How do you check mA going to a LED?

Hi Lunal_Tic,

I see that you're having problems determining exactly how you should be taking readings with your meter. I decided to draw up some simple diagrams to help you (and hopefully others) out.

This diagram shows how to connect the meter to measure the input voltage to the converter circuit (or Vin):


This one shows how to measure the output voltage to the LED (or Vout):


Now.. There are two different ways to measure the input and output current. The simple, less accurate way, and the slightly more complicated, more accurate way.

First, the simple way to measure the input current (Iin):


Now the simple way to measure the output current supplied to the LED (Iout):


Now.. The more complex way involves inserting a low value resistor in series with the circuit and measuring the voltage across the resistor. You can then use Ohm's Law to calculate the current from the voltage. The simplest way to do this is to use a 0.1ohm resistor, then multiply the number you see on the meter by 10.

For example, if you are using a 0.1-ohm resistor and you see "35mv" displayed on the meter, you can take (35*10) = 350, or 350mA.

To measure the current input to the converter (Iin) using this method, hook the meter up like this:


For the current output (Iout) to the LED, hook it up like this:


So...

I hope this helps you out,
pb

pbarrette-

Those diagrams should be a sticky at the top of the electronics forum - very nicely done!

One thing that might make things more clear - in the meter window, indicate that you're measuring V or A (since you labeled the socket for voltage measurement V mA) - that'll clear up that you're supposed to have the meter set to voltage or current.

Hi Evan,

Thanks..

The images have been updated per your suggestion. Incidentally, the figures listed in the meter window correspond to a real circuit of mine that I built using the LTC3490 boost IC.

To go one step further...

With these numbers, you can now determine the efficiency of the circuit at this input voltage. You simply take (Vout * Iout) / (Vin * Iin).

So, for this example:
(3.2V * .35A) / (2.41V * .51A)
or
(1.12 watts) / (1.2291 watts)
or
0.91

So, at this input voltage and current, the circuit is 91% efficient.

pb

Thanks guys this is really <great info> and helps a lot!

I guess that I have at least one if not 2 goofy converters. Was my assumption about the SYOH vs Q3J correct? i.e. The lower forward voltage should make for a more efficient run.

Thanks again,
-LT

Wow thanks!

That is very informative and you cant go wrong!

how'd i miss the updates to this thread.. fantastic stuff..and i agree. many many people can benefit from them because they are so exact and direct and to the point.

i just realized my URL early was goofed.. the correct URL is http://voltage2current.rouse.com

I developed a formula to calculate voltage from current in a Jbin luxeon and current from voltage...

The formula is I = (V-2.6)^2*.76

reality check.. plugging in the 3.41V above it comes out .341A.. very close to the .35A measured... is the emitter you measured a "Jbin" emitter?

some day i hope to develop a similar set of formulae for the H and K bin emitter, though it may become a moot point.. i'll just end up making one for K2s

The reason for developing the formula initially was just the convenience of being able to develop circuitry w/o having to do a breadboard real world test and it's helped a TON! I almost exclusively use J bin emitters so that's all i needed..

The reason it's very helpful to the john q public... is that disconnecting an emitter to measure the current is a real PITA in a working flashlight.. but measuring voltage is a cakewalk! I have found just like in the exampole above.. a 2.5% error.. but definitely in the ball park... it's been very helpful for getting decent estimates without breaking connections to emitters. (but only for Jbin at this point).

-awr

andrewwynn said:

The reason for developing the formula initially was just the convenience of being able to develop circuitry w/o having to do a breadboard real world test and it's helped a TON! I almost exclusively use J bin emitters so that's all i needed..

The reason it's very helpful to the john q public... is that disconnecting an emitter to measure the current is a real PITA in a working flashlight.. but measuring voltage is a cakewalk! I have found just like in the exampole above.. a 2.5% error.. but definitely in the ball park... it's been very helpful for getting decent estimates without breaking connections to emitters. (but only for Jbin at this point).

-awr

Click to expand...

While the formula is interesting and has uses, I can't see that it does much good when troubleshooting a circuit that is not working as expected. It can be used to predict the correct voltage that is to be expected at the LED, but if that voltage is in the ball-park and it still does not work right you have to break out the meters.

I'd like to hear of other ways to use that in trouble-shooting.

Daniel

This might be of use to you. LED converter weirdness

-LT

Lunal_Tic said:

This might be of use to you. LED converter weirdness

-LT

Click to expand...

Thanks, I'm fairly competent in trouble shooting basic circuits. I was interested in ways to leverage andrews formula to supplement conventional means.

Daniel

My bad, just trying to help. :shrug:

-LT

Lunal_Tic said:

My bad, just trying to help. :shrug:

-LT

Click to expand...

No bad involved. I like to help too. Besides, your post pointed out that I needed to make it clearer that I was looking for ways to apply Andrew's formulas in troubleshooting.

So it was your good!

Daniel

No worries, and thanks for your post in the other thread. It was very clear and helpful.

-LT

gadget_lover said:

Thanks, I'm fairly competent in trouble shooting basic circuits. I was interested in ways to leverage andrews formula to supplement conventional means.

Daniel

Click to expand...

The main purpose for the formula is that with Jbin emitters (99% of what i use)... i can do a lot of calculations by only knowing the voltage.. i also have the formula solved backwards to arrive at voltage by knowing current.. basically by knowing EITHER voltage OR current, i can estimate within like 10% all the unknowns.. voltage, power, current.. I use the formula in my spreadsheets when i want estimates of power consumption and to calculate the lumen output etc.

The handiest use is to estimate the current in a luxeon that is soldered in! for example.. if i measure 3.4V on my lux.. I would estimate 486mA and therefore 1.65W.

It is obviously only for 'ballpark' figures, but has worked great for me for about two years.

-awr

Oh.. I think i missed mentioning a fantastic coincidence..

MOST if not all constant current regulators will have nothing other than sense resistor that is between the LED- and the BAT- terminals.. so that sense resistor is pre-installed for you.. measure the mV for example between the LED- and the BAT- on a fatman driver and you will get 1/10th of the mA output viewed in mOhm exactly like described above... in other drivers, where the sense resistor is changed you have to calculate the voltage from the known Rsense i.e. you have to divide the mV reading by the resistance to get mA.. example.. if 0.08 ohm Rsense, just take the mV reading and divide by 0.08 vs 0.10).

-awr

pbarrette, you sir, are a saint. thx so much for that!

Just wondering, if I swap my test Lux I LED for a Lux V should I expect to see the current halve with the same converter? And in the reverse, if I've got a converter built for a Lux V but I test it on a Lux I will it read about double the current. i.e. 700mA to a Lux V would be about 1400mA to a Lux I?

-LT

that entirely depends on what kind of converter.. if it is the type typically used in LED.. the current will be exactly the same no matter what the load, as long as the input and output are within tolerance.

In the example above.. you take a 700mA luxV converter and run a Lux1 on it.. as long as the parameters are appropriate for operation it will be 700mA at roughly half the voltage. The converters job is to change the voltage to make up the required amount of current.

(possibly tricky in that case).. if it's a boost ckt you would need to be using on the order of 3V input.. but if a buck ckt it probably would work just fine, as long at the min. output is below 3.5V or so.

-awr