Question for the electronics gurus (you know who you are)

[Lemosteam]

My son and I are not understanding something.

We have a simple circuit.

A nine volt battery, with a 100 ohm resistor on the positive terminal to the anode of a small peanut LED (0402) encased in lacquer, and a wire from the cathode to the negative terminal of the battery. When we measure voltage between the battery + and the resistor to the negative on the battery, the the measured voltage is 5.4v. When measured AFTER resistor before the LED, to the negative on the battery the voltage was 3.06V. Could you please explain why the voltage isn’t 9V across the battery? When I measure the voltage across the battery without the attached circuit, it’s approximately 8V (which makes sense because it’s been discharging for a little while), but when the battery is connected to the diode-resistor circuit, the voltage across it drops to about 5.4V.

Thanks in advance!

[nickelplate759]

I'm no electronics wizard - there's a reason my entire career has been in software - but I think I know at least the resistor part of the puzzle.

Ohm's law states V=IR.  Your resistor is 100 Ohms.  The voltage differential across it is about 2.4V.  That suggests your resistor is seeing .024 Amps.  For power, W = VI :  0.24*2.4. = 0.576 Watts, which seems pretty reasonable.

Now, LED's have weird voltage drop characteristics that I don't understand, but I'll bet that explains the difference between 5.4V and 8 V.

[John]

I'm no electronics wizard - there's a reason my entire career has been in software - but I think I know at least the resistor part of the puzzle.

Ohm's law states V=IR.  Your resistor is 100 Ohms.  The voltage differential across it is about 2.4V.  That suggests your resistor is seeing .024 Amps.  For power, W = VI :  0.24*2.4. = 0.576 Watts, which seems pretty reasonable.

Now, LED's have weird voltage drop characteristics that I don't understand, but I'll bet that explains the difference between 5.4V and 8 V.

LEDs are basically a diode .. most diodes give you about 1V drop .. someone correct this if I'm wrong .. 

Mr. Google returned this ..

What is the voltage drop across an LED?
Typically, the forward voltage of an LED is between 1.8 and 3.3 volts. It varies by the color of the LED. A red LED typically drops around 1.7 to 2.0 volts, but since both voltage drop and light frequency increase with band gap, a blue LED may drop around 3 to 3.3 volts.

[Maletrain]

Lemosteam,

The short answer is that drawing current from a battery causes the battery voltage to drop, compared to what it reads when there is no current being drawn.

The longer answer is:

As I understand the question, you are measuring the voltage between the positive and negative contacts on the battery when there is nothing connected to the battery and getting 8 volts, and then connecting a resistor and LED in series between those contacts and finding that the battery voltage drops to 5.4 volts.

You also measured the voltage in the circuit after the resistor and got 3.06 volts.  So the voltage drop across the 100 ohm resistor is 5.4 - 3.06 = 2.34 volts.  That indicates that the current going through the circuit is 2.34 volts/ 100 ohms = 0.0234 amp.  That seems reasonable, considering that 0402 LEDs seem to run about 0.020 amps current.

So, your battery voltage is dropping by 8 - 5.4 = 2.6 volts when it is supplying 0.0234 amp of current.  That seems like a pretty weak battery to me.  Try it with a new battery, and I think you will see higher voltages. 

But wait, that strongr battery will also put a higher voltage across your LED, and that might cause trouble - LEDs subjected to too-high voltage promptly burn out.
 
With the battery you are using, the LED is getting that 3.06 volts across it, which is about right for that type of LED.  Since it will continue to pass only about 0.020 amp even at higher voltage, the voltage drop in the resistor will stay at about 2.34 volts, so a stronger battery will tend to make the voltage across the LED higher.  If you get a really strong 9.4 volt battery, the voltge across the LED could be up to 9.4 - 2.6 = 6.8 volts.  And the max voltage recommended for white LEDs like I think you are using is only 5 volts, some red ones are substantially less.  So, maybe best to do the experiment with a strong battery by using a higher value resistor.

The LED will continue to draw something near 0.0234 amp at different voltages, so using a 200 ohm resistor should produce 0.0234 amp x 200 ohms = 4.68  volts drop across the resistor and leave 9.4 - 4.68 = 4.72 across the LED.  That seems too close to 5 volts, because we are just using SWAG numbers here instead of the actual data sheet values for yout LED.  So, probably better to use a 250 ohm resistor and you get 5.85 volts drop, leaving 9.4 - 5.85 = 3.55 across the LED, which is close to what you have with the weak battery and the 100 ohm resistor.

There should always be some voltage drop between the battery terminals when comparing a loaded condition to an open-circuit (no load) condition.  Hw much drop depends on the size and condition of the battery. 

Another test you could do is to put 2 AAA batteries in series to get about 3 volts, and apply that to the LED with no resistor.  Look at the voltage drop across the pair of batteries.  Then add another pair of AAA batteries in parallel with the first pair, so that you basically double the capacity of the batteries feeding the LED, and see how the voltage drop is somewhat less.  (I can't predict how much less without getting the specs for the batteries and the LEDs).

[Point353]

See the graphs for battery terminal voltage versus operating time at various values of load current in the following datasheet:
https://www.duracell.com/wp-content/uploads/2016/03/MN1604_US_CT1.pdf

[peteski]

Yes a battery is non-ideal voltage source. Ideal voltage source would have zero internal resistance, so regardless of the current drawn from it, it would supply constant voltage.  A 9V battery has internal resistance, so the more current you draw from it the more its voltage drop.  The  9v battery has fairly high internal resistance (it is constructed of 6 small cells in series), so even if you shorted the terminals the current would be limited by its internal resistance.

I also highly discourage using a 100 ohm resistor in that circuit.The poor 0402 white LED is grossly overdriven with current.  I'm surprised that it has survived that long!

If the battery could supply steady 8V (Vss) and the white LEDs forward voltage (Vf) was 3V (which is typical for  white LEDs) then the current passed through the diode would be I=(Vss - Vf)/R or (8-3)/100=0.05A (or 50mA). For a typical 0402 LEDs the recommended current is 5mA and absolute maximum current is 20 or 25mA!  Even when the battery voltage (do to the heavy current being drawn by your circuit) the voltage drops to 5.4V then as you calculated the current is (5.4-3)/100=0.024A or 24mA.  That is  still quite high for a small 0402 LED.

Also, what (fresh) 9V battery produces only 8V in open circuit?  Most fresh alkaline 9V batteries have open circuit voltage of 9.4V, and standard inexpensive heavy-duty carbon-zinc batteries have it right around 9V.

[jagged ben]

The last part of Pete's post is probably it.  8V means a quite dead 9V battery. 

If so, the battery can't maintain anything close to nominal voltage with any current flowing.  A dead battery behaves far less like an ideal voltage source than a fresh/fully charged battery.   This would also explain why the LED hasn't fried, as Pete pointed out.

I suspect with a fresh battery, you won't see such a stark difference in voltage when the circuit is closed.  And you may also fry the LED.