i have implemented DAC controlled DC voltage source. I'm having trouble understanding physical reasons for voltage error due to load resistance change.

My circuit desing was taken from DAC manufacturer's datasheet ( http://www.ti.com/lit/ds/symlink/dac8811.pdf )

Here is schematics i use in my design VDAC schematics (The load is grounded and connected to Vout pin of the amplifier) This scheme works fine with arbitrary high load resistance values (i have tested it particularly on 1 MOhm load) but when load resistances get lower (50 or 75 Ohm) i get some transfer coefficient, which is diffrent for diffrent load.

Transfer coeff graph

So my question is - what are the reasons for this transfer coefficient to arise? And is there any ways i can change this schematic in order for it to work fine for any load resistance value?

  • \$\begingroup\$ How are you powering the output opamp? If the voltage is high maybe you see a offset because of self heating. 2 V across 50 ohm is 40 mA, times 12-2=10 it is 400 mW. Within spec, but enough to heat it a bit. \$\endgroup\$ – Vladimir Cravero Oct 13 '17 at 9:36
  • \$\begingroup\$ I'm powering my amplifier from +-10 V supply rails. But i can see your point, i've noticed that opamp chip heats quite a bit while working, but i didn't think this could cause such an error. I will try to test it. \$\endgroup\$ – Joker Oct 13 '17 at 9:48

The output resistance of your circuit is non-zero. This is what is causing your error. Notice, for instance, that with 50 ohm load and a set voltage of 500 mV, you have 10 mV error. With a 75 ohm load, you need an output voltage of 750 mV (approximately) to get the same output error of 10 mV. A coincidence? I think not.

Let us consider the circuit as a resistor ladder, with the set voltage as an ideal source, and the output resistance of the circuit as Rout. Now the error you measure should be the voltage drop over this resistor. If we measure the error voltage for different values of Rload, we find the curves match yours!


simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks for your answer! I have considered this aswell. I've tryed measuring this output resistance from transfer coefficient i've got and resistor divider equation, but i seem to get diffrent output resistance values for diffrent loads. Can that be the case (i mean does opamp output resistance vary with load) or it should stay constant for any load? \$\endgroup\$ – Joker Oct 13 '17 at 11:23
  • \$\begingroup\$ @JorenVaes: The feedback is connected to the same point as the load. You can add 10 Ω between the op-amp output and the feedback node and the op-amp will compensate. I don't think your answer is correct. \$\endgroup\$ – Transistor Oct 13 '17 at 17:27
  • \$\begingroup\$ Feedback lowers (or increases, in the case of a current-more output) the output resistance, but does not make it zero. Ofcourse, you can add 10 ohm and with will compensate, but it will not fully. There is also still the trace length and connection to whatever the OP uses as load. Contact resistance etc... \$\endgroup\$ – Joren Vaes Oct 13 '17 at 19:49

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