# Very simple circuit (LED and battery)

#### photastic27

Joined Apr 12, 2016
1
I'm asking you all a very easy question relative to what is usually posted on here. I have a lot of "book-knowledge" on circuits, etc. but not much practical knowledge. So I thank you for your patience... If I'm trying to make a circuit with a 3V battery (watch-coin battery) and a 5mm LED that withstands 3V as well (to make a small flashlight), and a switch too...
1) I don't need a resistor since the voltage drop around the loop = 0?
2) Why have I seen some people putting two batteries in similar devices? Does that double the voltage?
3) Let's say the the voltage of the battery was 6V, with the LED still being 3V. Would I then need a resistor to add in that extra voltage drop?

Once again, thanks for your patience and help.

#### shteii01

Joined Feb 19, 2010
4,644
2.
- Batteries in series, double the voltage.
- Batteries in parallel, double the current.

3.
The resistor is there to control the current.
The formula is: Voltage of the source (6 volts in your case) minus Forward Voltage of LED (3 volts in your case), then the difference is divided by the desired current (let us assume that you want 15 mA though the LED), the whole thing is equal to the value of the resistor you need in series with the LED.
$$R=\frac{V_S-V_f}{I_{LED}}$$
R=(6-3)/0.015=200 Ohm
So. If you want 15 mA to flow through your LED, you need to put 200 Ohm resistor in series with LED.

The current is what dictates the brightness of LED. If you want full brightness, plug the maximum current that LED can take into that formula. If you want dim LED, try 5 or 8 mA. 10 to 15 mA is normal, you get LED that is easy to see, but is not at full brightness. 20 to 22 mA is usually maximum.

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#### GopherT

Joined Nov 23, 2012
8,012
I'm asking you all a very easy question relative to what is usually posted on here. I have a lot of "book-knowledge" on circuits, etc. but not much practical knowledge. So I thank you for your patience... If I'm trying to make a circuit with a 3V battery (watch-coin battery) and a 5mm LED that withstands 3V as well (to make a small flashlight), and a switch too...
1) I don't need a resistor since the voltage drop around the loop = 0?
2) Why have I seen some people putting two batteries in similar devices? Does that double the voltage?
3) Let's say the the voltage of the battery was 6V, with the LED still being 3V. Would I then need a resistor to add in that extra voltage drop?

Once again, thanks for your patience and help.
In theory, a resistor or some current limiting circuit is needed to avoid infinite flow through the LED if battery voltage is higher than the LED forward voltage (they never match exactly so current limit would be needed - in theory). In reality, a battery is like a voltage source with a resistor in series (internal resistance). A little coin cell has plenty of internal resistance and is close enough in voltage that you don't need an external resistor. If you would have a nice lithium battery of any other size, still 3 volts, you will blow your LED - wear safety glasses when you do because it can be fun.

#### wayneh

Joined Sep 9, 2010
16,398
If you would have a nice lithium battery of any other size, still 3 volts, you will blow your LED...
That's not entirely true, if the voltage really does match up. A typical LED has a safe voltage range of under 0.5V, for instance 2.8V to 3.1V. The entire safe current range lies between those two voltages. It will barely come on at the lower voltage and will be rapidly approaching a damaging current level at the upper one. But if the battery is at 3V, it'll be fine without external current limiting, and the internal resistance of the battery could be very low. The diode itself will not pass a damaging current as long as the voltage stays below the critical level.

#### GopherT

Joined Nov 23, 2012
8,012
That's not entirely true, if the voltage really does match up. A typical LED has a safe voltage range of under 0.5V, for instance 2.8V to 3.1V. The entire safe current range lies between those two voltages. It will barely come on at the lower voltage and will be rapidly approaching a damaging current level at the upper one. But if the battery is at 3V, it'll be fine without external current limiting, and the internal resistance of the battery could be very low. The diode itself will not pass a damaging current as long as the voltage stays below the critical level.
So, by extension, can I assume that you are a believer that each LED in an array does not need its own current limiting resistor? Stated another way, you are saying, as long as the power source is within the possible range of LED's Vf (on the datasheet), that current will not all flow to the one diode with the lowest Vf and burn it out because each diode has internal current limiting?

#### Bernard

Joined Aug 7, 2008
5,555
I do not have to believe it, I see it over & over with white LED flashlights , 1,3. 5, 9, & 14 LED's in parallel on 3 AA or 3 AAA batteries. Before changing batteries on a 14 LED one noted that 2 LED were out, 3 were med bright, & remainder were dim. Put in 3 AAA's & all are bright-- the medium bright ones did not blow.

#### wayneh

Joined Sep 9, 2010
16,398
[responding to Gopher]

Nope, just that a single LED can be powered with a constant voltage as long as that constant voltage is controlled well and not too close to the upper edge. Once you introduce variance in LEDs, you get the Chinese flashlight effect.

#### tranzz4md

Joined Apr 10, 2015
310
Go to Sparkfun theyve got all the theory, practice, and calc stuff, and parts too for LED circuits.

If youve got an android get the electrodroid app.

Between those 2 youll learn a lot about LED circuits, especially about practice vs. theory.

#### dl324

Joined Mar 30, 2015
10,946
I do not have to believe it, I see it over & over with white LED flashlights , 1,3. 5, 9, & 14 LED's in parallel on 3 AA or 3 AAA batteries.
Those designs trade off reliability for cost savings. They don't care if current hogging causes a cascading failure. They want you to buy another one. If they have a warranty, they're counting on human nature to reduce warranty claims.

Most engineers would opt for reliability and pay the (reasonable) cost. But if you're selling a million LED flashlights annually, each penny reduction in BOM is \$10,000.

When I was a teen, we use to joke that a Cadillac was just a Chevy with lock washers...