PNP Open Collector output series resistor

Question

I have the following circuit to drive a logical output (the input is driven by an MCU):

partial schematic

The issue I have is that because of R4, when I want to draw a few mA from the output, the voltage drops. Right now I am trying to drive a 24V DC relay that requires 7 mA, but this is too much current and the voltage drops to ~10V and the relay does not open.

I need help with 2 things:

• First, what is the purpose of R4 and why was it designed to be 3.6k?

  My understanding is that it is meant to protect Q1 from a short to ground, but Q1 (BC857CLT1G) is rated for a maximum collector current of 0.1 A. So we could have used a 240 Ohm/2.4 W resistor.

• Second, would it be acceptable to reduce the resistance to 680 Ohms/1 W?

  This would allow a current of 7 mA while maintaining an output voltage of ~19.2V, which should be enough to activate the relay. In case of short to ground, the current would be 35 mA and the power dissipated 0.85 W.

For information, here is the circuit diagram of the relay:

relay internal and external connections

A1+ is connected to the output of the previous circuit and A2- is connected to ground.

Answer 1

Hard to say why 3k6 was chosen. Perhaps it was to protect against surges and ESD on an external connection. Perhaps it was intended for a lower output current. Anyway, your choice of a 680R resistor seems reasonable. If you want to protect against surges, ESD, etc with the 680R then you might place a capacitor, say 47uF from collector to ground.

• Benjamin T: The emitter is connected to the board 24V, which is connected to the outside-world 24V through a power supply stage which as several capacitors and diodes to protect from ESD and surges.
• scorpdaddy: Thanks Ben. My bad. Corrected.
• Benjamin T: OK, it makes more sense. However, we already have some PCBs produced, so the rework possibilities (e.g. adding components) are limited. At least, until the next production batch.
• scorpdaddy: Perhaps it is possible to rework them and attach an additional component with adhesive/staking - if the surge immunity is applicable in your application.

Answer 2

1) The value of R4 prevents burning out R4 with Pd=0.16W when active shorted to 0V or possibly limit current for a 6mA LED indicator.

2) You can reduce R4 to 0 then shunt with reverse diode. to drive 24V relay

3) You might also be able to drive from Q2 as a low side driver.

• with reverse clamp diode to 24V and coil's other side connected to 24V.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

• Benjamin T: I am not sure how your point #2 is supposed to work.
• Cristobol Polychronopolis: You want a reverse biased diode across the relay coil, because when you cut off the current to that coil, the magnetic field will collapse and drive a large spike in the other direction. The diode absorbs the spike and prevents damage to other components.
• Tony Stewart EE since 1975: A forward Zener also clamps faster and there are no other components
• Benjamin T: I have updated my question witch the circuit diagram of the relay. It seems there is already a flyback diode built-in.
• Tony Stewart EE since 1975: Then you only need to R value for LED (I would expect <=6mA) and R for coil and my answer still holds. And is far less than 100mA