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Components: 9 V battery (actual voltage is around 7 V). No resistor is used. Only red LEDs connected to a battery.

There are 2, 3, and 4 LEDs connected in these circuits with current readings. I can see that with 2 LEDs, I get 14.28 mA. With 3 LEDs, I get 3.19 mA, and with 4 LEDs I get 4.7 mA.

In real life, I know the answers to the current of these 3 circuits. But if I didn't have a multimeter, how would I calculate to find the current?

I'm aware of Ohm's law but I can't find an example that doesn't include a resistor, which is my case.

I'm a total beginner, so I appreciate any advice. Thanks!

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    \$\begingroup\$ You actually do need to apply Ohm's law, because the battery is not ideal and its non-ideality is mathematically modeled as a series resistor (the battery's "internal resistance"). You would have needed a multimeter (or spec sheet for the battery) to find it anyway. If there were no internal resistance, the LEDs would have failed from overcurrent right away. \$\endgroup\$ – nanofarad Oct 1 '20 at 18:15
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    \$\begingroup\$ Estimate the internal resistance of the battery (likely \$R_\text{BAT}\ge 2\:\Omega\$) and the bulk series resistance sum for your LED(s), \$R_\text{LEDS}\$, and form \$R_S=R_\text{BAT}+R_\text{LEDS}\$. Then compute \$I_T=\frac{\eta\,V_T}{R_S}\$. You already know \$V_\text{BAT}\$, but divide it by the number of series LEDs first and call that \$V_\text{LED}\$. Now just compute$$I_\text{LED} = I_T\cdot \operatorname{LambertW} \left( \frac{I_{\text{SAT}}}{I_T}\cdot e^{^{\left[\frac{V_\text{LED}}{\eta\:V_T}-\frac{I_{\text{SAT}}}{I_T}\right]}}\right)-I_{\text{SAT}}$$Exact, every time. \$\endgroup\$ – jonk Oct 1 '20 at 18:36
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    \$\begingroup\$ There's probably alot more resistance there than meets the eye. You have your battery contacts 'barely' making a contact, looks like they're just resting in place via gravity. Breadboards are notoriously finicky. Every metallic connection is an opportunity for more parasitic resistance. Plus your METER is intentionally adding some resistance ... There's a current-sense resistor internal to it (check your manual for its value). 2 LED's wired direct to a fresh 9V battery should have melted down pretty quick. \$\endgroup\$ – Kyle B Oct 1 '20 at 18:38
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    \$\begingroup\$ You'll need three parameter values for each LED, though: \$I_\text{SAT}\$, \$\eta\$, and \$R_\text{LED}\$. You can get these with three simple measurements. Get a current source that you can vary and a voltmeter. Put through three currents at, say, \$10\:\text{mA}\$, \$15\:\text{mA}\$, and \$20\:\text{mA}\$ and measure the voltage. This gives three equations and three unknowns, which you can readily solve. The simultaneous solution will be surprisingly accurate, too. \$\endgroup\$ – jonk Oct 1 '20 at 18:42
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    \$\begingroup\$ A 9 volt battery that measures 7 volts with no load is VERY dead - it will have a very high internal resistance, so the voltage at its terminals will vary widely depending on the load you apply. \$\endgroup\$ – Peter Bennett Oct 1 '20 at 18:50
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You actually do need to apply Ohm's law, because the battery is not ideal and its non-ideality can be mathematically modeled as a series resistor (the battery's "internal resistance"). If there were no internal resistance (i.e. if the battery were an ideal 9V source) the LEDs would have failed right away from massive overcurrent.

You would have needed a multimeter (or spec sheet for the battery) to find it anyway, and it so happens that your experiment gives us enough data to find it (with some assumptions).

If we assume that the LEDs are identical and well-modeled as a constant voltage drop, we can denote \$V_f\$ the forward voltage of one LED, and \$R_{batt}\$.

The results for 3 and 4 LEDs look a bit fishy since current increased when you went to four LEDs. I'll ignore the four-LED result as an outlier for the following calculation, but recommend that you try to investigate it further.

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Using the one- and two-LED circuits we can set up the following system of equations:

$$ 7\,[\text{V}] - 2V_f = 14.28\,[\text{mA}] \cdot R_{batt}$$ $$ 7\,[\text{V}] - 3V_f = 3.19\,[\text{mA}] \cdot R_{batt}$$

Solving, we get \$V_f \approx 2.1\,[\text{V}]\$, which is reasonable. However, we also find that \$R_{batt} \approx 192 \,[\Omega]\$ which is rather high. Presumably, the battery is already mostly discharged, or the currents are high enough that the simple model of a \$V = IR\$ resistor in series with the voltage source is no longer accurate due to electrochemical effects under heavy load.

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You can get a reasonably good approximation using a load-line chart.

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Figure 1. An LED load line chart for single LEDs running on a 5 V supply. Source: Loadline resistance graphic tool.

How it works:

  • Draw a line from your supply voltage on the X-axis, 5 V in Figure 1, ending at a point on the current axis. Mark that line with the resistance calculated from \$ R = \frac V I \$.
  • You can now decide what current you want through the chosen colour LED and find which resistor gives the closest match. For example if we want 10 mA through a green LED we can see that 270 Ω would be the closest.
  • The tool can be used in reverse to find the resulting current for a given resistor / LED combination.

In your case you have the added complexity of having multiple series LEDs. This means that you'd have to add the curves for each of your chosen LEDs. If I had to do this often I would consider a spreadsheet application to create the graphs and the loadlines to work it out - if I wanted a graphical tool.

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I'm aware of Ohm's law but I can't find an example that doesn't include a resistor, which is my case.

Normally, you should never drive a LED from a low-impedance constant-voltage source (battery, lab. type power supply, off-line adapter, etc.) without a resistor. Without a resistor, there'll be an extreme current flowing through the LEDs, which finally leads the LED to burn. So, there should always be a current-limiting resistor connected in series. Fortunately, in your circuit, the current-limiting resistor is the internal resistance of the battery (Remember that the internal resistance of an alkaline battery increases as the battery drains).

Here's the real schematic of your circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

I can see that with 2 LEDs, I get 14.28 mA. With 3 LEDs, I get 3.19 mA, and with 4 LEDs I get 4.7 mA.

Ohm's Law applies. Let \$\mathrm{V_F}\$ is the voltage drop for each LED (assuming they are the same brand/model and color) and \$\mathrm{n}\$ is the number of LEDs connected in series:

$$ \mathrm{ V_{BAT}=I_{LED}\cdot R_{int} + n \cdot V_F } $$

NOTE: I also have to say that the internal resistance of the ammeter (called burden or shunt) is connected to the circuit. The net resistance of the circuit may increase so this may change the behavior (i.e. LED current) a bit. But let's neglect this for now.


If you drive the circuit from a low-impedance source (examples are given above) then there should be a current-limiter resistor connected in series. So the circuit does not change but the series current-limiter resistor, \$R_s\$, comes in place of \$R_{int}\$. With the equation above, you can

  • either calculate the value of the series current-limiter resistor according to the desired LED current,
  • or calculate the LED current for a known value of the series current-limiter resistor.

Remember that the \$\mathrm{V_F}\$ of the corresponding LED(s) should be known. It may not be that easy to know that parameter exactly, but I remember that @Transistor has shared a nice graph for that purpose in one of his answers (Probably to your previous question).

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