Is it due to small value, or matched transistor assumption etc? II appreciate your channel, videos and your explanation. I understood so much with your lessons.
@@mehmetdurna3115 Alpha is the ratio between collector and emitter current. Emitter current is 2*Ib+Ic1 The implementation is usually on a single chip and the transistors are pretty much matched if close one to the other. __________________________________________
these linearization diodes, look like Voltage clamps. Never used them and it works just fine. it's still not very clear why you should use them, what benefit you get. Some new FET from Texas Instruments, released a couple of months ago has also a diode on the gate from the source, like the "circuit" on the right at 17:12. was the first time i saw i saw that, seams weird, like having the emitter power the base, seams illogical.
@@sambenyaakov Well i have build voltage controlled filter with them. amplifying a 10 V.p.p. from one the attenuated -20x, no distortion found, instead of CA3080. not Texas Instruments, but Analog devices. MAT12, SSM2212. seams weird, one would think that the output, emitter would also power the base "partially", like a feedback.
It's very interesting to know this "old" gems of electronics that I've never heard about before. Thanks.
👍😊🙏
Translinear circuits are always a exciting topic
Thanks
большое спасибо за ваш труд. всегда очень интересно смотреть ваши лекции. Потому что кроме функционала хотелось бы увидеть математику обеспечивающую функционал, но при этом не утонуть вы буквах.
Outstanding Job.
Thanks Hamid.
👍🙏
🙏😃👍
OTA are also stable with capacitive loads! That's been useful on a few of my projects, and a convenient way to tune feedback loops in converters with few external passives
Stable in open loop like any amplifier.
@@sambenyaakov I thought most VFB opamps need a series output resistor to add phase margin when driving capacitive loads as an integrator? I think TI has a good app note on VFB vs OTA and their respective integrator topologies.
@@benm4784 To stabilize an OP Amp in closed loop with a large capacitive load, you add a series resistor and take the feedback at the output of the Op Amp not from the cap. A transconductance is a current source, that is, has a large series resistance at the output. So if you try to operate it in CLOSED LOOP you can not not sample the signal before the Cap.
@@sambenyaakov Ah! I believe I follow, and this gives me some concepts to play with! Thanks!
Very Nice, could you please take composite amplifier too. one current feedback + voltage feedback . Current feedback is inside voltage feedback.
Not sure I follow. You mean to replace the transconductance amplifier?
They're also used in guitar 🎸 Pedals
Thanks for input.
Every bit of analog professional audio gear has OTA audio compressors in them
AS far as I know real professional audio, and stuff I work on, has everything digital. The units I mentioned have about 1% distortion for 100 mVpp input, not very professional to me. And BTW are the OTA you talk about one of the units I have pointed out? If not which?
Mackie SRM450 v2 uses numerous NJM13600M OTAs as variable gain stages and for variable frequency filters
@@sambenyaakov Behringer PMP6000 uses V13700 as compressors on the input to the class-D power amp stage.
Doubt 1% distortion would matter here because it's just going to speakers that add their own distortion anyway.
Soft 2nd and 3rd harmonic distortion is much more tolerable than hard high frequency effects such as class B crossover distortion.
Dear professor, thank you for an excellent video (as always).
I am wondering: Being an open loop topology, it should be easier to make this kind of amplifier working up to a very high frequency, in comparison to trying to do so with a classical global feedback of an op-amp.
That would make it a great application for a fast ADC preamplifier.
Unfortunately, those parts that you've shown have GBP only of 2 MHz. This suggests there might be some caveat there. Maybe the transistors are not fast enough or some parasitic capacitances (or inductances -- as it is all current driven inside) are impeding high speed operation.
Youneed a dedicated design to HF Also. open loop ,means inaccuracy in gain and distortion
The differential pair is indeed transconductance amplifier but the whole amplifier has a current to current gain
Indeed. Thanks for emphasizing this.
So it still could be used to condition a signal by changing the gain slightly rather than inducing a huge gain like a pre-amplification stage.
IN gain control you need a rather large amplitude range.
Dear Professor, thank you for the great video! I think there is a slight mistake at 19:15. When I worked out the equation, I found that Io = Iin * (IB/ID) (e.g. no factor of 2).
Thanks for checking the calculation. I think though that there might be a confusion as what is Io, at the collectors or after the current mirrors. The ratio between these two is 2.
Hello Professor. Could you please axplain why did we multiply (I_C1+2I_B) by the alpha factor to calculate the I_out?
Also, what is the rationale to assume the base currents of the Q1 and Q2 equal? Is it an approximation or a fact?
Please specify to which slide or minute you are referring.
Thank you @@sambenyaakov , for your response. I was referring the calculations in Slide #7 (around 5'47"), I could not understnad why we multiply the I_C1+2I_B by ALPHA (line 2 in the slide.)
And also in the same slide, what is the reason behind the assumption of identical base currents in the analysis. Is it due to small value, or matched transistor assumption etc?
II appreciate your channel, videos and your explanation. I understood so much with your lessons.
@@mehmetdurna3115 Alpha is the ratio between collector and emitter current. Emitter current is 2*Ib+Ic1
The implementation is usually on a single chip and the transistors are pretty much matched if close one to the other.
__________________________________________
these linearization diodes, look like Voltage clamps.
Never used them and it works just fine. it's still not very clear why you should use them, what benefit you get.
Some new FET from Texas Instruments, released a couple of months ago has also a diode on the gate from the source, like the "circuit" on the right at 17:12.
was the first time i saw i saw that, seams weird, like having the emitter power the base, seams illogical.
The benefit is linearization. Without the you get high distortion for larger signals. Please send the part number of the MOSFETs you mentioned?
@@sambenyaakov Well i have build voltage controlled filter with them. amplifying a 10 V.p.p. from one the attenuated -20x, no distortion found, instead of CA3080.
not Texas Instruments, but Analog devices.
MAT12, SSM2212. seams weird, one would think that the output, emitter would also power the base "partially", like a feedback.