# Would a voltage divider work to lower the output voltage in this case for this open collector chip?

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:

And the specs of the trigger input is:

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?:

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.

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:

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.

• Thanks! I dont have 47k but I think I can use 68k and 8.2k for R5 and R4. – cm64 Dec 6 '18 at 11:05
• An effective way to do this is to use a 5V1 Zener in place of R5. – henros Dec 6 '18 at 13:28

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

• Can I lower Vcc to 5V from 5.7V by a voltage divider? Would that work? – cm64 Dec 6 '18 at 10:23
• Yes that would work. it would be much the same as the second circuit, except using more parts. – Jasen Dec 6 '18 at 11:22