Electronic Circuits and Projects Forum

that potentiometer, labelled T - the one that is turned most clockwise - is probably the timer adjustment in question. If you alter this, does it alter the timeout period? This will probably be in series with another resistor. Turning it far clockwise takes the resistance to its minimum and the extra resistor is the residual 3 minutes. therefore you want to lower this fixed resistor. I would lower it, but not reduce it to zero (I dont like zero ohms in any circuit).

This is just a guess, but probably how it works. I have an en suite timer with the same problem. The above worked for me

Thanks for responding! That’s great if a change of resistor is all it needs. Yes the potentiometer labelled T is how you increase and decrease (clockwise) the time, the potentiometer labelled H is for adjusting the humidity level at which the fan will start automatically (I don’t want to affect that).

If you look at the second photo, its of the back of the board and the potentiometer T is now on the left at the bottom, any pointers?

...Thanks...

It looks as though R27 is a 12k4 resistor in series with the 'T' pot (you could confirm with a meter check). Perhaps you could connect another resistor (say 10k or even lower) in parallel with that and see if it reduces the time? In line with what simonbramble says I wouldn't reduce that resistance to zero though, unless you can be sure of not letting the magic smoke out.

12k4 is another way of writing 12.4k (it is done this way so you dont have to worry about missing a decimal point). Putting any resistor in parallel with the 12.4k WILL (and always will) bring down the overall resistance. Putting the same resistor in parallel with any resistor halves the value, so you can try putting in another 12k resistor (or similar). I suspect C10 is the capacitor it is charging, so if you were to probe anything, you need to probe the voltage across C10. personally I would just add a 10k (or 12k or something near to that value) across R27 and see what happens. Measure the resistance of the 10k before you do it, just to make sure it is not 10 Ohms.

I see how the R27 resistor is 12.4k ohms, is 12k4 another way it can be written down?

Yes. It's a convention where the decimal point is replaced by a multiplier (e.g. k, M, μ) for clarity, as points don't show up well.

where do I place the probes and what should I measure for?

Set your meter to measure Ohms, on the 20kOhm range, and put the probes on points 2 and 4. You should get a reading of ~ 12k4.

When you say "connect another resistor (say 10k or even lower) in parallel would that mean to remove (assuming it’s the correct one) the R27 resistor and put a lower one in its place?

No. Leave R27 as is and just connect another resistor between points 2 and 3 (or optionally between points 2 and 4 if your meter test gave the expected ~12k4 reading. This will test your soldering skills! Make sure you don't get any solder blobs causing short circuits.

That's reasonable, as there is presumably something else effectively connected across points 2,3. So, at your own risk, you could try temporarily connecting 10k between 2 and 4 (probably easier than between 2 and 3). In theory that should roughly halve the over-run time and you can work down from there with a lower value resistor. Unless you can post the full schematic of the control circuit, suck it and see seems the only approach.
But BE CAREFUL. Ensure mains power has been off for at least 30 secs before working on the circuit.

Well I put a 10k resistor between points 2 and 4 and it seems to add another minute to the time! Timing it properly the fan seems to actually average about 2 mins 20 sec of overrun as standard. With this resistor in place it runs for about 3 mins 20 sec.
The resistance of R27 is about 8k normally, when I added the 10k resistor the reading is about 4k4