This is my first proper circuit I've designed and I'm after some advice.

I need a 16 bit DAC, output 0-5V and I need to measure the resistance of a thermistor to 0.1mk accuracy. At the moment, I am using this circuit (left hand component is the AD5667R (DAC) in a MSOP package and the right is a ADS1252U (24 Bit ADC) in a SOIC 8 package): Circuit I am currently using,

However, I have read that a Wheatstone bridge is better than my design, why? How? What problems might I encounter?: Better Circuit

I would like some advice on the whole circuit. Ideally I need to replace the AD5667R with a chip with a larger package, SOIC 8 or larger, because it needs to be hand soldered by undergraduate physicists. I'm hoping some alternative design ideas might avoid the need to use that particular part.

The circuit is part of a high accuracy temperature stabilisation unit based on an Arduino, designed for optical clocks, if anyone is interested in this I've placed a link to a report I wrote on the first version in a code block below.

  • 1
    \$\begingroup\$ Even SOIC packages are going to be tough for people with limited soldering experience. What are you trying to accomplish? \$\endgroup\$ – Matt Young Aug 6 '15 at 12:26
  • \$\begingroup\$ Agreed, but with some practice they seem to manage. Whereas 0.65 pin pitch seems insurmountable for some. The dac is used for output to a peltier \$\endgroup\$ – Aaron Aug 6 '15 at 12:32
  • \$\begingroup\$ Have not read this carefully, but I don't see the need for the DAC at all. If your set temperature is fixed then the optimal value of Rt is the thermistor resistance and you can just use the ADC. It essentially is a Wheatstone bridge, just half of it is virtual (inside the ADC). \$\endgroup\$ – Spehro Pefhany Aug 6 '15 at 12:32
  • \$\begingroup\$ @SpehroPefhany, not quite. If handled optimally, much of the noise on the Wheatstone bridge will be common mode. I'd have to do the math, buy I suspect there might be better power supply rejection on the bridge, as well. \$\endgroup\$ – Scott Seidman Aug 6 '15 at 12:42
  • \$\begingroup\$ I like the question. I suggest it NOT be migrated, if such a decision comes up. \$\endgroup\$ – Scott Seidman Aug 6 '15 at 12:45

Okay, it looks to me like you're just using the DAC as a voltage reference as far as the front end goes. You're exciting the half-bridge with the (noisy) power supply. Yes, this is sub-optimal.

It would be better to have something like a Wheatstone bridge, but also to excite the bridge with a reference voltage rather than the power supply. Consider the below circuit. The 3.2:1 amplifier is adapted from the recommended reference circuit for the ADC. If you change the amplifier part number, it may not be happy with C2 so other changes may be necessary.


simulate this circuit – Schematic created using CircuitLab

You might want to buffer the outputs marked "To ADC" with a dual zero-drift op-amp**. If you are concerned about noise, the op-amps could be configured as a fully differential MFB filter, but that's probably not necessary since the signal levels are pretty high with a thermistor. You do have large currents floating around with the output driver, so layout is very important.

In particular keep the grounds for the R2 and Rth together with each other and well away from any high current circuitry.

You could use a network for R1 and R2, there are some nice parts available that track very well. You are looking for ~5ppm stability which is likely with two 0.1% resistors off the same reel, but you can have it guaranteed for more money. Of course the reference resistor is the most critical since it is being compared with the thermistor so the absolute resistance stability matters rather than just the ratio, so put your money there first. Even better, put it in the controlled area so temperature variations don't matter so much (but stability still does).

** The delta-sigma converter you are using does not have built-in buffer amplifiers and the input impedances (common mode and differential) are relatively low and vary with clock rate. See SBAA086 for more information.

  • 1
    \$\begingroup\$ If you're going to all the trouble of using an op amp to generate an excitation voltage, you may as well servo it like in figure 2.15 of analog.com/media/en/training-seminars/design-handbooks/… \$\endgroup\$ – Scott Seidman Aug 6 '15 at 15:31
  • \$\begingroup\$ @ScottSeidman Certainly could add two more op-amps and do that. 5ppm (my estimate) is about 50m\$\Omega\$ differential in the leadwires which is quite a bit unless connectors or really long wires at different temperatures are involved. Heck, it's a lab thing, why not! \$\endgroup\$ – Spehro Pefhany Aug 6 '15 at 15:48
  • \$\begingroup\$ Just one op amp would do the job, if you don't need the gain. \$\endgroup\$ – Scott Seidman Aug 6 '15 at 15:53
  • \$\begingroup\$ @ScottSeidman Not seeing that. The ADC should have the op-amp I showed, and it can't be shared if you want to do the servo thing, so I think you need one for each 'force' lead of the resistor or half bridge or bridge (and likely buffers for the 'sense' leads as I mentioned). \$\endgroup\$ – Spehro Pefhany Aug 6 '15 at 16:09

You're using a voltage divider. The Wheatstone bridge is simply TWO voltage dividers in parallel, with the non-sensor side just providing a baseline.

This provides two main advantages. First, the Wheatstone bridge facilitates a differential voltage measurement, as opposed to a single-ended measurement. If you're careful with your wiring, this means that much of the noise on the positive and negative ADC inputs will be common mode noise, and the common-mode rejection of the ADC can attenuate it. Second, The differential nature of the input signal of the Wheatstone Bridge makes it easier to amplify (either with ADC options, if they are available, or with an external amplifier), helping you to more easily use the whole dynamic range of your ADC.

As a fine point, the bridge provides you with the opportunity to servo-drive the excitation voltage if you use the right kind of amplifier, which can really help when there are long cables between your sensor and your circuit. The mechanism is shown in figure 2.15 of http://www.analog.com/media/en/training-seminars/design-handbooks/49470200sscsect2.PDF, and works on the principle that there is essentially zero current in the sense line, so it makes no difference how long the cable is.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.