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Identify resistor help please.

stan smythe

New Member
If it is possible can someone identify what I need to order to replace this resistor (arrowed)
I have tried googling but still at a loss.
Image:
res.jpg.
 

ronsimpson

Well-Known Member
Most Helpful Member
It is a burned up resistor!
With little information that is all I can say.
What product?

Good picture!
 

stan smythe

New Member
I wasnt sure if it would be identifiable. It is from the plug-in charger that a torch sits on.
The green line was a capacitor that blew out, I am not in to electronics but thought I would give it a go in trying to repair it.
I am ok with mechanical stuff but electronics baffles me.

res2.jpg
 

gophert

Well-Known Member
Most Helpful Member
It looks like your green line used to hold a 220nF (0.22uF) capacitor good for 400v. The rest of your board shows the wite component line below the actual component. The white line for your resistor appears to be quite damaged, possibly the damage was caused when the capacitor fell off of the board.
 

gophert

Well-Known Member
Most Helpful Member
It is also possible that your lack/charred component is an inductor. Any printing on the board to indicate what is next to the empty capacitor label?
 

KeepItSimpleStupid

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Most Helpful Member
I tend to agree with the resistor designation. Since it's so lose to the diodes it's probably wire wound and effectively acts as a non-user-replacement fuse. Metal oxide/flameproof resistors are used for this purpose too. That one should probably have been raised off the PCB.
 

ronsimpson

Well-Known Member
Most Helpful Member
I have a bad idea!
This looks like a transformerless battery charger.
Bottom right, 4 diodes looks like a full wave bridge.
Why 4.7uF 400 volt? I don't know why 400Volts.
My monitor is not showing the colors well. What are the resistors? 1meg ohm 1 or 2 watt???
Look at the RC parts of the circuit.
Left side: 0.47uf & 1meg resistor in parallel. and 0.56uf & 1 meg
Top right, unknown cap & 1 meg resistor.
Right center: 0.22uF 400 volts. & 1 meg resistor. (burned so who knows what value)
-------------
Bad idea:
I think all 4 resistors are the same. Could be wrong. It happened one last year.

Gophert and KeepItSimple, What resistor 1meg or maybe 10meg? I think 1 meg.
----------------------------------
474 for a capacitor = 47 0000 pF or 0.47uF
Resistor; Brown = 1, Black = 0, Green = 5; 10 00000 = 1,000,000 ohms = 1,000k ohms = 1 meg ohms.
 

AnalogKid

Well-Known Member
Most Helpful Member
What you have is a non-isolated off-line, low current power supply. The large film capacitors probably are part of a voltage reduction or current limiting circuit.

The missing cap probably was a film type like the other reddish-brown ones on the board. What is printed on the smaller one in the upper right corner?

Since everything else has the component value in the legend, look under the burned resistor. It might be a box with letters/numbers inside. For example, 1R0 is a 1.0 ohm, 22R is 22 ohms, 1K0 is a.0 Kohm, etc. The letter is both the multiplier and the decimal point location.

Visually inspect the fuse in the upper left corner (or test it with a meter).

Gold star for the nice clear photos. Any way to get a similar view of the bottom side of the board?

The cap is rated 400 V because peak AC out of the bridge (the four rectifier diodes) is 340 V. I'm assuming you're in the U.K because of your last name and "torch". If the two big caps are there to drop the line voltage, there is no voltage drop across them if the load is disconnected; full line voltage appears across everything downstream.

Based on the relative size of the diodes, I think the two 1 M resistors are 1/4 W; 1/2 W at the largest.

13001 is a small NPN transistor, 400 V, 300 mA, recommended for high speed switch and high voltage applications.

I think the burned resistor is smaller than 1 M. At 240 Vac, a 1 M resistor dissipates only 60 mW, not nearly enough to fry it. At worst case 340 Vdc across the filter cap, it still comes to only 115 mW.

ak
 
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KeepItSimpleStupid

Well-Known Member
Most Helpful Member
Well, it could have been meant to be a 22,0000 pf capacitor, or replaced by a bleeder resistor if the component is across the capacitor.

Bottom side of the board may help if that's the case.
 

Nigel Goodwin

Super Moderator
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I would suggest it's a 1Meg resistor, like the other three on the board - I imagine they are placed across each of the capacitors to discharge them when unplugged. The 0.22uF capacitor failing has probably caused that one to cook.

As suggested by AnalogKid, a picture of the bottom would should where the resistors connect - which is probably across each capacitor.
 

ronsimpson

Well-Known Member
Most Helpful Member
Here is an example:
There are 4 diodes not D1.
What I don't understand why internet schematics use 800V or 1000V diodes when low voltage will work just fine. And why some use 400V capacitors when 25V will work.

1554925314439.png
 

stan smythe

New Member
Chris58, yes that was my original post.
I will take an image of the bottom and post it.

Visually inspect the fuse in the upper left corner (or test it with a meter).
The fuse gives me continuity on a multi meter.

I'm assuming you're in the U.K
Yes UK, the item in question is probably cheap chinese, I will add an image of the label.

What is printed on the smaller one in the upper right corner?
CBB28 125OV 526J

resa.jpgresb.jpgresc.jpg
 

AnalogKid

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What I don't understand why internet schematics use 800V or 1000V diodes when low voltage will work just fine.
A good rule of thumb for long term reliability (and a requiement in many MIL/Telecom circles) is that components run at 50% or less of their critical ratings. 340 Vac peak = 800 V diode.

For reasons no one has ever explained to me, film capacitors, especially X and Y rated, get a special dispensation from this rule. The term I've seen is "self-healing" ...

ak
 

ronsimpson

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Most Helpful Member
A good rule of thumb for long term reliability (and a requiement in many MIL/Telecom circles) is that components run at 50% or less of their critical ratings. 340 Vac peak = 800 V diode.
Most transformerless power supplies only apply 5 to 20 volts on the diodes. They do not see the line voltage at all.

X and Y rated
Most capacitors are used with DC and are rated for DC.
Some caps like X and Y caps are intended for AC.
I see that many AC capacitors for line voltage are rated at one voltage if line to line and a different voltage if line to ground. Then they have a "pulse voltage" of several times the line voltage.
Example: "Intended for 120/220 use" Y=500V, X=400V pulse 2kV. I think the cap is designed to work with 500 or 400 volts at 50/60hz but will handle 2kv for short times. My testing says it will hold 2kv at DC.

With capacitors there are many ways to get into trouble. Say 60hz, 220Vac there is current flow. Now at 60hz 440Vac the current is 2x more. The cap will heat based on current flow. So the Y & Y caps are rated in part for current flow caused by AC voltage and frequency. There is a different effect where the voltage causes insulation breakdown.

Wima capacitors have good articles on AC capacitor use.
----edited----
A "X" capacitor might be rated at 2k or 5kV but you can't say 2kv because some one will try to put 2kV ac on it. (even if you understand average verses peak voltage) So marked on the side of the capacitor; is the rated line voltage at line frequency.
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
Most transformerless power supplies only apply 5 to 20 volts on the diodes. They do not see the line voltage at all.
The reasons are simple, there's pretty well no difference in cost, so you only stock the higher voltage versions, then you can use them anywhere.

I've never stocked the lower voltage versions of any diodes, it seems a pointless exercise?.
 

AnalogKid

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Still don't see how a 1 M resistor can get so hot that it fries itself and scorches the board through to the bottom side. 1 W of heat in a 1/4 W body would take 1000 V RMS or DC.

ak
 

AnalogKid

Well-Known Member
Most Helpful Member
I've never stocked the lower voltage versions of any diodes, it seems a pointless exercise?.
Today, agree. However ...

A lot of my parts are from "back in the day", including 1N400x diodes with gold-plated leads (individually packaged in glassine envelopes!). The higher voltage ones have higher junction capacitance, higher Vf, and back then they cost enough more to make it a decision.

ak
 
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