Resistors – How to Counteract Pull-Up Resistor Increasing Low Side of Signal

The answer depends on what you want the "default" configuration to be. For example, say you have a down-stream N-channel MOSFET, and you want it default off. Then you would use a pull-down resistor to ensure this behavior if the input becomes high impedance.

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On the other hand, suppose you have an upstream P-channel MOSFET, and want it default off. This time a pull up resistor is required to create this behavior.

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There's also the alternative case where you want a device to be default-on, in which case the above two cases would be reversed (pull-up for the N-channel MOSFET, pull-down for the P-channel MOSFET).

A few other considerations:

1. I2C lines specify pull-up resistors because devices are "expected" to have an open-drain to ground, and thus need some way to raise the line potential.

2. Analog comparators are usually configured as open-drain devices, and thus also need pull up resistors to get a high potential output.

3. You may draw more current using pullup/pulldown resistors, depending on what's hooked to the input/output.

4. Either configuration could works equally well in your application (i.e. there's no significant advantage one way or the other).

... And any number of very application-specific reasons why one configuration may be preferred.

PLC 24 V digital inputs come in two varieties:

• Current sink requires a 24 V feed to power them on. (I'm assuming that this is what you're using.)
• Current source requires a connection to ground to sink the current sourced by the input.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Open collector output to current sinking PLC input.

Assuming that you are using a current sinking input, the best way to do this would be to use a converter unit or industrial opto-isolator but using a pull-up resistor is not uncommon. The values in the calculation below came from the PLC spec in another question I answered. Yours should be similar but it would be worth checking to ensure reliable information.

R4 value

The PLC is guaranteed to turn on at 3 mA / 11 V on the input. Let's give it 12 V to make the maths easier. R4 then has to drop 12 V at 3 mA. \$R = \frac {V}{I} = \frac {12}{3m} = 4~k\Omega \$. The next lower value is 3k9.

Q2 current

Now check the max current through the fan transistor shown as 30 mA max. When Q2 is on R4 will be have 24 V across it. \$ I_{max} = \frac {V}{R} = \frac {24}{4k} = 6~mA \$. We're fine. In fact we could decrease R4 to 3k3 for extra PLC margin if we wanted.

R4 power dissipation

Finally check the power rating of R4. Worst case will be if the fan stops with Q2 on. \$ P = VI = 24 \cdot 6m = 150~mW \$. A 1/4 W resistor would suffice.

Re-run the calculations for your components and you should be able to make it work.

If, by any chance, you have a current sourcing input then just make a direct connection to your open-collector transistor.