[time-nuts] DAC resistors

Lux, Jim (337C) james.p.lux at jpl.nasa.gov
Tue Oct 6 22:29:01 UTC 2009


> -----Original Message-----
> From: time-nuts-bounces at febo.com [mailto:time-nuts-bounces at febo.com] On Behalf Of Neville Michie
> Sent: Tuesday, October 06, 2009 3:19 PM
> To: Discussion of precise time and frequency measurement
> Subject: [time-nuts] DAC resistors
> 
> Hi,
> I am constructing a phase meter to monitor the phase creep of clocks.
> It consists of a BCD counter counting say microseconds that has its
> count strobed into a latch by a pulse from the clock.
> The Latch drives a DAC which drives a pen recorder and an analogue
> data logger.
> Now I am familiar with R - 2R networks, and that method is used on
> each decade
> but the resistors that combine the decades in a 10:1 ratio are the
> problem.
> I have an approximate value, and I will probably have to trim them to
> eliminate
> digital errors later.
> But I can not find a reference anywhere to how to calculate the correct
> resistors or even a working example except for an old Analog Devices
> data sheet which seems to use a different structure, by reducing the
> supply voltage of each decade.


Do you have a summing amplifier?  What about resistive dividers in a 1:10:100 ratio..

Say your r/2r network is designed to dump into a 1K ohm load to ground?  If you replace the 1K with 1K in parallel with 10K, the current through the 10K will be 1/10th that into the 1K (actually you need to pick 1.1K and 11K or whatever works out to give you 1K in parallel).  The 1K goes to ground, the 10K goes into the current summing junction of your opamp (which is at virtual ground)





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