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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Regulator Question Oct 25^th, 2009, 9:23pm hi, when i was reading a novel regulator called "replica compensated supply regulator" i got the following question. in the figure i attached if op amps have different offsets then whats the output? lets assume A1 has 10mv offsets and A2 has 20mv... Thanks, Rajasekhar.
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Berti Community Fellow Offline Posts: 356	Re: Regulator Question Reply #1 - Oct 26^th, 2009, 4:35am Your figure shows two unity-gain buffers driving a common output. Can you please give some more details how this "novel" supply regulator should work? Can you please give more details on the paper(?) where this circuit is proposed? Regards
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HdrChopper Community Fellow Offline Posts: 493	Re: Regulator Question Reply #2 - Oct 26^th, 2009, 4:58am Yes, please. Could provide more details on this circuit? I figure out the final output voltage would be 15mv assuming exactly the same output impedance for both opamps. Most probably the loop gain will significantly drop in order to achieve such operating condition. Best Tosei
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Mayank

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Re: Regulator Question
Reply #3 - Oct 26^th, 2009, 6:43am

Hi ,
Isn't this circuit like connecting 700mV to o/p through two identical Unity-Gain Buffer Paths.....Hence, making it similar to a circuit with 700mV connected to o/p through a Single Unity-Gain Buffer But DOUBLE the Drive of the above used opamp

Hence if two opamps are exactly identical in all other respects except for different offsets, I think it should settle at avg off (700-10) & (700-20) ie 685 mV

Now, the question is, Am I Correct ?? Tongue

thanx,
Mayank.

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raja.cedt

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Re: Regulator Question
Reply #4 - Oct 26^th, 2009, 7:37am

hi,
first thanks for your reply guyes. you can see fig7 in the following paper.
Actually i desinged this circuit in office, but that time i didn't get this doubt. so i thought and for this will come through systematic offset. lets say if both are same then both has +5mv,_5mv systematic offset.hemce 685mv(any one agree?).
lets if A1 is faster than A2 then what might be the ans.

Please download pap from here http://www.sendspace.com/file/l3ev7n

Sorry for this complicated question

Thanks,
rajasekhar.

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buddypoor Community Fellow Offline Posts: 529 Bremen, Germany	Re: Regulator Question Reply #5 - Oct 26^th, 2009, 8:21am Hi rajasekhar, may I jump into the discussion ? In Fig. 7 of your paper I see - Two opamp outputs which are ADDED , and - Two OTA outputs which are connected, but no circuits in which two opamp outputs are tied together ! Or am I wrong ? Regards
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loose-electron

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Re: Regulator Question
Reply #6 - Oct 26^th, 2009, 11:56am

The circuit does not make sense at the level of analysis shown.

Two ideal op-amps (note the word "IDEAL" here) can not be connected together at the output.

If they are OTA's thats a different story.

If they are non-ideal amplifiers then it can be analyzed at the transistor level to get an answer.

This reminds me of the problem where you connect a logic inverters input to its output. At that level of abstraction it can not be analyzed. You need to go down one level, to the transistor level, and then it can be easily figured out.

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buddypoor Community Fellow Offline Posts: 529 Bremen, Germany	Re: Regulator Question Reply #7 - Oct 26^th, 2009, 3:23pm Yes, I agree with the above arguments. The circuit with opamps makes no sense. Probably, rajasekhar has mixed opamps with OTAs.
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HdrChopper

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Re: Regulator Question
Reply #8 - Oct 26^th, 2009, 5:32pm

loose-electron wrote on Oct 26^th, 2009, 11:56am:

The circuit does not make sense at the level of analysis shown.

Two ideal op-amps (note the word "IDEAL" here) can not be connected together at the output.

Hi Jerry,

I mostly agree with you. However, if we starting point are two (in principle) opamps with different input referred offset, then those two opamps have to be different (since their op point will not be the same) regardless the used topology.

Again assuming non-ideal opamps and that the output/input impedances are not very different and finite (they will due to the different internal unbalances) a first order approx will be that the overall offset will be about +15m for one of them and -15 for the other one.
If those impedances are significantly different the offset will be split according to such "voltage divider" set by the output impedances.

By my point is that from the very first moment it is stated one amp has 10m of input referred offset while the second one has 20mv, then the opamps are not ideal and in principle they could be connected as shown.

Best
Tosei

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loose-electron

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Re: Regulator Question
Reply #9 - Oct 26^th, 2009, 9:13pm

Assumptions are dangerous things to make in the engineering business. �I agree that if assumptions are mad you can get to all kinds of interesting conclusions. I think the original question needs to be �re-stated with a bit more detail and perhaps at one level deeper in detail to get anything meaningful.

Just for you entertainment - Take a look at the attached schematics. Neither, in my opinion are "well enough defined" to determine whats going on. Two ideal voltage sources can't connect to each other. As well, an inverter, fed back on itself, needs to be defined one level deeper in order to really know whats going on.

There are other examples.

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raja.cedt

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Re: Regulator Question
Reply #10 - Oct 26^th, 2009, 9:56pm

hi,
thanks for your inputs and sorry for for the name opamp instead of OTA. Now can any one assume it is OTA and tell me the ans.

@Tosei: hi i didn't understand that impedance division concept? could you please explain more

@loose-electron: thanks for your entertainment ckts, regarding 2nd one i don't know ans, every body saying that there is no solution for that. whats your ans?

Thanks,
rajasekhar.

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Mayank

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Re: Regulator Question
Reply #11 - Oct 26^th, 2009, 9:57pm

OTA wire Anding(or o/p short) makes sense because currents can get added up at a node...Two Ideal Voltage Sources on the other hand if connected in parallel will make no sense, until and unless they are non-ideal in which case as buddy said, you can analyse them through their internal resistances (in this case --> o/p res of the opamps) , & as raja said they were identical in all respects except for offsets, 685mV should be the case.

But anyways , i dont see any point how this replica scheme acts as voltage regulator.

Jerry, Just out of curiosity, If you connect an Inverter's o/p to i/p node & as you said, going one level deeper & Taking into account some propagation delay tau, It will still oscillate right ? [Although i think the waveform will be mostly triangular, coz it wont hv much time to saturate and maintain rail voltages for long] Correct ??

regards,
Mayank.

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Regulator Question Reply #12 - Oct 26^th, 2009, 10:02pm hi mayank, it's OTA not Op amp i am sorry for that, inverter connected back to back don't oscillate because at max you will get 90 deg in the loop or otherwise if you think transistor equivalent it will stuck at the trip point of the inverter. Thanks, Rajasekhar.
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Mayank

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Re: Regulator Question
Reply #13 - Oct 26^th, 2009, 10:08pm

hi raja,
Yeah, i was confused that it will get stuck at the trip point itself...But can you explain why it will get stuck at trip point when it doesnt in the case of 3 inverter chain ?? I mean, there's always a possibility, but like in simulations we give some initial conditions, what if we give some initial condition here also ??

Also, why only 90 degrees phsae shift ?? it's an inverter -- a class AB common source -- should give 180 deg phase shift ??

regards,
Mayank.

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raja.cedt

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Re: Regulator Question
Reply #14 - Oct 26^th, 2009, 10:27pm

hi Mayank,
in case of single stage inverter you will have single cap at the output and there are no other caps so how can you expect more than 90 deg you can refer razaavi ring oscillator section for more clarification and mean while you told it's a class AB stage thats why it has to give 180 deg, i didn't understand this.
in case of single inverter you will have one operating point , where as in case of 3 stage oscillator you are correct there is a chance of staying at the trip point but it has enough loop gain, so any noise in the ckt will get you the right output.

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