do you know these resistors?

Hello Bob and welcome to the Electronics Point forums

I'm pretty sure the schematic is wrong in at least one place. Did you draw it up yourself? The relay coil is connected across Q1's collector and emitter, but it should be connected between Q1's collector and the positive supply voltage from D1's cathode, I think.

R3 should be around 220 ohms and should be a fusible resistor that is designed to go open-circuit quickly, without burning the circuit board and without catching fire, on an overload. If the circuit board has been damaged by overheating of either of the resistors, you need to remove all traces of charring before applying power to the board, otherwise leakage currents may flow in the charred sections and cause future problems.

R7 I can't suggest a value because that part of the schematic seems to be wrong.

Please feel free to post photos of both sides of the board for us to check against your schematic.

Is SW1 the wall switch that starts the fan running?

You should check all of the diodes, and both transistors. The relay coil should have a diode connected across it to protect Q1 from back EMF when the relay is switched OFF.

Edit: Harald, SNAP!

 

Of course Bob. I'm sure we can contain our excitement until the 20th

Yes, I would assume D3 is also a 1N4006.

It doesn't matter if Circuit Wizard doesn't know about the BCxxxL versions, does it? As long as you know what the pinouts of the actual transistors are, so you can draw them connected correctly on the schematic diagram.

 

OK, I've re-traced the schematic and you were right. It's an unusual design but I think I understand how it works. The way you arranged your schematic made it hard to follow.

But there are a number of discrepancies between the circuit references on your schematic and the references you've marked on the first photo in post #6. All or most of the capacitors disagree, and at least one resistor disagrees too.

Please go over the whole board and correct the circuit references in the photo and/or on your schematic. Please also check the values. Please add complete part information on the schematic - voltages for the electrolytics, part number for the relay, and anything else that you can find. Then repost the photo and the schematic.

Then please go through your notes in post #1, post #4 and post #6, and repeat them in your next post with the correct circuit references.

When you've done that, I'll post my schematic and a circuit description.

 

That schematic is wrong; you've used the wrong pinouts for the transistors. The L suffix devices have the Japanese pinout. Looking at the flat side, with the leads pointing downwards, the leads are, from left to right, emitter, collector, base.

I've copied the component values and references from your post, checked against the schematic.

The cap next to the BC212LB is C5, right? Your first schematic says it's 1 nF, which sounds right (it would be marked "102") but your latest post, and the schematic attached to it, say that it's 1 µF. Can you check this please.

There seem to be quite a few missing component references: C1, D5, D6, R4, R5, R6, R10. Can you confirm this?

Are the component references marked on the board, or have you assigned them yourself?

Can you confirm that D1 is a 1N400x like D2, D3 and D4? Can you see the markings on it?

I still need to know all the markings on the relay so I can estimate an appropriate value for R7.

 

Thanks Bob. Just two more things.

First, I can't find the original data sheet for that relay. I need to know the coil resistance. Can you measure it? With R7 removed, it should measure correctly in-circuit, but just in case Q1 is damaged, you might want to remove the solder from either one of the relay's coil pins to disconnect it from its track, and measure just between the pins.

Second, can you draw the external circuit, showing the incoming mains supply, the fan, the wall switch, and the four connectors on the board please.

 

OK. I think R7 should be about 560 ohms, for example http://www.digikey.com/product-detail/en/CFR-50JB-52-560R/560H-ND or http://uk.farnell.com/welwyn/mfr4-560rfi/resistor-metal-film-560-ohm-500mw/dp/1833360 and R1 should be about 330 ohms 0.5W fusible - for example, http://www.digikey.com/product-detail/en/FRM-50JR-52-330R/330DTCT-ND. Farnell don't have a 330 ohm fusible resistor but 470 ohms will do: http://uk.farnell.com/te-connectivity-neohm/frn50j470r-s/resistor-fusible-470r-5-0-5w/dp/1898508

The most likely cause of R1's failure would be the failure of C4, and I strongly suggest that you replace C4, even if it measures OK. A suitable replacement is the EPCOS B32922C3224K which is available from Digikey: http://www.digikey.com/product-detail/en/B32922C3224K289/495-4897-1-ND and Farnell: http://uk.farnell.com/epcos/b32922c3224k189/capacitor-film-0-22uf-10-rad/dp/2367352

A better quality alternative for a slightly higher cost is the EPCOS B32912A3224K which is also available from Digikey: http://www.digikey.com/product-detail/en/B32912A3224K/495-3970-ND and Farnell: http://uk.farnell.com/epcos/b32912a3224k/capacitor-pp-film-0-22uf-330vac/dp/2098616

For safety, it is wise to connect a "bleed resistor" across C4, to discharge it when power is removed. There is no bleed resistor in this design but you can easily add one. I suggest adding two 100k 0.25~0.5W metal film resistors in series, on the underside of the board between the terminals of C4. Suitable resistors would be http://www.digikey.com/product-detail/en/RNF12FTD100K/RNF12FTD100KCT-ND and http://uk.farnell.com/multicomp/mcmf0w2ff1003a10/resistor-metal-film-100kohm-500mw/dp/1126841


The following image is a simulation of the important current paths in the circuit, using a relay coil resistance of 4608 ohms (a 500 mW coil), the value you measured on the original relay on your board.



The simulation shows that the average voltage across the relay coil is about 41V. This is pretty low for a 48V relay and I would recommend replacing the relay with one with a 360 mW coil (coil resistance 6400 ohms) such as a Panasonic JS1 series relay:
From Digikey (USA): http://www.digikey.com/product-detail/en/JS1-F-48V-F/255-2889-ND/2177080
From Farnell (UK): http://uk.farnell.com/panasonic-ew/js1-b-48v-ft/relay-pcb-spco-10a-48vdc/dp/2095679

The JS1 series has the same pin layout as the one in your board, but you'll need to check the dimensions and pin distances to be sure it will fit.

These are pretty cheap and using a relay with a more sensitive coil will improve safety. Variations in AC mains voltage will affect the relay coil voltage, and if the relay does not pull in firmly, the contacts can arc, causing damage and possibly fire. Using a 360 mW coil gives an average coil voltage of about 53V which is within the relay's specifications and safer than 41V.

Unfortunately, changing to a sensitive coil relay will increase the voltages in the circuit, and the 10 µF electrolytic, and the transistors, will be run uncomfortably close to their voltage limits. You should replace the 10 µF electrolytic with one rated for 100V, and the transistors with types rated for 100V and reasonably high gain, such as:

For the NPN: ZTX694B, available from Digikey: http://www.digikey.com/product-detail/en/ZTX694B/ZTX694B-ND and Farnell: http://uk.farnell.com/diodes-inc/ztx694b/transistor-npn-e-line/dp/9525610

For the PNP: ZTX795A, available from Digikey: http://www.digikey.com/product-detail/en/ZTX795A/ZTX795A-ND or NTE129P, available from Farnell: http://uk.farnell.com/nte-electronics/nte129p/bipolar-transistor-pnp-80v-to-237/dp/4524159

These transistors all have different pinouts from the originals.


The next image is the schematic redrawn to it's easier to follow and explain.



I've drawn the AC mains supply, the wall switch, and the fan on the right side of the diagram, as I think they should be connected. If I'm wrong about this, the following circuit description won't be exactly right.

The rail across the bottom, marked "0V", is the "zero volt" rail - the reference rail. It is not "ground", and you shouldn't use the ground symbol for it. It is connected to the Neutral connection from the AC mains, which is grounded at the switchboard, but it is not ground and should not use any kind of ground symbol.


Here is how I believe the circuit works.

With the circuit idle, and the wall switch OFF, there is 230V AC RMS (relative to the circuit's 0V rail) on CN2 but since RL1 is de-energised, it doesn't go anywhere. The circuit starts up when the wall switch is turned on.

When 230VAC appears on CN3 from the wall switch, C3 starts to charge up on every alternate half-cycle of the AC mains waveform. There are two paths for this charging current: through C4, R1 and D2 (D1 is also involved), and through R2 and D3.

C4, D1 and D2 form a "capacitor fed" power supply that charges C3 from the mains supply. R1 is included for safety; if C4 fails, R1 will overheat very quickly and go open-circuit, protecting the rest of the circuit from damage.

This capacitor fed power supply is a common way that small circuits can be powered from the mains. It avoids the size, weight, and cost of a transformer, but it provides no isolation, and is only suitable when the whole circuit can be kept isolated, and when its load current is small (this circuit draws only about 10 mA). The method is often used to power LEDs directly from the mains in space-sensitive and cost-sensitive applications.

Capacitor fed power supplies are described in many places on the web, but often not very well. Also, most designs use full-wave rectification, whereas this board uses half-wave rectification. Here is one explanation that isn't too bad: http://www.dos4ever.com/TiT/TiT.html (search for Figure 24).

So current through C4 via D2 charges C3, but current through R2 and D3 also charges C3 while the switch is closed. D3 also clamps the peak voltage at point B in the circuit to slightly higher than the positive supply rail (marked V+).

D4 pulls point A up to roughly the V+ rail voltage as well, so C2 is discharged; there is no significant voltage across it. R9 and R12 form a voltage divider that sets point C to about 40% of the V+ voltage.

Remember, in this description, voltages at single points in the circuit are measured relative to the 0V rail at the bottom of the schematic.

Q2's base is therefore fed (via R11) from a higher voltage than its emitter, which causes Q2 to turn OFF (not conducting). R8 pulls Q1's base to 0V and Q1 turns OFF as well. Current flows through R7 and through the relay coil.

D7 is needed to protect Q2 against reverse base-emitter voltage, which can permanently damage the transistor if it exceeds about 7V. No significant reverse current can flow from R11 through D7.

When C3 has charged up to around 40V, there will be enough voltage across the relay coil for the relay to pull in. When this happens, its contact changes over from the way it's shown on the schematic to the normally open contact, which is fed from mains Phase directly. This feeds the mains to the fan and into C4 as before; the wall switch is now not part of the circuit.

But the wall switch continues to affect the voltage at point A due to current through R2 which produces a voltage that's clamped to the V+ rail and holds point A near the V+ rail voltage. This keeps Q2 and Q1 OFF, so the relay remains ON.

The circuit remains in this state while the wall switch is ON.

When the wall switch is turned OFF, the fan and the circuit remain powered (because the relay is active), but current can no longer flow through R2 and D4 into point A, so C2 starts charging up slowly, due to current flowing through R3 and VR1. As it charges up, the voltage at point A falls.

The voltage at point A falls at a rate determined by the capacitance of C2 and the total resistance of R3 plus VR1 at its present setting. Once this voltage has fallen below about 40% of the V+ voltage, as set by R9 and R12, Q2 now has forward bias, because its base is negative relative to its emitter, so it starts to conduct.

Conduction in Q2 quickly reaches the point where enough voltage appears across R8 for Q1 to start to turn ON. When this happens, the V+ rail voltage starts to fall, because significant current is being drawn from it via R7 and Q1. Since C2 has about 60% of the original V+ voltage across it, this dropping V+ voltage causes the voltage at point A to drop further, turning Q2 on harder and therefore saturating Q1.

This change happens quickly and Q1 pulls its collector down to 0V, removing almost all of the voltage at point B and across the relay coil, causing the relay to drop out. The relatively high current through R7 and Q1 discharges C3 fairly quickly, and since the relay has returned to the position shown in the schematic and the wall switch is OFF, there is no longer any mains voltage feeding into C4.

The V+ rail collapses fairly quickly. Q2 and Q1 remain ON until the rail voltage is so low that when Q1 turns OFF, there is not enough rail voltage available to activate the relay coil. The V+ rail then discharges all the way to zero through R9 and R12 and the circuit is idle once again, ready to be reactivated when the wall switch is turned ON.

Normally, relay coils have a diode connected across them to protect the driving device from "back EMF" when it turns OFF and the magnetic field in the relay coil collapses and produces a voltage spike because of its inductive nature. That diode isn't needed in this circuit because the current to the relay coil is interrupted by Q1 shorting the coil out. When Q1 finally turns OFF, there is almost no energy available in the V+ rail to have any effect.

The V+ voltage during operation is mainly determined by the relay coil's resistance in series with R7. The V+ voltage is not regulated or limited. The two current sources (C4 and R2) provide a reasonably accurately controlled amount of current onto the rail, and the resistance of R7 plus the relay coil determines how much voltage appears on V+ according to Ohm's Law.

I'm not sure what C5 does. It will help protect Q2's base-emitter junction from any spikes, and it may play a part in ensuring that Q2 and Q1 remain ON as the V+ rail is falling towards 0V. It's important that they remain ON during this time; if they turn OFF too quickly, there might still be enough voltage on the V+ rail to make the relay pull in.

 

Cool

 

No. The voltage at V+ is around 50V maximum. A 63V rated cap is fine.

That all sounds right and makes sense except for the Ic and Ib figures. But don't worry about that.

The LB suffix devices have a minimum current gain of 240, so you should use those if possible.

Did you replace C4? (That's the 0.22 µF 250VAC capacitor.) Even if it measures OK, it could be faulty. I gave a suggested replacement in the second paragraph in post #13 on this thread.

Edit: I think I forgot to mention this before, but you should get a few fusible resistors in case there's some other problem and the next one fails too.

 

No, I don't think using a BC182L instead of a BC182LB (or BC212L instead of a BC212LB) would cause a problem.

This is what I suggest.

1. Replace the four black diodes (D1~4) with 1N4006 or 1N4007. They may measure OK but still be faulty. I doubt that the small diode will be damaged but you could replace it too.

2. Replace C2 and C3. They may have been damaged if one or more diodes were faulty. Make sure they're the right way round. On the first photo in post #6, C2 should have its negative side upwards (towards the BC212) and C3 should be the reverse - negative towards the board edge.

3. Replace the transistors again. I've had a look for suitable alternatives. The important specifications are Vce(max) >= 60V, Ic(max) >= 100 mA, gain as high as possible (but not a Darlington transistor).

You mentioned Farnell and used the "pined" symbol in your last post so I assume you're in the UK. Here are the most suitable replacements available from http://www.farnell.co.uk.

PNP (BC212LB replacement):
NTE NTE129P: Vce(max)=80V; gain=300; package TO-237 (TO-92 with short metal tab at top); price 1.62 for 1-up quantity: http://uk.farnell.com/nte-electronics/nte129p/bipolar-transistor-pnp-80v-to-237/dp/4524159

NPN (BC182LB replacement):
MPS8099RLRAG: Vce(max)=80V; gain=300; package TO-92; price 0.20 for 1-up quantity: http://uk.farnell.com/on-semiconductor/mps8099rlrag/bipolar-transistor-npn-80v-to-92/dp/1273221

Download the data sheets for those transistors and check out the pinout very carefully. Check whether the diagram shows the top or bottom view. I would mark the E, B, and C next to the holes on the top of the circuit board. And take a few photos so I can check them before you power it up.

At some point (maybe when all the components are removed ready for replacement), clean both sides of the board using a toothbrush and/or cotton swab with some solvent - isopropyl alcohol (tape head cleaner) is best, or CRC IPA cleaner or any solvent cleaner that doesn't leave residue.