I created a I/V stage with amplifier for my DAC project. An excerpt of the circuit can be seen in the attached picture in the "Device under test" part. The DAC has a 3.9mA pp current output, which I convert to voltage using a high precision resistor (R302).

I want to draw an approximate bode plot of the circuit using my function generator, whose output is a voltage. I have studied some literature, and I know that using opamps and transistors is possible to create very precise voltage controlled current sources, but before proceeding with building another device, I tried to find a easier solution by myself.

What I need is a current which is more or less the same my DAC will give as output (circa 4mA). My idea was to just use the Ohm law V = RI with the intuition (which may be wrong, since I am quite a beginner) that if I put the low input resistor in series with a much bigger one (R301 in the schematic), the input current will be dictated almost only by the latter. In other words, the small resistor almost disappears when put in series with a much bigger one. This way, I could generate the needed current by just putting this big resistor in series with the input of my circuit, and trimming the output voltage of my functon generator.

Could it be a good way to go? enter image description here


This is a pretty bad idea. How bad, depends on the DAC itself, and how its current output linearity varies with output voltage. Really, the I-V stage is the small resistor itself. Too large, and the DAC output voltage is large enough to affect linearity. Too small, and the resulting voltage will be swamped by noise.

As the DAC is the main unknown here, testing the I-V stage with a function generator won't tell you very much useful in terms of audio performance.

Better would be to use a current-input amplifier (like the traditional opamp configuration) but I gather you don,t really want to use an opamp in this project.

Fortunately there is an alternative : the common-base or common-gate amplifier has a very low input impedance (that is, the input voltage is essentially independent of input current) and high voltage gain : the ADC output current is developed in full across the anode resistor. So use that - in its vacuum tube form, it's called "grounded grid".

Easiest way would be to set up the vacuum tube stage as a normal self-biased stage (cathode resistor to ground) but with the grid literally grounded (hence the name) and AC couple the DAC output into the cathode.

(There are other tricks you could play if you can arrange the right DC conditions, e.g. if the DAC output was referenced to 2.5V and you needed a grid voltage of -2.5V wrt cathode, you could get things to work quite nicely. But I'd start with AC coupling into the cathode as a proof of concept.

  • \$\begingroup\$ The idea came to my mind primarily because the DAC datasheet reports the following text: "current output is equivalent to a differential (+ or -) voltage source in series with a 781.25Ω resistor". Because of my lack of experience I am not fully able to understand all the concepts you've written. I will have to re-read your answer with some books in front of me... Just for my better understanding, are you suggesting to use a common-base amplifier to generate the test current? \$\endgroup\$ – Enrico May 3 '18 at 10:32
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    \$\begingroup\$ No, I'm suggesting you use a grounded grid amplifier (= common base only cooler ... well. actually hotter) as your I-V conversion stage. (Your approach to generating a test signal is reasonably OK, it's just that it's not IMO a very useful test) \$\endgroup\$ – Brian Drummond May 3 '18 at 12:15
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    \$\begingroup\$ Ok now I understand. So instead of the 10Ohm resistor, If I understand well? \$\endgroup\$ – Enrico May 3 '18 at 12:16
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    \$\begingroup\$ Right ... the 10 ohm resistor will kind of work, but it's not he best approach unless you only care about simplicity not quality. \$\endgroup\$ – Brian Drummond May 3 '18 at 12:20

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