I'm working on a cooling fan rotation supervision system, and I'm planning to use my old 4-wire Intel E18764 12 Vdc cooling fan to test it (as I don't have access to the 24 Vdc actual fan I'm going to eventually use) but I'm not much of an electronics guy so I was worrying if I'm doing it right. I need ~24 Vdc signal from the open collector tachometer pin for my PLC digital input. The 12 Vdc source for the fan is at the moment actually a 9 Vdc battery. Can I do it like this with a pull-up resistor?

Intel cooling fan "model" In my fancy-pancy LTspice model it seems to work, but I don't know the actual characteristics of the internal transistor of the fan so I don't trust it (nor my skills in using the program).

  • \$\begingroup\$ To be safe, I would pull up to 12V, and then feed the output into a level shifter - which could be as simple as another transistor with pull up to 24V. \$\endgroup\$ Jun 26 '16 at 14:38
  • \$\begingroup\$ My answer is "yeah probably but maybe not". Just depends on how expedient you want to be and how much you care about your 12V fan. \$\endgroup\$
    – Daniel
    Jun 26 '16 at 16:30
  • \$\begingroup\$ Well I'm not emotionally attached to it so it won't be a disaster if I fry it \$\endgroup\$
    – stmas
    Jun 26 '16 at 17:41
  • \$\begingroup\$ What kind of input are you sending this to? A regular digital input will most likely not be fast enough... You will probably need to send it to a high speed counter input, most of which can tolerate anything from 5 to 24VDC. \$\endgroup\$
    – R Drast
    Jun 27 '16 at 12:41
  • \$\begingroup\$ The input has 3ms scan time, which i know is not enough, but it suffices for this test. 1 kHz counter input is ordered! \$\endgroup\$
    – stmas
    Jun 27 '16 at 13:30

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.

  • \$\begingroup\$ Thank you alot! This gave me courage to try, and it is indeed working! Needs a bit tuning, as I used a resistor that just happened to be at hand, some 4.7k one \$\endgroup\$
    – stmas
    Jun 27 '16 at 7:51
  • \$\begingroup\$ Simple test: disconnect the fan (or stop it with transistor off) and just power the PLC input with the resistor. Measure the voltage on the input. Compare this with the PLC input spec. and if it's a bit low you need to decrease the resistor value. If you share the make and part number we can help. \$\endgroup\$
    – Transistor
    Jun 27 '16 at 7:58

Not to discount the fine answers that you already have, but you may not need to do anything.

The tachometer output is open-collector, and is "supposed" to be good up to 30 volts (but no guarantees). So you could hook it up directly and attach a pullup resistor.

It also means that some solutions may not work, as the tacho output cannot source any voltage.

---- Ok, while I was typing, Transistor has posted a more thorough version of my suggestion.

  • \$\begingroup\$ Thank you! Where can I find this info that it should be ok till 30V? \$\endgroup\$
    – stmas
    Jun 27 '16 at 7:52
  • \$\begingroup\$ I highly doubt that a 12V fan will have its tachometer output good to 30V. \$\endgroup\$ Aug 26 '16 at 1:11
  • \$\begingroup\$ @user2943160 - The output doesn't go to 30 volts. The output is open-collector. The breakdown voltage is above 30 volts, so it is capable of sinking 30 volts. \$\endgroup\$
    – Mark
    Aug 27 '16 at 5:53

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