Resistance across DC input? - Page 1
Forum section: electronics / beginners

author: jj5
author_rank: Unknown
timestamp: August 21, 2023, 12:06:55 pm
content: 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.

author: Grandchuck
author_rank: Super Contributor
timestamp: August 21, 2023, 12:24:06 pm
content: A fixed load on the charger is perhaps a bleeder. You could try 1.5 k ohms or something in that neighborhood.

author: Ian.M
author_rank: Super Contributor
timestamp: August 21, 2023, 12:33:09 pm
content: 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.

author: jj5
author_rank: Unknown
timestamp: August 21, 2023, 12:40:29 pm
content: Quote from: Ian.M on August 21, 2023, 12:33:09 pmIt'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.Quote from: Ian.M on August 21, 2023, 12:33:09 pmI'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.Quote from: Ian.M on August 21, 2023, 12:33:09 pmYes but such a polarity protection diode wouldn't be connected across the two input voltage terminals, just in series with one of them, no?

author: Vovk_Z
author_rank: Super Contributor
timestamp: August 21, 2023, 12:41:25 pm
content: I guess that was a decoupling capacitor, C1. It shorted and burned. Any 0.1-1.0 uF should help I guess.

author: jj5
author_rank: Unknown
timestamp: August 21, 2023, 12:46:02 pm
content: Quote from: Vovk_Z on August 21, 2023, 12:41:25 pmThe 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.

author: Ian.M
author_rank: Super Contributor
timestamp: August 21, 2023, 12:55:24 pm
content: 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.

author: jj5
author_rank: Unknown
timestamp: August 21, 2023, 01:37:33 pm
content: Quote from: Ian.M on August 21, 2023, 12:55:24 pmHere '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...

author: Ian.M
author_rank: Super Contributor
timestamp: August 21, 2023, 01:42:52 pm
content: That does look like a capacitor. Its certainly not a diode.

author: jj5
author_rank: Unknown
timestamp: August 21, 2023, 01:52:43 pm
content: Quote from: Ian.M on August 21, 2023, 01:42:52 pmIf it's a capacitor, what is its purpose? Perhaps it's just a 1.5k bleeder as mentioned above?

author: Ian.M
author_rank: Super Contributor
timestamp: August 21, 2023, 01:55:13 pm
content: Decoupling, as Vovk_Z commented

author: jj5
author_rank: Unknown
timestamp: August 21, 2023, 02:04:09 pm
content: Quote from: Ian.M on August 21, 2023, 01:55:13 pmOkay, 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.

author: Vovk_Z
author_rank: Super Contributor
timestamp: August 21, 2023, 07:03:34 pm
content: Quote from: jj5 on August 21, 2023, 12:46:02 pmBut I see U1, D4, R14 numbers?

author: Vovk_Z
author_rank: Super Contributor
timestamp: August 21, 2023, 07:07:06 pm
content: Quote from: jj5 on August 21, 2023, 01:52:43 pmThere is non-zero possibility it is a resistor. But a possibility that it is a decoupling capacitor is much higher.

author: Vovk_Z
author_rank: Super Contributor
timestamp: August 21, 2023, 07:10:33 pm
content: Quote from: jj5 on August 21, 2023, 12:46:02 pmSo 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.

author: radiolistener
author_rank: Super Contributor
timestamp: August 23, 2023, 12:51:38 am
content: Quote from: jj5 on August 21, 2023, 12:06:55 pmNo, 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.