The only architectures that our professor taught us to design are OTA, CS stage and then Miller Opamp, so I thought that opamp was the right choice. Is there another kind of amplifier that should be used with large cap loads?Why do you need a two-stage miller-compensated opamp if you have such a large capacitive load?
I was thinking of using a simple R-2R ladder made with resistors. The other choice would be the other topology that we saw that was the charge redistribution, but the problem is that I don't know how to design it and how to make it work, since we didn't do any design for it and I can't find good sources on how to learn to do it (neither books nor online)What kind of DAC topology do you envision?
Oh, I think I understood. So having a gain > 4096 would gain up the DAC output, giving a 0-4095 volts scale, right?The gain is given by the required 12-bit resolution. To meet a 4096-level accuracy, your gain has to be at least higher than that (give-or-take).
Thanks for that tip, I'll take a look at it!The 1-% settling error within 1 us gives you the bandwidth and phase-margin requirements. Yet again, with the 10-us load you might not have to care that much about phase margin and can assume a single-pole system. A standard e^(t w0) calculation will suffice to give you the required pole location.
And that much of a current (that's really high) is strange for an application like the one I described before, right?Slew rate would be given by current (power) over load capacitance and dictated by the 1-% requirement too. But assuming you need to toggle full swing in 1us would require you to output an SR at 0.99*Vdd/1us = Iout / 10 uF => Iout > 18 A (which is huge!) per opamp. So that puts another requirement on your architecture. (One specific architecture does not suffer from slew-rate limitations). Or your specification is wrong/incomplete. There is for example not clear to me why they specify in us when your sample rate is in ms.
Perhaps you're already aware of all-in-one DAC chips available to convert given amounts of bits. 12-bits normal cost higher than $10. And the experimental stage may consume several chips.I was thinking of using a simple R-2R ladder made with resistors.
Dear jjx,
Thanks for your reply, I'll try to answer to everything. I forgot to specify that the project is given from a company, reviewed by the professor and then sent to us. For this project in particular, the application would be the use of the DAC in implantable devices used for the monitoring of bio-electric signals from the brain and for the electrical stimulation of the CNS.
The only architectures that our professor taught us to design are OTA, CS stage and then Miller Opamp, so I thought that opamp was the right choice. Is there another kind of amplifier that should be used with large cap loads?
I was thinking of using a simple R-2R ladder made with resistors. The other choice would be the other topology that we saw that was the charge redistribution, but the problem is that I don't know how to design it and how to make it work, since we didn't do any design for it and I can't find good sources on how to learn to do it (neither books nor online)
Oh, I think I understood. So having a gain > 4096 would gain up the DAC output, giving a 0-4095 volts scale, right?
Thanks for that tip, I'll take a look at it!
And that much of a current (that's really high) is strange for an application like the one I described before, right?
Ok, so there's also a possibility that the specs are not complete. I wanted to ask about that to my professor but he's not famous for being availablethat's why I'm trying to look as much as I can around the internet.
Dear jjx,
- "... and for the electrical stimulation of the CNS ..."
- Ok, so either your DACs are for stimulation or for reference to the ADCs. Nevertheless, 10 uF sounds large in both cases...
- And it might also be that they have an idea of what they need in an electrical way, but not fully sure of how it is implemented.
- And yes, that will fry the nerves quite well and give seizure.
- Here again, remember that response time of the cns is in the order of ms. µs does not make much sense tbh.
- " ... Is there another kind of amplifier that should be used with large cap loads? "
- Miller-compensated opamps are used to move the dominant pole inside the OTA and typically used for low(ish) capacitive loads (unless you add an additional source follower inbetween). Folded cascode would do in your case, I would guess. Beware of resistive load in the electrodes to the nerves.
- " ... using a simple R-2R ladder made with resistors ... "
- R-2R is dead simple to design, so probably an OK choice with the output buffer. You can drive it with current sources and also scale currents to obtain thermometer coding (in case you would have had a linearity requirement on your DAC).
- A so called current-steering DAC does not need a buffer nor suffer from slew rate. But has limited swing and requires resistive termination (for your case).
- "... So having a gain > 4096 would gain up the DAC output, giving a 0-4095 volts scale, right ..."
- Yes. The closed-loop accuracy of the opamp would be 1/(1+1/A). A caveat here is once again that they have not specified DAC linearity and thus requirement might not be that strict to resolute the whole DAC range properly
- my professor but he's not famous for being available
that's
- Haha
That's a very common phenomena among professors
No.Considering a single pole system, my DAC will reach its full output voltage after 5τ, which is the time constant of my system.
Maybe you want your tau to be 0.2ms. (Did not check whether this makes sense). Then this has nothing to do with "1 sample per second". It will be 0.2s independent of sampling rate.Putting into relation the 5 tau and the update rate of a sample for every second, it is clear that 5 tau is equal to 1 ms; so, my tau will be equal to (1/5)ms=0.2ms. From the τ value, I can first find the omega as 1/τ and multiplying by 2pi, the cutoff frequency (and so the BW of my opamp), which will be equal to ~32kHz.
Where does the 1us come from? Why 10uF load? This is a huge load and ususlly can´t be driven to get a settling time of 1us.• Settling time (1%) of analog out = 1us, with capacitive load of 10uF
The value of 1us is given from the specs that the company gave to us. The text that gave to us says:Where does the 1us come from? Why 10uF load? This is a huge load and ususlly can´t be driven to get a settling time of 1us.
No, the bandwidth of you OPAMP is high (and should be), the bandwidth of the filter is 788Hz or 32kHz.
Oh ok, now I understand. But now the question is, how can I derive the other design parameters from the specs they gave to me? I really have no idea now, I thought that considering the 1Ksample/s spec was the right way, but now I'm confused.Maybe you want your tau to be 0.2ms. (Did not check whether this makes sense). Then this has nothing to do with "1 sample per second". It will be 0.2s independent of sampling rate.
fc = 1/(2 * Pi * tau) .. then fc = 788Hz.
In either case 788Hz as well as 32kHz violates nyquist. As long as you are aware of this .. everything is fine.
Ok and after selecting the allowed error, how can you derive a design parameter from it?No.
1) The DAC output will be fast
2) the filter output will be delayed
3) tau is "The time constant"! How can 5 tau be "the time constant of your system"? What does this mean?
Settling time: there always will be an error. You have to define the "allowed" error first to calculate how many tau this takes.
as a rule of thumb:
1tau --> 30%
2tau --> 10%
3tau --> 3%
4tau --> 1%
5tau --> 0.3%
6tau --> 0.1%
Not I did talk about a filter and an OPAMP. I just referred to your informations.When you talk about the filter, are you talking about the filter behaviour of the opamp? Because there's no filter involved in this project, I should only use the R-2R ladder and design an opamp (and use it in an inverting configuration) which will be at the output of the ladder.
Nyquist says that you need more than twice the sampling frequency than the analog frequency of interest.Another thing that's not clear to me is: how are the two freqeuency values violating Nyquist sampling theorem?
The task is not clear to me.Ok and after selecting the allowed error, how can you derive a design parameter from it?
Not I did talk about a filter and an OPAMP. I just referred to your informations.
"...the cutoff frequency (and so the BW of my opamp), which will be equal to ~32kHz."
--> "Cutoff frequency" clearly describes a filter. also: "single pole", "time constant", "tau"
Nyquist says that you need more than twice the sampling frequency than the analog frequency of interest.
Thus 1ksmpl/s results in less than 500Hz analog frequency.
If you violate it just means you get overtones, switching noise, and you can´t get a clean waveform reconstruction.
Again: if you aware of it - and fine with it - there is no problem.
From my experience: (more than 2 decades of professinal indusrial equippment develeopment)
* I never had the requirement for 1us settling time with a 1ksmpl/s sampling rate. There might be applications for this .. but for the kajority of applications it makes not much sense.
The other way round: using an ADC in "undersampling" technique (which also violates nyquist) i sued quite reguralely. But still you need to know what happens - especially that you get "alias frequencies".
The task is not clear to me.
Where does your tau, 5 tau, ans 0.2ms come from? Ist it you or the requirement.
When I see a "settling time of 1us to 1%" then for me it means:
1% error --> 4 tau. 4 tau = 1us. 1 tau = 250ns.
I don´t see any reference to the 1ksmpl/s here. It could be 10 smpl/s or 10k smpl/s and still tau is 250ns.
Yes, it is at least a 1 pole system.Alright, clear. But isn't the opamp I have to design, in this particular case, acting like a single pole system?
No, in no way. I think you are doing a good job.Am I considering everything in the wrong way?
Honestly: I don´t see that the one depends on each the other ... in your case, since it is overruled by the 1us settling time.But how does one use the update/sample rate and analog frequency as a useful info to put it into the opamp design?
Yes, it is at least a 1 pole system.
But usually not that "adjustable" one with a cutoff of 788Hz or 32kHz.
I differ between
* a bare OPAMP (it´s cutoff usually is in the low Hz.. while having high gian)
* a OPAMP_circuit (used to amplify and not as a dedicated filter) it´s cutoff is due to it´s feedback much higher (while the gain is much lower)
* an OPAMP filter circuit (here external parts are used to determine the cutoff frequency)
Now when you say "OPAMP" which one of the above do you mean?
No, in no way. I think you are doing a good job.
And it´s quite normal to have a different view as a student than an enigneer with decades of experience. Nothing wrong with the one or the other.
****
1 tau = 36.8%, 2 tau = 13.5%, 3 tau = 4.98%, 4 tau = 1.83%, 5 tau = 0.67% .. so your values are closer than my "over the thumb".
****
Honestly: I don´t see that the one depends on each the other ... in your case, since it is overruled by the 1us settling time.
It maybe more relates to the digital interface ...
Hopfully others contribute to give you more assistance.
Klaus
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