Resistance across DC input?

[jj5]

Hi there. I'm having a go at repairing a friend's electric hair trimmer, but I've made a mess of it and knocked off some of the surface mount components. The part of the circuit that I'm looking at presently is this:



Those terminals marked V+ and V- are the DC input from the charger. Is that likely to be a resistor that's supposed to be across those terminals? I don't understand why there is a component there at all frankly, connecting the +ve to the -ve, why do that? I can only imagine this is some mega-ohm rated component there for some reason? Is it safety related?

My question is: what would a typical value be for a resistor in this place? I need to find a suitable part to replace it.

[Grandchuck]

A fixed load on the charger is perhaps a bleeder.  You could try 1.5 k ohms or something in that neighborhood.

[Ian.M]

Other possibilities include a capacitor, a TVS diode for transient overvoltage protection, or a reverse biased diode for reverse polarity protection.  The designator next to the silkscreen outline would give you a clue if you could see all of it.  I can't see from the photo if its obscured by flux or other glop, or if its missing due to board damage.   If it begins with 'R' its a resistor as you suspect, otherwise 'C' for capacitor or 'D' for diode.   

What type of power jack does it use?  Reverse polarity protection diodes are most common if the jack is a barrel connector due to the lack of standardisation for their polarity.

[jj5]

Other possibilities include a capacitor, a TVS diode for transient overvoltage protection, or a reverse biased diode for reverse polarity protection.  The designator next to the silkscreen outline would give you a clue if you could see all of it.  I can't see from the photo if its obscured by flux or other glop, or if its missing due to board damage.

It's not glop, it's missing due to board damage. I haven't used a soldering iron for 30 years (just getting back into the swing of things) and I managed to damage the board when removing the batteries, which were soldered in. Today I learned that I should take photos of the circuit board before I take my soldering iron to it.

What type of power jack does it use?

I'm not sure of the DC voltage or power, but it uses a weird two-prong input the likes of which I have never seen before. It's not a standard DC barrel input.

Reverse polarity protection diodes are most common if the jack is a barrel connector due to the lack of standardisation for their polarity.

Yes but such a polarity protection diode wouldn't be connected across the two input voltage terminals, just in series with one of them, no?

[Vovk_Z]

I guess that was a decoupling capacitor, C1. It shorted and burned. Any 0.1-1.0 uF should help I guess. 

[jj5]

I guess that was a decoupling capacitor, C1. It shorted and burned. Any 0.1-1.0 uF should help I guess.

The component numbers are four characters, a letter and then three digits, so it's not C1. I suspect this component is actually a resistor. The damage wasn't due to a short, it was due to very poor soldering when I was desoldering the batteries. I still have the component on my bench. It's tiny and I can't get a good reading of it through my multimeter.

[Ian.M]

As the DC jack isn't a barrel type, its unlikely to be a reverse polarity protection diode.

Reversed biassed shunt diodes are quite common for reverse polarity protection when the additional voltage drop of a series diode cant be tolerated. 

Post a picture of the loose component.  Resistors are usually a white ceramic substrate, black on one side, but capacitors are usually brown or tan in color.

[jj5]

Post a picture of the loose component.

Here 'tis:



That's an AUD$2 coin for context. Not sure if that's brown/tan that we're seeing? Tinning seems to be missing on the right hand side...

[Ian.M]

That does look like a capacitor.   Its certainly not a diode.

[jj5]

That does look like a capacitor.

If it's a capacitor, what is its purpose? Perhaps it's just a 1.5k bleeder as mentioned above?

[Ian.M]

Decoupling, as Vovk_Z commented

[jj5]

Decoupling, as Vovk_Z commented

Okay, cool, thanks. I will give that a try. I will solder in a 0.1 uF (or similar) capacitor and see if I can fix this thing. Thanks again.

[Vovk_Z]

The component numbers are four characters, a letter and then three digits, so it's not C1.

But I see U1, D4, R14 numbers?

[Vovk_Z]

If it's a capacitor, what is its purpose? Perhaps it's just a 1.5k bleeder as mentioned above?

There is non-zero possibility it is a resistor. But a possibility that it is a decoupling capacitor is much higher.

[Vovk_Z]

The damage wasn't due to a short, it was due to very poor soldering when I was desoldering the batteries.

So there was an unstable contact when you was soldering it. It could lead to voltage spikes across power lines. It could bring a ceramic capacitor to die and to make a short circuit instead of himself.

[radiolistener]

Those terminals marked V+ and V- are the DC input from the charger. Is that likely to be a resistor that's supposed to be across those terminals? I don't understand why there is a component there at all frankly, connecting the +ve to the -ve, why do that? I can only imagine this is some mega-ohm rated component there for some reason? Is it safety related?

No, there is no sense to put resistor in parallel to power supply. Most of all, this is a capacitor. Such capacitors often used to filter RFI.