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I have two outputs of a voltage source that do ±10 V with 16 bit. This leads to a theoretical resolution of a bit step of 0.3 mV. My goal is to have more resolution by using a voltage divider of 100 on the second output and combining the two. This would give ±10 V with 3 μV theoretical resolution. This voltage will be used to charge up a capacitance <10pF. The adder will be nicely shielded and will be sitting at a temperature of 4 Kelvin.

The only problem is that I am seeking is adding the two voltages WITHOUT using active components, as transistors add noise and there is only a thermal budget of about 0.1 mW (less is better).

The whole passive network doesn't have to be perfect at all, though. The only object is to have more voltage resolution in the end.

I played around with resistors in LTspice but didn't succeed yet. I would be happy to get some new ideas.

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    \$\begingroup\$ 16 bit: is that the effective number of bits (ENOB), or what is the actual resolution, at the temperature the DAC operates at (which probably is not 4K); assuming the DAC itself sits at room temperature, it's almost certain its output will be noisy; Johnson-Nyquist noise is pretty unforgiven for anything that's got non-zero impedance. \$\endgroup\$ Aug 19 at 10:09
  • \$\begingroup\$ you are right of cause, I am writing "theoretical resolution" because of that. Have to see where the noise level is. The DAC sits at room temperature. However all cabling at this temperature is very short and the lines are heavily filtered and thermalized at 4K. \$\endgroup\$
    – Daniel
    Aug 19 at 11:39
  • \$\begingroup\$ the noise of the cabling is nearly irrelevant for typical high-bitdepth DACs, as the source impedance is what gets multiplied with the temperature and Boltzman constant. \$\endgroup\$ Aug 19 at 12:14
  • \$\begingroup\$ anyway, adding is just subtracting with a different sign, so consider the voltage divider simply a ratiometric adder. \$\endgroup\$ Aug 19 at 12:15
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    \$\begingroup\$ What bandwidth you need and what sampling rate you have? How about using high sampling rate with dithering and low-pass filter the output? \$\endgroup\$
    – Justme
    Aug 19 at 15:33

3 Answers 3

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It's not possible to exactly do addition with voltages sources as you describe, but you can do a weighted average, which should fulfill your requirement.

schematic

simulate this circuit – Schematic created using CircuitLab

Vout = (V1/R1 + V2/R2) * (R1 || R2)

If the resistors are equal, we have an output of (V1 + V2)/2.

If the ratio is, say, 10,000:1 then we have

Vout = (V1 + V2/10000)/1.0001

The higher the ratio, the less effect resistor tolerances have (and one would expect some change in resistor values from 300K to 4K temperature). A high ratio will also minimize the power dissipation in the resistors for a given output resistance, which is:

\$P_D = \frac{(V1-V2)^2}{R1+R2}\$.

Output resistance, of course, is just R1||R2 = \$\frac{R_1 \cdot R_2}{R_1 + R_2}\$

To the resistor tolerances you also have to add the electrical resistance of the wires that enter the cryostat. Typically they are relatively high resistance to minimize heat loss (since electrical resistance generally follows thermal resistance). That's going to result in some uncertainty and perhaps variability in the ratio, especially wrt the lower resistance. If you can add a wire or two and do force/sense on the lower resistance ADC you can eliminate that effect.

You might want to calculate the actual J-N noise of the combining resistors over the BW that actually makes it into the cryostat before doing this. Compared to the ADC output noise. It could well be that combining them at room temperature is better, say with a 10 ohm resistor and a 100k ohm resistor.

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  • \$\begingroup\$ thanks for the solution - life can be so simple. I'll check how good the DACs are and if dividing at roomtemperature makes sense. \$\endgroup\$
    – Daniel
    Aug 19 at 12:59
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    \$\begingroup\$ @Daniel btw, this is exactly the solution from my answer. I'm thankful Spehro wrote it out so nicely! \$\endgroup\$ Aug 19 at 13:24
  • \$\begingroup\$ Of course, this unbuffered solution means that one must pay close attention to how it's loaded. \$\endgroup\$ Aug 19 at 13:54
  • \$\begingroup\$ as far as I see the J-N voltage noise is proportional to \$\sqrt{ T \cdot R} \$. Could you please explain how a reduction of a factor of 10 in R is sufficient? I suppose it's 100. \$\endgroup\$
    – Daniel
    Aug 19 at 14:58
  • \$\begingroup\$ @Daniel Yes, you're right. \$\endgroup\$ Aug 19 at 15:03
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The solution is to not add, but divide the difference between the two voltages; instead of dividing the second DAC down, you could just do the voltage divider between the two outputs.

Another note: this sounds like your DACs are rather high-grade, so chances are you're operating them with an external voltage reference. Your second DAC could generate the voltage reference for the first!

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  • \$\begingroup\$ Unfortunatly, the DACs have the same reference. \$\endgroup\$
    – Daniel
    Aug 19 at 13:15
  • \$\begingroup\$ excellent, so you disconnect one and use the other as reference of that. Makes much more sense than trying to add voltages. \$\endgroup\$ Aug 19 at 13:23
  • \$\begingroup\$ sorry I meant to say that they cannot be disconnected as they are in the same device. \$\endgroup\$
    – Daniel
    Aug 19 at 13:30
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0.1 mW and 10 V means using an R of over 1 MΩ. To combine the signals with 100:1 weights, connect the 'main' DAC to the output with 10 kΩ and the 'fine' DAC to the same output with 1 MΩ. Connect the ground of each DAC together.

Note you will have higher resolution, but not higher accuracy ==> using 100:1 will mean that there will be overlap between the settings for a desired output.

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