# Testing Volts and Amps in my Circuit

Discussion in 'General Electronics Discussion' started by Dustin Smith, Jan 2, 2013.

1. ### Dustin Smith

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Jun 27, 2012
My question is, why are my reading so close together?
I was under the impression that if I used 1 white LED rated at 3.6V, 20mA, and added another of the same in parallel, that it would double the current. But this is not the case.

The Circuit..

1-8 white LEDs in parallel with the resistance controlled by a potentiometer set to 70 ohms. (70 ohms being the recommendation by an LED wiz calculator at 5.0V)

I took readings across the LEDs for voltage and after the power supply board for the current. So I believe the current readings are what the potentiometer and LED(s) were using and not included the power supply. Please let me know if I'm measuring incorrectly.

# of LED mA V
8 29.85 2.89
7 29.73 2.9
6 29.39 2.93
5 28.69 2.96
4 28.51 3.0
3 27.74 3.05
2 24.65 3.26
1 23.54 3.45

And to add a further note, I did a test of the current in AC where the power is being supplied and I got a current reading of 8 LEDs of only 27mA. It's less than what my reading said I was using in DC with only the LEDs and the potentiometer! I'm lost.

2. ### BobK

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Jan 5, 2010
If you have a single LED and resistor, then add a second LED in parallel, the current will increase but not double. This is because the increased current through the resitor lowers the voltage across the two LEDs, and LEDs current is very sensitive to the voltage across them.

Bob

3. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Yeah, my guess would have been that the current would increase by less than 10% going from 1 LED to 2, with that decreasing toward zero as the number increases.

This is due to 2 effects:

1) the series resistor will limit the change in current for any load (even one with a normal linear resistance)

2) LEDs are not a linear device, they have a Vf that varies relatively little with Vf.

A way to do this would be to reduce the 70 ohm resistor as you added LEDs. So with 2 it would be 70/2 ohms, with three 70/3, etc.

HOWEVER that is a bad thing to do because placing LEDs directly in parallel is asking for trouble (thermal runaway) unless they are in very intimate contact and their junction temperatures are very similar.

The correct way of doing this is to either place the LEDs in series (and reduce the resistor value to maintain the same current, OR to place each LED and its own resistor in parallel, OR do both of these in combination.

It may be useful for you to take a look at this:

https://www.electronicspoint.com/got-question-driving-leds-another-work-progress-t228474.html

Section 2 directly addresses your problem (I believe it is the first "don't do this" image). Sections 0 and 1 provide some more background.

4. ### Dustin Smith

52
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Jun 27, 2012
Bob, so let me see if I understand this correctly. Let's say I put a White LED of 3Vf at 20mA with 2 AA batteries (fully charged and placed in series giving me 3V). Then it should be drawing around 20mA (as long as the voltage holds out). Then, if I added another in parallel, it should then just about double the current, because there is no resistor to limit the current anymore. Correct?
Of course, this would go against running the LEDs on a current source as I understand it, but I'm just trying to understand, not wanting to do it in practical application.

Steve, thanks for the redirect to the LED page you posted. I have in fact viewed it before, but apparently I didn't remember enough of what I read. Thanks for the review. Bringing me to a new question or maybe more.

How frequent is thermal runaway?

Also, from what I can understand, the best way for me to drive a simple low powered LED or array, would be to use a linear voltage regulator along with a resistor for each series of LEDs, and put those series in parallel. Then I would be driving my LEDs from a current source with a regulated voltage to boot. Does this match what I learned from your lecture?

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
No, they may draw 5mA, or 20mA, or 100mA, You just can't tell. A few tenths of a volt either way can easily change the current by a factor of two (or even more).

Worse than that, as the LED heats up slightly, its Vf at a particular current goes down. This causes the current at the fixed voltage to rise. This generates more heat, which lowers the Vf still more, which allows more current, then more heat, more current, more heat, until *pop* -- it dies.

Well, if they were identical, yes. And then both would eventually fail (maybe)

No, it indicates why you want a current source. Firstly, you can specify the current (which is what is important). If this current heats up the LED (and it will) the Vf drops, but the constant current source compensates and keeps the current the same. Bevcause the voltage across the LED has fallen, the power dissipated in it also falls. This means the device cools. The end result is that the temperature stabilises at some (hopefully save) value

I would venture to suggest that it is the chief cause for failures in LED Christmas lights (where many LEDs are often placed in series/parallel without individual resistors).

It is also noticable wherever you see a LED light made up of a number of smaller LEDs. A really good example is the early strip lights that were placed in the spoilers of cars. You often see that the brightness of these vary greatly from LED to LED. This is again. almost certainly caused by thermal runaway hogging current in some LEDs or damaging others.

It should NEVER happen in a well designed circuit.

A voltage regulator and a resistor does not make (an ideal) current source but it makes something with characteristics that approximate a current source over a small change in the load. Even though it is quite sensitive to changes in the line voltage (hence the regulator) even that can often be tolerated.

The key in this instance is a resistor which will swamp the thermal effects in the LED. Generally having the resistor drop a couple of volts is more than sufficient at low power levels.

6. ### Dustin Smith

52
0
Jun 27, 2012
Thanks for your detailed replies Steve. I seem to have misused my words in one of the quotes. I meant that it would go against the correct way of using a current source to drive LEDs. But I've got the answers needed from my last post, thanks. However, your detailed replies give me new questions.

Steve "No, they may draw 5mA, or 20mA, or 100mA, You just can't tell. A few tenths of a volt either way can easily change the current by a factor of two (or even more)."
(sorry, I don't know how to use the quote thingy)

Why is it that you can't tell what they will draw? Is it because there is no resistor in the circuit? I've tested LEDs and found they are very close to their Data sheets, so I don't understand your reply unless you're referring to the fact there is no current source driving the LED. If this is the case, then I understand.

Steve "A voltage regulator and a resistor does not make (an ideal) current source but it makes something with characteristics that approximate a current source over a small change in the load."

If this is not an ideal current source, then I don't understand your lecture paper. The only two indications of how to get a current source that I remember reading about was 1. Through a resistor, and two, through a link that shows 4 circuits that all use voltage regulators with resistors (and capacitors which I believe they stated are not mandatory). So could you please help me understand the best known way to get an ideal current source?

P.S. I hope you don't feel I'm attacking you, I'm just prodding your brain for my knowledgeable gain, he he.

7. ### BobK

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Jan 5, 2010
Bob again, doing tag team with Steve

The characteristic Vf in an LED datasheet is only the typical value. Often they give minimum and maximum values as well, indicating a range of voltages at which the voltage might be the stated current. This is due to the variation from one device to another of the same type. For instance, if your datasheet says Vf min =3.0V typ=3.3V max=3.6V and you design for the typical, the one that has a Vf of 3.0V might easily draw double the desired current at 3.3V. The other thing to remember is that the voltage / current curve is exponential, not linear, thus a small change in the Vf (say 10%) does not mean a 10% change in the current, it might mean a 100% change in the current.

So this is where using a higher voltage and a resistor come in. If you design for the 3.3V typical and use a 6V supply and a resistor, the resistor will drop 2.7V. If you happen to get one of the LEDs that has the miniumum 3.0V, then the resistor will have to drop 3.0V instead of 2.7V. But resistors ARE linear, so this will mean only a 10% increase in current, which is probably not going to smoke your LED. The resistor is essenially providing negative feedback, if the current is above what you designed for, it ends up lowering the voltage from your design voltage and thus keeping the current in check.

A current source does this even better with negative feedback and amplification, so that any small error it the current will be corrected completely, not partially like the resistor above.

A voltage regulator can be used to make such a current source. Look at the datasheet for the LM317 adjustable regulator to see how this is done.

Bob

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Let me add that an LM317 and a resistor does indeed make a current source, but this is because the circuit *is* a current source.

When I said a voltage regulator and a resistor does not make a current source, it means that simply sticking a resistor in series with a voltage regulator does not produce one.

Clearly I need to number the figures in my LED tome, because referring you to some of the diagrams would be good here...

Bob's discussion of the relationship between Vf and If are illustrated in Image 0.3. This image shows the exponential relationship (i.e. a small change in Vf produces large and varying changes in If, or alternately, the Vf remains relatively constant with large changes in If). Also this diagram illustrates that the actual curve may vary slightly between devices of the same type, but significantly between devices of different types.

Image 3.2 shows how to wire an LM317 as a current source.

What is perhaps missing from the thread is a graph showing the effect of changes in Vf and V on If and power dissipated for different circuit configurations.

9. ### Dustin Smith

52
0
Jun 27, 2012
Ahh, so when the graph at 0.3 says random LEDs from brand A, it means random LEDs of the same rating and color, and not 3 different ratings/colors. Yes, the picture illustrates this point when you are looking at it in the right way.

The LM317 has many family members. But I think I found that LM117HV/LM317H will do what I want and more!

Bob and Steve, Thanks for advancing my education in electronics. I've learned a lot, but not enough. I'm sure I'll be returning when I find that 1+1=1 and I start scratching my head again.

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
Good point. I'll review the text.

It's a pleasure