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Given an N-bit R2R DAC, with input voltages of e.g. 0-3.3 volts, what is the the best way (signal integrity and quality maintained) to attenuate its output voltage to a range e.g. 0-.7 volts for a VGA signal.

Is it as simple as using a voltage divider, or would something like an opamp be more appropriate?

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    \$\begingroup\$ depends... how much drive strength do you need? what is the load impedance? \$\endgroup\$
    – user16222
    Commented Jun 19, 2018 at 17:41
  • \$\begingroup\$ VGA would be 75 ohms, unless someone forgot to terminate it. Essentially, one either designs the ladder for the load impedance, or you need a suitable amplifier (video amplifier is a category) with less than unity gain or with an output divider including the load impedance. Also, given likely AGC it may not matter, as long as electrical maximums are not exceeded. \$\endgroup\$
    – Chris Stratton
    Commented Jun 19, 2018 at 17:44
  • \$\begingroup\$ @jonRB current draw would never exceed 40 mA, and load impedence would be 75 ohms. For the purpose of making this question/answer more useful for everyone, I would be interested in the answer for high vs low current draws and high vs low load impedances. \$\endgroup\$
    – Connor Spangler
    Commented Jun 19, 2018 at 17:45
  • \$\begingroup\$ @ChrisStratton even if designing for the load impedance, how would you factor in the specific output voltage? I understand how to design an R2R DAC with a max output equivalent to the input level, but not with a specific lower gain. \$\endgroup\$
    – Connor Spangler
    Commented Jun 19, 2018 at 17:48
  • \$\begingroup\$ @Answoquest - Conceptually one way you could do it would be to design your DAC with some additional high order bits that you only ever drive as zero - ie, go from 8 to 10 bits and now your maximum is 1/4 of the previous voltage. Then you could simplify the redundant components into a smaller number. I believe if you study the result you'll be able to see how you could then use different values to create a gain fraction which is not a power of two, but I haven't actually done it on the back of an envelope. \$\endgroup\$
    – Chris Stratton
    Commented Jun 19, 2018 at 17:51

1 Answer 1

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The R-2R ladder has a fixed output impedance of R. If you shunt it with a resistor Rx you can attenuate the output equally for all codes. The output impedance will obviously be R||Rx.

You thus have two degrees of freedom that allow you to set the output impedance and output amplitude independently.

In this case, if you set R = 353 ohms and Rx = 95.2 ohms you will get the desired characteristics (full scale output voltage of 0.7V with 3.3V supply and 75 ohms output Z).

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  • \$\begingroup\$ Could you please clarify R||Rx? \$\endgroup\$
    – Batperson
    Commented Jun 19, 2018 at 21:52
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    \$\begingroup\$ The parallel combination of R and Rx = R*Rx/(R+Rx) \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jun 19, 2018 at 22:26
  • \$\begingroup\$ That makes sense, thanks! I'm having trouble putting together equations to calculate values for R and Rx... We know (R*Rx)*(R+Rx) = 75, and Vout_max = 3.3*(7/8) = 2.89 V. We need Vout_max = 0.70 V. We have a system of two unknowns here, so do I just have to trial and error values? \$\endgroup\$
    – Connor Spangler
    Commented Jun 19, 2018 at 22:51
  • \$\begingroup\$ For clarification, this would be for a 3-bit DAC specifically, but I'm also interested in developing an equation (set) for n-bit DACs. \$\endgroup\$
    – Connor Spangler
    Commented Jun 19, 2018 at 23:25
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    \$\begingroup\$ You should solve the system of two equations. The number of bits doesn't matter except that you need to adjust it a little if you care about the difference of one LSB. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jun 19, 2018 at 23:40

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