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I want to interface the output Vo of this chip to a trigger input of a DAQ device. Here below is the chip which is open collector output so There will be a resistor between Vcc and Vo:

enter image description here

And the specs of the trigger input is:

enter image description here

So Vo will be coupled to the trigger input.

In my case the Vo is around 5.6 or 5.7V so exceeding 5.5V max trigger input a bit. Would a voltage divider using two resistors betwweb Vcc and Vo below like 820 Ohm and 6.8k work fine?:

enter image description here

I really don't want to use an extra buffer.

ı mean in my case Vcc is 5.7V. So is it better to use voltage divider at Vcc or between Bcc and Vo as in my second diagram.

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2 Answers 2

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Like you propose it will not work, when the open collector output is off you will still get 5.7 V at \$V_o\$

What you need is to add a resistor between \$V_o\$ and ground:

schematic

simulate this circuit – Schematic created using CircuitLab

Left: circuit you proposed

Right: circuit I propose.

Note that the only change from your first circuit (the one with only one resistor near \$V_o\$) is adding a 47 k ohm resistor between \$V_o\$ and ground.

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  • \$\begingroup\$ Thanks! I dont have 47k but I think I can use 68k and 8.2k for R5 and R4. \$\endgroup\$
    – cm64
    Commented Dec 6, 2018 at 11:05
  • \$\begingroup\$ An effective way to do this is to use a 5V1 Zener in place of R5. \$\endgroup\$
    – user131342
    Commented Dec 6, 2018 at 13:28
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No, it would raise the "LOW" voltage of Vo from GND to R2/(R2+R3)*Vcc. In the "HIGH" case, Vo would still be VCC, as it gets pulled up to VCC.

edit:

you could create a Voltage divider between GND and Vcc, and then use a third resistor to pull Vo up to that level, but it still feels ugly, imho

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  • \$\begingroup\$ Can I lower Vcc to 5V from 5.7V by a voltage divider? Would that work? \$\endgroup\$
    – cm64
    Commented Dec 6, 2018 at 10:23
  • \$\begingroup\$ Yes that would work. it would be much the same as the second circuit, except using more parts. \$\endgroup\$ Commented Dec 6, 2018 at 11:22

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