A zener is only a two terminal regulating device. but the way it connects to Q4 that makes no sense?
Max.

 

Also Q2 and Q3 connections/polarities do not look right, a PNP should have the emitter at +ve.
Max.

 

I think it is a diode of some sort, so therefor a zener and its properties fits the bill here.

I can only go by using a multi meter in continuity mode and resistance mode to work out these connections. I'm pretty sure the connections are correct, but if i have replaced them with the correct parts is another thing. I did my best with the markings i had. This board has caused me some right headaches with reading the device markings, the damn thing is covered in PCB Underfill that much that i've had to scrape away the material and use rubbing alcohol to soften the 'glue'.

 

So if i just short the cap out, with a wire or with a resistor?

The trace thats broken is very thin, however i will use something substantial to handle the current from the 240v motor.

I know this is some basic principles but i just wanted to double check i was safe to put the soldering iron to the copper with this capacitor on board after AC power.

A fully charged mains rectifier/reservoir capacitor can spark erode a decent chunk off the end of the screwdriver used to short it. In "do as I say, not as I do style" - you should use eye protection because tiny globules of molten metal fly off.

Sometimes you can salvage the inrush current limiting NTC thermistor from scrap equipment. It has significant resistance at room temperature and reduces as current flow through it causes heating.

Failing that - an ordinary resistor can be used to discharge the capacitor slowly and safely, anything from 100R to 1k is close enough, but it needs a couple of W rating.

 

Thank you for the information Ian, was getting a little confused on what size resistor would be ideal for this situation. Low resistance, High power rating. Thanks!

 

Thank you for the information Ian, was getting a little confused on what size resistor would be ideal for this situation. Low resistance, High power rating. Thanks!

You could go as low as 10R if you like. A 2.5W vitreous would be good because the glass coating insulates it from the fingers holding it.

 

And when calculating the wattage needed, you would use ohms law, right? So for instance i take the highest voltage it could be in the cap, so in this case 400v. Say i wanted to use a 100R resistor, 400/100 = 4A of current, so 4A * 400V = 1600 - 1.6A.

Is my working correct?

Also, if i wanted to add a light bulb to this to see when its drained, would you just use a resistor value as if you wasn't using a bulk then just add a bulk in series to the resistor? I've been reading that due to the resistance characteristics of a normal light bulb you don't need a resistor, not too sure on this

 

Wattage rating is basically a rating of how hot the resistor will get under relatively constant conditions. For discharging a capacitor, you can exceed the labeled wattage rating considerably, as the time should be short. And as the capacitor voltage drops, so will the current.

John

 

Okay, thank you for, much appreciated!

 

Wattage rating is basically a rating of how hot the resistor will get under relatively constant conditions. For discharging a capacitor, you can exceed the labeled wattage rating considerably, as the time should be short. And as the capacitor voltage drops, so will the current.

John

Pretty much same here - unless its a seriously huge electrolytic, a 10R 2.5W should do the trick.

Vitreous resistors provide some insulation to hold it by - many types of power resistor do not.

 

I would have though 10R was low, all you have to do is connect a even a 1k- 2k and just wait a short period until the charge dissipates, often these are permanently connect across the cap in circuit.
Max.,

 

I would have though 10R was low, all you have to do is connect a even a 1k- 2k and just wait a short period until the charge dissipates, often these are permanently connect across the cap in circuit.
Max.,

But no one was suggesting 10R permanently across the cap in circuit - just dab it on the terminals, that's all the time it takes to discharge.

And the 2k you suggested would be dangerously low for permanently across a mains rectifier reservoir.

 

My numbers come out to .2secs for a 100μf with 300vdc and a 2k resistor?
Max.

 

Well it may have gotten off topic of the original issue i had, but i'm very grateful of the learning outcome you guys have given me regarding the discharge of caps.

Thank you!

 

Yes, I thought we had covered that on page one!
Max.

 

While we touched on it briefly, this time around there were a little more given information, more depth. I know feel confident in the resistor choice for a given application.

I'd still like to know how to go about doing it with a normal household lightbulb, visual indication that the cap is drained that way too.

 

Max, you will find Dyson shifted manufacturing to China quite some time ago ( cheep labour but still an expensive product)

 

And when calculating the wattage needed, you would use ohms law, right? So for instance i take the highest voltage it could be in the cap, so in this case 400v. Say i wanted to use a 100R resistor, 400/100 = 4A of current, so
4A * 400V = 1600 - 1.6A.

Is my working correct?

Minor point as most seem to have understood what you meant. But. As the esteemed @WBahn is wont to point out - units, units, units!
4A * 400V is NOT equal 1600 -1.6A

You don't specify the units of 1600, but they should be Watts. Then what in the world are you trying to subtract Amps from Watts? No can do! Personally, I don't know why you are attempting to subtract anything at all.

Since it does not make sense, what if you meant the minus sign to be an equal sign. I'd argue that 1600 something (it must be Watts, because that's what the left side equates to) cannot be 1.6A. The right side of the equation could make sense if the units for 1600 were mA, BUT where did they come from?

I've gone on and on to illustrate the confusions that occur if your units are missing please don't take it too personally.

 

While we touched on it briefly, this time around there were a little more given information, more depth. I know feel confident in the resistor choice for a given application.

I'd still like to know how to go about doing it with a normal household lightbulb, visual indication that the cap is drained that way too.

When I did servicing full time, I used to keep a light bulb with a pair of leads soldered on. It was manly used to bridge fuses that had blown violently so I could power up without further damage and take measurements.

The bulb was usually ready to hand and often got used for dumping reservoir caps. A NTC inrush thermistor was better if one was to hand - but TBH: if neither was handy, I'd just poke it with the pointy end of a screwdriver.

 

Minor point as most seem to have understood what you meant. But. As the esteemed @WBahn is wont to point out - units, units, units!
4A * 400V is NOT equal 1600 -1.6A

You don't specify the units of 1600, but they should be Watts. Then what in the world are you trying to subtract Amps from Watts? No can do! Personally, I don't know why you are attempting to subtract anything at all.

Since it does not make sense, what if you meant the minus sign to be an equal sign. I'd argue that 1600 something (it must be Watts, because that's what the left side equates to) cannot be 1.6A. The right side of the equation could make sense if the units for 1600 were mA, BUT where did they come from?

I've gone on and on to illustrate the confusions that occur if your units are missing please don't take it too personally.

You're right. I'm sorry. I did mean to use a = rather than a minus sign, i don't normally do that sort of thing as i'm anal about little things like this. Thank you for pointing this out to me. So it would be 1600mW which is 1.6W, is that correct now?