# Basic parallel LED question

Discussion in 'General Electronics Chat' started by cabers, Jul 30, 2013.

1. ### cabers Thread Starter New Member

Jul 30, 2013
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Hi all, new here and grateful of this wonderful group of helpful brainiacs!

To jump right into it, I have a project that involves surface mount LEDs that i would like to power via 2 cr3032 (cmos type) batteries. These have a voltage of only 3v so it looks like I am stuck wiring the LEDs in parallel, given that the batteries will be in series.

Could anyone tell me how many LEDs I could power for a reasonable amount of time (~4hrs) if i choose to use .5mA/3v LEDs? also for bonus points could you tell me what value resistors I would need to limit the current (if any)? Formulas are more than welcome, as id love to be able to understand this beyond this one application- i just havent been able to turn up a solid explanation on google...

Thank you in advance for any and all input!
cabers

2. ### MrChips Moderator

Oct 2, 2009
12,652
3,461
You only need two formulas:

1) Ohm's Law: I = V/R

2) Battery Capacity = Ampere-Hours = I(amps) x time(hours)

3. ### wayneh Expert

Sep 9, 2010
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3,256
Are you sure of the 0.5mA? That's low, and that won't be terribly bright.

Applying the formulas is a little more complicated. The LEDs have a forward voltage drop specification. You subtract that from the supply voltage, and then you apply ohm's law to drop any residual volts across the resistor at the targeted current flow. An additional complication is that the battery has an internal resistance as well, and will limit current to some degree depending on the current level compared to the battery capacity. This is often not an important factor but it is relevant when the battery voltage is barely enough to light the LED.

In other words, you can't finalize the resistor until you have detailed information on the LED and the battery.

4. ### cabers Thread Starter New Member

Jul 30, 2013
10
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Ok, so if I have this right, there is something that's is referred to as a foe ard drop voltage that is seperate from the forward voltage? And this drop cannot be avoided by wiring in parallel?
Sorry for the lack of understanding
Cabers

5. ### LDC3 Active Member

Apr 27, 2013
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The "foe ard drop voltage" and "forward voltage" are the same thing. Different people use different terms for the same thing, For diodes and LEDs, there is an impedance from the device that appears as a voltage drop instead of resistance. For your LEDs, the drop in voltage for current flowing in the forward direction is 3V. It is much higher in the reverse direction.

6. ### cabers Thread Starter New Member

Jul 30, 2013
10
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So I can only expect to light one led with a forward voltage of 3v of the battery I'm using is a 3v button cell?

7. ### LDC3 Active Member

Apr 27, 2013
920
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No, because you need to limit the current through the LED or you will quickly discharge the battery and overheat the LED. Also, the 3V is an indication of the forward voltage, it can vary substantially. You would need at least 4V for the power supply and a resistor to limit the current, about 100Ω. On the up side you can have several parallel resistors and LEDs if your batteries can supply the current for each network.
Do you have the datasheet or the exact LEDs you are using so I can tell you exactly what you need.

8. ### wayneh Expert

Sep 9, 2010
12,405
3,256
The situation might work because of the battery's internal resistance and the slight positive influence of current on Vf of the diode. I think LED-lit key fobs are simple battery + LED, no resistor needed. But I wouldn't want to rely on this without testing. Use a resistor to start, and see what happens to LED current as you reduce the value. It might not light at all even with no resistor.

9. ### cabers Thread Starter New Member

Jul 30, 2013
10
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thanks for the input, i guess i have a bit of reading to do. i was under the impression that for components in parallel you divide the total current draw between the number of like components. and when in series you divide the total voltage of the supply by the number of like components. anyhow, heres the datasheet: www.hebeiltd.com.cn/led.datasheet/PLCC2LW6CT.pdf

another thing: ive been told it unwise to stack batteries in parallel (unbalancing effect?) but would that situation work here (using two 3v watch batteries) to effectively boost my voltage to 6v?

10. ### LDC3 Active Member

Apr 27, 2013
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OK, your LEDs have a typical forward working current of 25mA with a typical forward voltage drop of 3.2V. The manufacturer should reject any LEDs with a forward voltage drop less than 3.0V or greater than 3.4V. Since you want to have several strings, you could probably drop the current down to 15mA and still have the LEDs fairly bright.

I found this current source that you might want to incorporate, especially if you use a low-drop out voltage regulator.

11. ### BobTPH Active Member

Jun 5, 2013
807
121
A CR2032 battery has about 200mA hour capacity. So at the very best, you would get 8 hours driving one LED at 25mA. In reality, you will probably not get half of that, because CR2032 is designed to supply only 0.5mA, not 25mA. In fact, you might not even be able to get 25mA out of it. Running multiple LEDS will, of course shorten the life.

If you use the two batteries in series (not parallel as you suggested) you will be starting with 6V. You could then calculate a resistor to get 25mA as follows:

voltage across resistor is 6V - 3.2V = 2.8V

V = I R
R = V / I
R = 2.8 / 0.025 = 112 Ohms, use 120, the nearest standard value.

Bob

12. ### LDC3 Active Member

Apr 27, 2013
920
160
Since you need more current than what a CR2032 can produce, you can try the CR123A type of battery. It is also a lithium battery and about 1/2 the size of a AAA cell. It can provide up to 1.5A continuously.

13. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
The CR3032 battery is nominally 3V, 500mAh, 0.2mA coin cell, so putting two of them in series will give you about 6V. Here is Panasonic's datasheet for their offering:

http://www.panasonic.com/industrial/includes/pdf/Panasonic_Lithium_CR3032.pdf

Notice that even under a light draw of under 500μA the battery voltage starts out below 3V (~2.9V) at room temperature (20°C)

I don't know where you got the ".5mA/3v LEDs" numbers from. I've seen LEDs in the 2mA range, but haven't seen any below that (but I haven't looked very hard, either). I would think that 5mA to 10mA would be more typical, but 2mA would probably be a working minimum (unless, of course, I'm wrong and sub-milliamp LEDs are available).

The LED datasheet you posted in a latter post shows a maximum current of 25mA, so that would be consistent with an operating current in the 10mA to 20mA range. Let's use 10mA. At 10mA the datasheet shows the forward voltage as being 3.15V. Since the specs allow for 3.0V to 2.4V at 20mA, we should allow for the same ±0.2V here, so 2.95V to 3.35V. Let's give ourselves a bit of a margin on top of this and call it 2.9V to 3.4V. As long as you are planning on operating around room temperature, that is probably good enough.

The bottom two charts on the battery page make it pretty clear that this battery is not intended for the kind of load you are considering. The middle charge (battery voltage at half capacity versis load resistance) only shows load up to 30μA. Assuming the same trend continues (and this may be overly optimistic), the output voltage at 10mA will probably be down around 2.4V to 2.5V.

Worse, the bottom chart shows that the battery capacity starts dropping of sharply above currents in excess of about 1.5mA. By the time you get to 10mA, you may easily be well below half capacity.

I would recommend using a different cell. If you want it to last for 4 hours and you need 10mA per LED, then you need a battery that has a usable capacity of 40mAh/LED, call it 50mAh/LED. If you use the CR3032 you will probably ge the life that you need out of it if you power just one LED and perhaps two, but if you try to draw 30mA or more you may be so far out on the graphs that you waste nearly all the energy in heating the battery

Another option to consider is using a joule-thief type circuit to draw as much energy out of the battery as possible. That might allow you to run from a single cell, but it won't solve the current draw problem (will actually make it worse), so again you are probably best to choose a difference battery type.

14. ### cabers Thread Starter New Member

Jul 30, 2013
10
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Thank you all so much for the helpful input! I didnt plan out my power source other than, "hey, ive got a bunch of these lying around" the cheapo got the better of me. thanks for the reccomendation for that 1.5mA watch batt. - i think ill order a few and experiment!
Thank you all again!
Cabers

15. ### cabers Thread Starter New Member

Jul 30, 2013
10
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Oh just looked at the images for a cr123a and realized that its cylindrical like a AAA. anyone know of any button cell batteries that can supply more than 50mA for an extended period?

16. ### BobTPH Active Member

Jun 5, 2013
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If an extended period is measured in minutes, sure.

Bob

17. ### ian field Distinguished Member

Oct 27, 2012
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792
Many discount stores sell keyfob lights with 2x CR2016 directly driving a blue/white LED, there's no current limit resistor - it relies on the cells internal resistance to protect the LED. The battery doesn't last all that long, and isn't worth replacing as it didn't do the LED much good when it was fresh.

The 2x CR20xx cells only gives you a few volts headroom, so even if you use a current limiting resistor - it will be small, and the brightness very sensitive to the discharge curve.

18. ### WBahn Moderator

Mar 31, 2012
18,093
4,920

Don't make us read between the lines, compile information from multiple posts, or resort to crystal balls and mind reading.

So, you want button cell batteries that can supply 50mA for at least four hours. If you find a button cell battery that can supply 50mA for at least four hours will that meet your requirements? If not, what will?

Coin cells typically have high internal resistance -- a few dozen ohms -- which severely limit peak draw. In many cases your short circuit current isn't going to be 50mA, since that requires only a 30Ω internal resistance.

These cells are generally intended for applications that draw tens of microamps and only consume milliamp-scale currents in short bursts and frequently use capacitors to provide for these transient current needs. So you are asking them to continuously deliver 1000 times the current they are intended to.

It appears that you are likely operating so far outside the intended envelope that you will need to buy a bunch of batteries and simply test them.