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I want to use this 16-channel analog multiplexer/demultiplexer (Datasheet: 4067) to pass the output current of a stimulation circuit through different channels. The output current is a square wave with a maximum of 3mA and 20KHz. I looked through the datasheet and I can't find a maximum current.

Does this mean I can pass whatever current form I want through the 16 outputs? Should I treat this IC simply as a few ideal switches?

Also, I assume the maximum output voltage would be the supply voltage. Am I right?

If there is anything important that I haven't taken into consideration, I'd appreciate it if you point it out.

EDIT: I need to both source and sink current, with voltage varying from -3V to 3V. I'm not sure if this IC works with double supply. Should I switch to another IC? Can I pass the current in both directions?

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  • \$\begingroup\$ Yes, it passes current in both directions - otherwise it would be a multiplexer or a demultiplexer, but not both. And although it doesn't use a split supply as such, in your application just use a negative supply for GND and positive for Vcc, making sure not to exceed the 10V maximum total. You may need to level shift the control signals. \$\endgroup\$ – Finbarr Sep 1 '19 at 9:05
  • \$\begingroup\$ what amount of cleanliness will you need? suppose there is 0.1 volt of trash on the power rails, and 0.05 of that trash is injected on the signal you are passing thru? \$\endgroup\$ – analogsystemsrf Sep 1 '19 at 16:35
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I looked through the datasheet and I can't find a maximum current.

Look again. In particular, look at section 8 ("Limiting Values"). The absolute maximum switched current is 25 mA.

Also, I assume the maximum output voltage would be the supply voltage. Am I right?

Correct. The input should stay between GND and VCC. Higher/lower inputs will be clamped to the rails, but you must not drive more than 20 mA through those clamping diodes. (If there is any possibility of an input outside the rails, use a series resistor to limit that current.)

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  • \$\begingroup\$ I didn't realize that's what switched current meant. In the same row it says "Switch current V SW = -0.5 V to V CC + 0.5 V". It looks like I can't use it with double supply. I need negative voltages at the output as well. Should I switch to another IC? \$\endgroup\$ – Mah Sep 1 '19 at 8:01
  • \$\begingroup\$ Please update the question with all the details(Voltage levels, swing etc), so that the answers will help you and also give justice for the time spent by experts in answering \$\endgroup\$ – User323693 Sep 1 '19 at 8:12
  • \$\begingroup\$ @Umar Please let me know if there's any more extra info that I need to provide. \$\endgroup\$ – Mah Sep 1 '19 at 8:21
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You can use that chip to switch 3mA at +/-3V, however it’s not all that ideal. On resistance limits are specified in the datasheet, and note that they vary significantly with voltage (Fig 9) so there will be significant distortion introduced, maybe a few percent to less than 10%.

You would need to use dual supplies within the maximum 10V supply total (eg +5/-5) (and more than the total signal excursions) and since that chip does not have built-in level shifting you would likely need to shift all the digital control signals to approximately the total supply range.

It would be easier to use two 74HC4051 chips which do have the level shifting functions incorporated, assuming your control signals are normal 0/5V logic signals. You might need to add a single inverter to control the /E line. It also has lower resistance than your proposed chip.

There are also much better (performance and protection as well as voltage capability) analog switch products than the 74xx series, such as the ADG1606, with commensurately higher prices. Unless you are cost-constrained you may be better off looking at one of the many such products.

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  • \$\begingroup\$ Thank you for a thorough answer. "you would likely need to shift all the digital control signals to approximately the total supply range." Do you mean that considering a supply of +/-5V, "0" would be -5V and "1" would be +5V? \$\endgroup\$ – Mah Sep 1 '19 at 12:42
  • \$\begingroup\$ Correct, so you would need a circuit to convert 0/5V to -5V/5V. \$\endgroup\$ – Spehro Pefhany Sep 1 '19 at 14:20

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