Extractor fan - a bit of circuit help, please

[David]

Hopefully you can see this.

There should be a shonky hand drawn circuit diagram and three photographs.

From the diagram I can see most of what is going on (I think) but I am puzzled by the black wires to the indicator light.

I can see that the black wire from the control panel goes to the circuit board, then links through a couple of resistors and the light (neon I think) on to the white wire which goes straight back to the -ve terminal on the choc block wiring centre on the motor housing.

So it seems logical that if buttons 1,2 or 3 are depressed then there is a common link via the black wire to the fan power light and then back to mains.

This in turn suggests that the light coming on just indicates that the switch is functioning, not that any power is reaching the motor.

However I am currently baffled by the other black wire. On the picture of the back of the circuit board (it is the blob of solder with the diagonal top like a pointy hat in the top middle/left of the picture) it doesn't seem to link with any other component.

Assuming that the black wire from the control panel is there to feed the fan power light, then that just leaves two wires to actually make the fan run; blue and red.

Is it reasonable to assume that the logic for a 3 speed fan is 1, 2, 1+2? This at least gives me a basis for testing.

It doesn't seem logical that the black wire carries the "power on" indicator to the fan power light and also magically always lights the light. Hang on, what if the black wire is power for speed 1, and the sequence is 1 (black), 1+2 (say black+red), 1+2+3 (black+red+blue)?

The design may be messing with my head by having the rear terminals opposite the front switches but not being connected via those switches.

Anyway, some testing to be done slowly and carefully before I can decide if power is getting through to the motor.

Any advice and views on the general wiring most welcome.

I haven't yet opened up the motor housing itself to confirm that there is no wiring issue there.

First plan is to test between +ve in (2nd terminal on central block) and terminals 3, 4, and 5 to see if voltages appear there when expected by pushing in the switches.

Cheers

Dave R

[David]

Stage 1 of testing:

If any of the 3 buttons for the fan are pressed then 240V appear on all 3 wires.

This does not seem logical so I must suspect a fault in the switches (but if so surely the fan would just run at full speed) or a fault in the fan motor somewhere.

Next step is to disconnect the red, blue and black wires into the fan and see what presents on the choc block with the motor out of circuit.

Cheers

Dave R

[David]

Stage 2 of testing:

Switch 1 pressed - power on the blue wire

Switch 2 pressed - power on the red wire

Switch 3 pressed - power on the black wire AND the indicator light comes on.

So there seems to be something going on in the motor assemble which puts power on all three switch lines when any one is given power.

I am now completely baffled as to how the indicator lamp works because testing the switch mechanism in isolation only powers up the light when the black switch is closed. Does something feed power back down the black wire when the other two switches are pressed?

I suppose that could work; black wire permanently connected to somewhere at the motor end so it will back feed when red or blue are live. However I can't visualise how that would give a third speed when only the black wire was live. Unless there is a diode or two in there somewhere.

Next step is to isolate the capacitor just in case that is causing the problems.

Looking more like a motor fault at the moment.

Cheers

Dave R

[David]

Disconnecting the capacitor made no change to the behaviour (I reconnected the motor wires as well).

Looking under the fan part of the motor it looks as though everything disappears inside the motor assembly.

Enough for one day; back for the final push tomorrow.

It is looking more and more as if the motor is kaput.

Cheers

Dave R

[Mike Humphrey]

It looks to me like it links via the square pad to the bottom of the resistor. That makes the circuit:

Black ---- R2 ---- (Black)------ Neon ------ R1 ----- White

So that means that the black wire from the switch passes power to the motor through R2, and also powers the neon/R1 as an indicator. Since the light must be on at all speeds, the black to the motor is also on at all speeds, and the other two wires must control the speed.

It also looks like the PCB is designed to have the switch mounted on it, using the square pads - the three bottom pads, not connected to anything, are labelled for purple, red and blue wires, and brown above it - the colours connected to the switch. So this PCB was designed for a different model and re-used. One of the black wires is connected to a pad labelled orange, which may have something to do with the orange wire at the capacitor.

I'm struggling to make sense of the overall circuit though. The motor has a permanent live on the brown wire, but I can't see how neutral gets to it.

[tabbypurr]

do yourself a favour and use an image host that isn't so packed with crap.

[Dave W]

You don't say but I take it the motor doesn't go.

I have redrawn your circuit with wire colours shown:

I take it your central rectangle is a connection block with the top row connected to the bottom row or terminals.

The two blacks are connected together on the board with the neon indicator.

The Lights On/Off button is a push-push type, and the OFF button releases buttons 1,2 & 3. The red, blue and black wires go to tappings on the motor winding, so that only one tapping gets the mains on it at any time. Because all tappings go to the same winding, when the motor is running there will be a voltage between the black and common brown which powers the neon indicator.

As you say that any button makes all three wires 240V, I guess there is an open circuit somewhere in the common brown connection, outside or inside the motor.

[David]

The last bit is straightforward; the circuit is completed via the coloured wires and the control switch. Via the purple wire back to the control block.

The connection of the black wire is a puzzle because it has to be live when any button of (1) (2) (3) is pushed in to make the indicator light come on, but the connection through the switch is only made when the (3) button is pushed. See last post for the results when the wires are disconnected from the motor. Testing without the motor in circuit only makes the light come on when button (3) is pressed.

So the black wire must both control the speed - red and blue aren't powered but black is when (3) is selected - and also in some way feed power back from the motor when (1) and (2) are selected to make the indicator light come on. Without activating the winding for fastest speed which comes on when black is powered via the switch. It must use directional electricity! :-)

Thanks for the comments.

Cheers

Dave R

[David]

Doh!!

Coffee hasn't kicked in yet!

Ummm........yes, that is a bit of a puzzle.

I've just double checked and the -ve mains terminal (1) just seems to have the grey (which goes to the light and presumably completes the circuit via the brown wire to (6) then brown wire to the switch and then the purple wire back to (2)), and the white which goes to the indicator light board.

Power to the motor seems to go (2) to the motor then back via (3), (4) or (5) to the switch. The only route back seems to be via the white wire from the indicator light board to (1).

I was assuming the white was an extra return to allow the indicator light to work but it seems to be the only return from the motor and switch wiring. The brown at the other end of the switch can't be involved because then the lights would be on at any time the fan was running; that is, you couldn't run the fan with the lights off (which IIRC you could).

So logically (?) the return path must be through the white wire on the indicator board. How it both lights the light AND carries full power for the fan is something I haven't worked out yet. Presumably, like the black black connection on the board it is hidden inside somewhere.

Plus there is still the puzzle about the black connection and to the board and the (3) push button. Testing shows that pushing the (3) button makes a circuit to the (5) connector on the choc block via the indicator light board. I can see how that would work if the white was just a return (on the other side of resistors) to make the indicator light work but having one black wire to (5) on the choc block which seems to provide a connection to the motor for speed 3, a connection to the indicator light for speeds 1 and 2, and presumably a return from the motor for speeds 1 and 2 seems a puzzle.

If the black wire from the switches to the fan power light board wasn't there, or if it was always live whichever switch was pressed I could guess that it was just a return wire and the three speeds were blue, red, blue+ red.

However a circuit is made to (5) when the 3rd switch is pressed, but not to (3) or (4).

Logically, at the motor, brown provides the +ve and blue and red provide the -ve directly through the switch, and black does a couple of things; switch to fan power light to -ve (to complete the circuit) which is logical if you just have two windings and "something else" between the motor, terminal (5) and the fan power light board.

At this point I am baffled.

Perhaps I have a fault on the (3) switch and instead of connecting blue and red together it connects permanently to black? This might explain the voltage on all 3 terminals (3, 4, 5) when any button is pushed in.

However this still does not explain why there is a black wire which returns from the motor to the fan power light board.

So I still don't know if I have a motor fault or a switch fault.

Cheers

Dave R

[Dave W]

See my reply yesterday showing my idea of the circuit.

[David]

Thanks for the diagram.

It seems to reflect what I have drawn, but still does not explain the role of the black wires. That is, the one from the motor to the neon which seems to be both a feed back from the motor to SW3 and also a feed back from the motor to the Neon.

It shows the only return path from the motor to be via two 47K resistors and a neon. Assuming that the black wire from the motor is the return path then how does SW3 work?

Cheers

Dave R

[Dave W]

When SW3 is pushed it applies 240V to the neon board via a black wire, where it is joined to the other black wire which goes to the motor.

I'm afraid I made a silly mistake with my diagram, as all inputs are shown as going to Live via various paths. Probably the brown wire from the motor does not go to the Live terminal near it as I hastily assumed, but to Neutral.

[David]

Thanks for responding. :-)

Your routing of the brown wire from Live to the motor is absolutely correct.

I think the area of doubt is your linking of the purple wire (also +ve) to all the switches. This detail is concealed within the switch panel.

I suspect that the purple wire only serves the light switch. This is then a straightforward circuit of:

Live -> purple -> switch -> brown -> (6) -> grey -> Lights -> Neutral

As you have pointed out, the big issue is how the fan connection completes. Live goes to the fan via the brown lead.

The only route back to Neutral is via the switch panel, then the Neon, then the white lead.

The only apparent route from the switch panel to the Neon is via the black lead.

So far so good, but what role does the black lead from the motor via (5) to the Neon fulfil?

More testing is required but the logic still escapes me.

Cheers

Dave R

[David]

I should have said that the routed back could be either via the black lead from the motor to the Neon, or the black lead from the switch panel to the Neon.

I just can't work out the logic of the black wires which seem to be both a route through the Neon to Neutral and a route through Switch 3 to {apparently nothing}.

Doing my head in!

Cheers

Dave R

[Dave W]

As I said in my last post, the motor brown lead does not go to Live, but to Neutral. You need to look at the back of your connection panel in the centre of your diagram, or measure the continuity. SW3 goes to the same common purple wire as SW1 and SW2. The black wire from it goes to the neon board and to the motor via the other black wire from the neon board.

[David]

Yes, you are correct.

I checked the original diagram several times and still got it wrong. Sigh!

So that just leaves me to work out how the neon lights when speeds 1 and 2 are selected.

I can see how it works with speed 3, as there is a connection to the neon and to the motor. However there must be something making the light come on when speeds 1 and 2 are selected.

Thanks again

Dave R

[Roger Hayter]

It may be that the three live fan connections are all taps on the same motor winding.

[Dave W]

As I told him on my first reply

[Roger Hayter]

I know! I wanted to keep it simple and direct - otherwise I would have referred back.

[David]

However my brain is still struggling a bit.

For that to work (that is, for power to go to 1/3, 2/3 or the full motor) I am thinking (and decades ago in A Level Physics I tended to think of electricity flow as akin to water flow) that electricity will flow between live and neutral (purple to the switch and brown to the motor).

Which ever route is open (like three water taps) the electricity will take that route but water pressure will be there at all three taps.

Presumably in that case there will always be a voltage on the black wire when any switch is closed which will light the neon, but only a full current flow to the purple when the switch (3) is closed.

Hmmm.....thinking about in water terms seems to have helped again.

So having a voltage on all three connections from the motor is to be expected because they are all connected to the same winding, albeit different parts.

In turn this suggests that the switch is behaving as expected and the voltage readings on the central connectors are as expected.

The only remaining issue is that the motor will not spin.

I am assuming that even if the capacitor is banjaxed the motor should run with an initial spin?

Which it doesn't.

Thanks very much for the patient explaining.

Cheers

Dave R