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I have a device I'm building that uses a single high-precision current source (driven by a DAC for various arbitrary waveforms) to drive one of 128 channels. The current level will be on the order of 20uA

Building 128 redundant current sources is expensive and unnecessary, and a typical 74595/596 shift register won't work because I need to "pass through" the current source to the channel.

Current idea is to use 128 triacs switched by either an fpga (with many pins) or daisy chained 595's and a micro, connected to the common current source. Only one will be open at a time. As the current source and the controlling device are all off the same reference I don't believe I will need to use optoisolators although that will be determined when prototyping

I have 3 questions.

  1. Is there a standard way to do this that I'm unaware of
  2. Will using triacs with such a low current source cause issues with noise or switching
  3. Are there low power triacs in small form factors (preferably smd)
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  • \$\begingroup\$ What sort of voltages are you dealing with? Is there any reason a "tree" of standard CMOS analog mux/demux chips couldn't do the job? \$\endgroup\$
    – Dave Tweed
    Commented Jun 7, 2013 at 3:34
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    \$\begingroup\$ The "old standby" analog switches seem to have leakage currents in the uA range, probably too high for this application. But I found 1:8 analog mux parts at Maxim and Linear with leakage in the 1 - 10 nA range---is there any reason not to use a tree of those? \$\endgroup\$
    – The Photon
    Commented Jun 7, 2013 at 3:41
  • \$\begingroup\$ Is there a DC level associated with the signal you are injecting? What value is it? Is it important to have this DC level maintained i.e. does the current source generate a precision dc level for the "channel"? What is the peak to peak AC level superimposed on the DC level? What is the input impedance characteristics of the channel? \$\endgroup\$
    – Andy aka
    Commented Jun 7, 2013 at 9:16
  • \$\begingroup\$ If it's really a current source, the on-resistance of the analog switch is irrelevant, only the leakages. Interesting spec. \$\endgroup\$
    – markrages
    Commented Jun 7, 2013 at 17:17
  • \$\begingroup\$ DC level is ideally zero or close to it, the current source will just be DAC with a mosfet and opamp similar to the one described in the AD circuit note analog.com/en/circuits-from-the-lab/cn0151/vc.html . I don't have the expected voltage levels available yet, but I suspect they are on the order of a few volts. The current will be used for impedance measurements across biological tissues and will be recorded and processed on the other end, so noise should be manageable, but moderate levels can be eliminated in signal processing on the other end. \$\endgroup\$
    – crasic
    Commented Jun 8, 2013 at 22:04

2 Answers 2

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How about something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Essentially a Pass Gate (aka Transmission Gate, aka an Analog Switch) per channel. That's an NMOS and a PMOS transistor and you'd want to ensure the were complements of one another, and you'd need 256 control signals, 128 NMOS transistors, and 128 PMOS transistors, but I imagine this would probably be your lowest cost approach if you really need to do what you've described. You'd obviously need to incorporate some sort of FGPA or CPLD or MCU with a lot of I/O expansion to get the control signals, and a custom PCB of course.

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  • \$\begingroup\$ If the current is unidirectional, only one gender of transistoris required. \$\endgroup\$
    – markrages
    Commented Jun 7, 2013 at 2:42
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I2C or SPI based GPIO chips can expand the I/O capability of an arbitrary MCU a whole lot. If you can bit bang the protocol on a number of pins, you can even do this in parallel. The question is then how to mux a weak current source without introducing noise. A handful of MCP23017 can provide logic-level switch signals (16 each.) Use a low-noise bipolar transistor or P-channel MOSFET per switched output, and I think you're good.

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