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Shunt Resistors

hawkmoon45

Jan 1, 2013
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Hi Guys, having been an electrician for the last 7 years, I thought it was time I understood more about electronics, thus I am asking you guys for help.
My current situation is that I bought some purple fairy lights from Hong Kong which seemed to work on a untransformed system. The little box controlling it all had basically a bridge rectifier and 3 transistors to separate circuits and a chip which controlled the flashing sequence between them.
My idea was that I could use them to create mood lighting in my daughter’s bedroom. My first thought was to separate the 3 circuits and put 1 inside the translucent wardrobe doors, one behind the body length mirror and one behind the bed head.
I stated on the Job and was never happy – No transformer? Tested the output voltage and found it to be around 220V DC. Not Good
This is where my problems started, and maybe should have asked for your help first.
100 fairy lights – 33 per circuit
I looked on line and used a LED calculator to eventually work out that I could run them on a 24VDC power supply
The parameters I needed to enter were: -
1. Supply Voltage
2. Voltage drop across each LED
3. Desired LED current
4. How many LED’s connected
1 and 4 can be easy, but when you have no facts about the LED’s 3 is a hard one, I settled at 15 mA. 20 was recommended but I stayed on the side of caution, the different resistors I used did not seem to work any better, if I went for 20 mA.
TO THE POINT - I had no way to test the amp’s used, I had a volt meter and could see the difference between 17 and 28v by adjusting the pot on the power supply.
I have a 0 to 1 mA amp meter but it is too low to use, I tried it.
I know that I can use shunt resistors to view many scales using this amp meter:-
0 to 1 mA MU 45 Class 2.5
I would like to think I have a test bed in future and be able to look at what the voltage is and what the circuit current is, it would make it so much easier.
Questions
1. Can I buy standard shunt resistors for this?
2. Can I make my own? I have 0.8. 0.56, 0.4, & 0.315mm insulated transformer wire in 500g rolls. (I would expect to test up to maximum 2A)
3. How do LED relate to resistance?
I would be Greateful for any advice
Hawkmoon
 

Harald Kapp

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In order to know which shunt resistor to use, you'd have to know either the resistance of the ammeter or the voltage drop across it at any given current and calculate Ri from these values.
By conecting a shunt resistor the current is divided inversely proportional to the resistor values or the instrument (Ri) and the Shunt (Rs):
- Ishunt/Iinstrument = Ri/Rs
Since also
- Iinstrument+Ishunt=Itotal
holds, you can compute Rs for any total current. Depending on the result, you may have to "build" your own shunt by combining two or more standard resistors. Or you can roll your own from wire, but unless the resistance is very ow, you may end up using a lot of wire. Besides, using a wirewound resistor may give you trouble when measuring AC due to the inductance.

May I suggest a quick fix for your original problem: determin the current through the LEDs? Measure the voltage across the series resistor that's already in place. Then I=V/R.

In additin I recommend you buy a multimeter that has a better selection of measuring ranges. Unless you're into precision electronics, a cheap instrument (I guess $5-10) can do a good job here.

3. How do LED relate to resistance?
Only indirectly. Iled= (Vsupply-Vled)/R. However, an LED has a nonlinear chcracteristic and Vled=f(Iled). You have to look up Vled for any given current in the datasheet. Don't expect too exact coincidence between theory and practice since Vled is typically not a controlled parameter and can vary widely (this should be no problem at all for your purposes).


Happy new year,
Harald
 
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hawkmoon45

Jan 1, 2013
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Shunt Resistors and Multi Meters

Thanks Harald for the reply, I will in the following days find out the facts that I should have given in the first place and what you suggested
1. Resistance of the amp meter.
2. Voltage drop across the amp meter.
I understand what your point is, we create 2 resistors in parallel to achieve a 1/10X, 1/100x 1/1000x etc. flow through the amp meter.
For all I doubt I would want to measure AC as most electronics work on DC, and the heart of my test box I want to make would be 24V DC, I had considered inductance. My thoughts were that if I needed to use 3000mm of 0.8mm wire, I would bend it in the middle and wind it with 2 parallel wirers that would oppose each other, the + end and the – end finishing in the same place. After all other than batteries, you may get some inductance from the wave form of other than really good power supplies.
As to Multi meters, I have 4. My best a Fluke, belongs at work where I need it most, my life depends upon it, my second is a starter level Fluke, Good, but cannot measure amps, I keep this at home, where I am doing my hobby work in question. My 3rd is what I got for achieving the best results in the first year of my apprenticeship. It does all including temperature – but not very well. This is where I draw a line, for hobby electronics buy a cheap Multi meter as long as you are working with Extra Low Voltage (Sub household mains) But if you work with the likes of 220V AC and higher Always, Always use a reliable brand. Fluke are one of the best. Do not buy as my 3rd and 4th a cheap Chinese multi meter – your life depends on it.
Thanks again Gary
 
Last edited:

tedstruk

Jan 7, 2012
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notes from knowitalls

Of course you would know that lots of guys used shunts successfully, but my teach said it would be better just to arc a hot line across it and fry the whole thing, cause thats really what your doing with a shunt anyway. Shunts are a bad idea if they aren't designed into a circuit. thats because ciruits aren't desigend to handle a special voltage or amperage, they are designed to work within a certain value. Special lights made in china, taiwan, japan, korea, all share the same incommon. they were designed to work for a while and then be replaced, not reworked.

sorry.... anybody can cut two wires and hook them together, and it might work for a while, but somewhere along the line, something that was not designed in series is going to go out...if its a parrallel circuit, and massive flashings, chasings and blinkings are really hard to do in series, (but of course, who knows about the oriental mind...personally I think they are all in series).
 

(*steve*)

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Hawkmoon45, the easiest way of measuring current in a circuit like this (a heap of LEDs in series) is to measure the voltage across a resistor (this is essentially what a shunt does).

If you're trying to measure the current through the LEDs (I'm not exactly sure what you're trying to measure) and there is a resistor in series with them, measure the voltage across it. Use Ohm's law to determine the current.

Note that since the voltage is almost certainly pulsed (probably rectified AC) then the reading you get will be some sort of average. This may be close enough to the ballpark to give you average current.

I'm really not sure what tedstruk was trying to say. I do gather he has a poor impression of Chinese technical design (but he's out by at least 30 years on Japan).

Typically these lights (Christmas decorations?) employ one or more strings of LEDs with a very large number of LEDs in series and a heap of resistors to spread out the heat. They're not very efficient, and prone to failure if operated at above their rated voltage.

Also tedstruk, what about building radios at night saves so many lives?
 

CDRIVE

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May 8, 2012
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I'm really not sure what tedstruk was trying to say.

Also tedstruk, what about building radios at night saves so many lives?

I'm disappointed. I was hoping someone would translate the entire post, especially this. :confused:

Chris
 
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