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Analog Devices quizzes
KE5FX:
I've been trying out the new o1-pro model by feeding it some of the brainteasers from ADI's marketing department. These have the advantage of being made up on the spot or drawn from material that either wouldn't appear verbatim in the training corpus or wouldn't be well-connected to the questions being asked.
Previous models haven't done very well, but o1-pro is showing some improvement. It got the latest question right the first time, but that was an easy one. I didn't think it would handle the previous question as well as it did:
You have to log in to see the answers provided by readers but they largely echoed the reasoning above. Most people got the sign issue right off the bat, but overlooked the input polarity.
Technically the polarity objection is questionable because there's no reason why Vref has to be positive, but... anyway, some reasonably impressive work from HAL on this one, hand-waving about a "non-inverting integrator" aside:
KE5FX:
Another quiz:
Answers highlighted in green (noting that cut/paste mangles the actual math expressions badly):
--- Quote ---Step-by-step Reasoning and Answers
We have a two-stage op-amp circuit as shown. Both op-amps are ideal (infinite gain, infinite input impedance, zero output impedance, and inputs draw no current). Each op-amp is powered from ±12 V rails.
Notation:
Let the source voltage be VSVS (or VSRCVSRC).
The input node where VSVS is applied will be called Node A.
The node at the first op-amp’s inverting input will be Node B.
The output of the first op-amp will be Node C.
The node at the second op-amp’s inverting input will be Node D.
The output of the second op-amp will be Node E.
From the schematic:
Node A is connected to VSVS and to Node B through a 10 kΩ resistor.
Node B is the inverting input of the first op-amp, with the non-inverting input at ground (0 V). A 40 kΩ resistor from C to B provides feedback for the first stage.
The first stage output (Node C) goes through a 10 kΩ resistor to Node D, the inverting input of the second op-amp (also with non-inverting input at ground).
The second stage has a 20 kΩ feedback resistor from E to D.
Finally, from Node E (the second op-amp output), a resistor R connects back to Node A, creating an overall closed-loop configuration.
Ideal Op-Amp Conditions (Linear Mode):
For each op-amp, since the non-inverting input is at 0 V (virtual ground), the inverting input node will also be at approximately 0 V (assuming the op-amp is not saturated). Thus:
Node B ≈ 0 V
Node D ≈ 0 V
1. Input Impedance Seen from the Source VSVS for Given R Values
First, let’s find the relationship between outputs and input.
First Stage Analysis:
At Node B (inverting input of first op-amp):
Voltage at B is 0 V (virtual ground).
Currents into Node B: From A to B through 10 kΩ: IAB=VS−010 kΩ=VS10 kΩIAB=10kΩVS−0=10kΩVS
From C to B through 40 kΩ: ICB=VC−040 kΩ=VC40 kΩICB=40kΩVC−0=40kΩVC
No current enters the op-amp input itself. Thus, KCL at Node B:
10kΩVS+40kΩVC=0.
Multiply through by 40 kΩ:
4VS+VC=0⟹VC=−4VS.
So the first op-amp output is VC=−4VS.
Second Stage Analysis:
At Node D (second op-amp inverting input at 0 V):
Current from C to D through 10 kΩ: ICD=VC−010 kΩ=VC10 kΩ ICD=10kΩVC−0=10kΩVC
Current from E to D through 20 kΩ: IED=VE−020 kΩ=VE20 kΩ IED=20kΩVE−0=20kΩVE
KCL at Node D:
VC10 kΩ+VE20 kΩ=0.
10kΩVC+20kΩVE=0.
Multiply through by 20 kΩ:
2VC+VE=0 ⟹ VE=−2VC.
But VC=−4VSVC=−4VS, so:
VE=−2(−4VS)=8VS.
Therefore, the second op-amp output is VE=8VS.
Overall Node A Equation:
At Node A, we have the source VSVS, a 10 kΩ resistor to Node B (0 V), and the resistor R to Node E (8V_{S}). The source current ISIS supplied by VSVS is:
IS=(current through 10 kΩ)+(current through R)
Current through 10 kΩ from A to B:
I10k=VS−010 kΩ=VS10 kΩ.
I10k=10kΩVS−0=10kΩVS.
Current through R from A to E:
IR=VS−VE R=VS−8VSR=−7VSR.
IR=RVS−VE = RVS−8VS = R−7VS.
Thus:
IS=VS10 kΩ−7VSR.
IS=10kΩVS−R7VS.
Factor out VSVS:
IS=VS(110 kΩ−7R).
IS=VS(10kΩ1−R7).
The input impedance seen by the source is:
Zin=VSIS=1110 kΩ−7R.
Zin=ISVS=10kΩ1−R71.
Simplify the denominator:
110 kΩ−7R=R−70 kΩ10 kΩ⋅R.
10kΩ1−R7=10kΩ⋅RR−70kΩ.
Hence:
Zin=10 kΩ⋅RR−70 kΩ.
Zin=R−70kΩ10kΩ⋅R.
For R=120 kΩR=120kΩ:
Zin=10 000×120 000120 000−70 000=1.2×10950 000=24 000Ω=24 kΩ.
Zin=120000−7000010000×120000=500001.2×109=24000Ω=24kΩ.
For R=30 kΩR=30kΩ:
Zin=10 000×30 00030 000−70 000=300 000 000−40 000=−7 500Ω=−7.5 kΩ.
Zin=30000−7000010000×30000=−40000300000000=−7500Ω=−7.5kΩ.
A negative input impedance means the circuit is actually "feeding back" power into the source under those conditions.
Answer for Question 1:
For R=120 kΩ: Zin=24 kΩ
For R=30 kΩ: Zin=−7.5 kΩ (negative input impedance).
2. Find R to Cancel the Current (I=0) from the Source
For no current from the source:
IS=VS(110 kΩ−7R)=0.
IS=VS(10kΩ1−R7)=0.
For any nonzero VSVS, we must have:
110 kΩ−7R=0 ⟹ 7R=110 kΩ.
10kΩ1−R7=0⟹R7=10kΩ1.
Therefore:
R=7×10kΩ=70kΩ
Answer for Question 2: R=70 kΩ..
3. Voltage Range for VSVS to Keep Op-Amps in Linear Mode
We must ensure neither op-amp saturates. Each op-amp saturates if its output tries to exceed ±12 V (the supply rails).
From first stage: VC=−4VSVC=−4VS.
To avoid saturation:
−12 V<−4VS<+12 V.
−12V<−4VS<+12V.
Divide by -4 (reversing inequalities):
+3 V>VS>−3 V ⟹ −3 V<VS<+3 V.
+3V>VS>−3V⟹−3V<VS<+3V.
From second stage: VE=8VSVE=8VS. To avoid saturation:
−12 V<8VS<+12 V.
−12V<8VS<+12V.
Divide by 8:
−1.5 V<VS<+1.5 V.
−1.5V<VS<+1.5V.
The second stage is more restrictive since ±1.5 V is a smaller range than ±3 V.
Answer for Question 3: To keep both op-amps in linear mode:
−1.5 V<VS<+1.5 V.
4. Outputs of the 2 Op-Amps When VS=+2 V
From our formulas:
VC=−4VS=−4×(+2 V)=−8 V.
VC=−4VS=−4×(+2V)=−8V.
VE=8VS=8×(+2 V)=+16 V.
VE=8VS=8×(+2V)=+16V.
However, the second op-amp cannot produce +16 V since its supply is ±12 V. It will saturate at about +12 V. In reality, the second op-amp output will saturate near +12 V, not reach +16 V. The first op-amp’s output of -8 V is within the ±12 V range, so the first stage is not saturated by itself. The second stage, however, is definitely saturated with VS=+2 VVS=+2V.
Answer for Question 4:
First op-amp output (Node C) ideally would be -8 V.
Second op-amp output (Node E) ideally would be +16 V, but in practice it saturates at about +12 V due to the supply rails.
These results fully answer the given questions.
--- End quote ---
4/4 correct, despite the awkward diagram and question phrasing.
They are charging $200/month for access to this model, and it may end up being worth it. It's demonstrating valid reasoning here, not something obtainable from a "stochastic parrot", a Markov model, a lucky guess, or (for that matter) Google Gemini Pro 1.5, which went 0/4.
tom66:
Certainly very impressive, and somewhat scary. $200 a month is a lot cheaper than my salary! Can it synthesise outputs too, give it a novel problem (one clearly not present in its training data) and get a reasonable output?
AndyC_772:
It's certainly reached the point of being useful, and I fully intend to become an engineer who takes advantage of and uses tools like this, rather than one who is replaced by them. We certainly shouldn't sit idly by and fail to keep up to date on what these AI tools can and can't do - especially when they can already do so much.
A while ago I challenged ChatGPT to tell me the differences between two similar looking, competing ADCs from different manufacturers. It failed miserably, but I do hold out hope that this is an area in which these models will improve.
As it happens I've spent a happy day exchanging C code fragments with ChatGPT and getting it to check for things like arithmetic overflows, division by zero and the like. It's been helpful, and I must confess to being quite chuffed at being told at one point:
--- Quote ---Benefits:
Avoids undefined behavior (out-of-bounds access).
Keeps the logic compact and readable.
Works seamlessly without requiring any additional checks elsewhere.
✅ I fully recommend this implementation. It’s clean, efficient, and robust.
--- End quote ---
KE5FX:
Yep, if fishing for compliments from a robot doesn't feel like the future, I don't know what does. :-DD
--- Quote from: tom66 on December 17, 2024, 05:49:55 pm ---Certainly very impressive, and somewhat scary. $200 a month is a lot cheaper than my salary! Can it synthesise outputs too, give it a novel problem (one clearly not present in its training data) and get a reasonable output?
--- End quote ---
This qualifies as a novel problem, I think, given that the idea was to act as a brainteaser, and given that it can't possibly be in the training corpus. If you read the solution threads, they are full of responses from people who are (by appearances anyway) working EEs or at least EE students. Not all of them get the answer right, and some of them get pretty argumentative.
One excellent benchmark is the geometry quiz from last May. Most respondents got it wrong, and some got downright pissy about it. Previous OpenAI models didn't come close to getting it, and o1-pro didn't get it either, but it used the same flawed reasoning as some of the humans to get the same wrong answers. If I argue with it enough, I can convince it of the correct answer. I wouldn't be surprised if the next major release has no trouble with it.
IMHO even the $200/month model is still a work in progress, and we are still a long way from being able to trust these models implicitly. But you can treat it like an intern or junior employee who is very good at design planning, sanity checks and general brainstorming.
I used to double-check ChatGPT results against Google Gemini, but there's no point doing that with o1-pro. It has no peers right now.
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