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I have a circuit powered by 9V battery, I'm using two 2V LEDs. If I connect the LEDs in series with 10k Ohm resistor, the voltage before resistor is 9V, after the resistor it's 3.3V and after the first LED it's 1.65V. Both LEDs are dim, since they are at voltage lower than they are supposed to. If I connect them in parallel from the resistor, after the resistor it's 1.65V for both of them, they are still dim.

How is this voltage difference possible? I am aware of the Ohm's law, but it still doesn't make any sense to me. In my mind, a resistor can't change it's resistance, it's 10k Ohms no matter what.

Q: How is it, that I get different voltage after the resistor, depending on how the LEDs are wired? Why isn't resistor's behavior constant?

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  • \$\begingroup\$ The 9 V battery has an equivalent resistor of ~50 ohm in series with it. If you repeat your experiment with a stiff/ideal 9 V source, the phenomenon goes away. \$\endgroup\$ – winny Jul 16 '16 at 17:00
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    \$\begingroup\$ so in both cases the voltage drop across each LED is 1.65, this is so obvious since 1.65 is the forward voltage needed to turn the LED on, in case you want a higher light intensity you should lower the 10k ohms series current limiting resistor \$\endgroup\$ – Elbehery Jul 16 '16 at 17:11
  • \$\begingroup\$ This doesn't seem like a phenomenon to me, I probably just lack basic knowledge. I read that in series, voltage differs in a circuit, while in parallel, current differs. All I'm wondering is - why? I imagine electrons running through resistor and it doesn't make any sense to me that depending on how load is wired in a circuit, the resistor applies different resistance on the electrons. \$\endgroup\$ – masiton Jul 16 '16 at 17:12
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    \$\begingroup\$ @masiton that series/parallel theory is only true for simple resistive circuits. LEDs don't play by those rules. \$\endgroup\$ – brhans Jul 16 '16 at 17:19
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    \$\begingroup\$ I removed the picture, because it was inaccurate. Both circuits use two LEDs. The series connected LED circuit has 3.3V after the resistor and 1.65V after first LED, the parallel circuit has 1.65V behind the resistor and applied to both LEDs. \$\endgroup\$ – masiton Jul 16 '16 at 21:05
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) Just a resistor. (b) Series LEDs. (c) Parallel LEDs.

Q: How is it, that I get different voltage after the resistor, depending on how the LEDs are wired? Why isn't resistor's behavior constant?

First consider the resistor on its own - Figure 1a. From Ohm's Law we can calculate the current through it as \$ I = \frac {V}{R} = \frac {9}{10k} = 0.9~mA \$. This is the maximum we can get and adding LEDs will reduce it.

Looking at Figure 1b you can see that the same current goes through all the elements in the circuit.

enter image description here

Figure 2. A typical LED graph of current vs forward voltage. Note that even at tiny currents (< 1 mA) the forward voltage drop doesn't change much.

We can see from Figure 2 that the 1.65 V reading you are getting across each of the LEDs is correct.

Now what happens when you connect the LEDs in parallel? We will still be limited to a maximum of 0.9 mA through the resistor but now this has to split two-ways through D3 and D4. They will get about 0.45 mA each. Checking back to Figure 2 we can see that they will still drop about 1.6 V or so.


Note that adding the LEDs reduces the voltage across the resistor and therefore reduces the current available. For Figure 2b the actual current will be \$ \frac {9-3.3}{10k} = 0.57~mA \$. For Figure 2c it will be \$ \frac {9-1.65}{10k} = 0.735~mA \$ which helps a bit when the current is shared.


[From OP's comment:]Going back to analogies, I often see circuit being described as tube, current is the water in it, pump is voltage and garden hose is the resistor, lowering the voltage and limiting the current.

The water analogy is not great, but let's try. To resemble an electrical circuit the water will have to flow in a closed loop such as a hot water central heating system. The hot water pump raises the pressure at the outfeed of the pump (the battery raises the voltage). We feed through the pipes (the resistor) to two check valves which piped in series or in parallel.

enter image description here

Figure 2. The check-valve is the plumber's diode. Water can only flow one way. A certain pressure is required to overcome the spring. This causes a pressure drop across the valve but once the valve is open the pressure drop doesn't change much even at high currents. Source: Modified from Gentec image.

Now let's put some numbers on our system. The pump raises the pressure to 9. (Think psi, bar, pascals, inches of mercury or whatever you like.) The valves will open at 1.65. If we pipe them in series as in Figure 1b it should be clear that if the valves open the pressure reading at the top of D2 will be 1.65 and above D1 it will be 3.3. If we connect them as shown in Figure 1c they will both open when the pressure reaches 1.65.

1) Why, when in a series, doesn't first LED consume 2V and the second LED is not left with just the rest?

LEDs and resistors don't "consume" voltage in the same way your hose doesn't consume pressure. Voltage is dropped or "voltage decreases" across the resistor or LED in the same way that pressure losses occur in the water system.

2) When you say across the resistor, you mean this is the current inside it, but after it, it gets lowered by the LEDs themselves, because they are also, in fact, resistors, and that's why they get 0.45mA each, instead of the 0.57mA?

While they have resistance we do not call them resistors because they are non-linear just as the check-valve is. Also the current doesn't "get lowered" because what leaves the battery on one terminal must come back on the other.

I hope that helps.

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  • \$\begingroup\$ Thank you for explanation. Going back to analogies, I often see circuit being described as tube, current is the water in it, pump is voltage and garden hose is the resistor, lowering the voltage and limiting the current. 1) Why, when in a series, doesn't first LED consume 2V and the second LED is not left with just the rest? 2) When you say across the resistor, you mean this is the current inside it, but after it, it gets lowered by the LEDs themselves, because they are also, in fact, resistors, and that's why they get 0.45mA each, instead of the 0.57mA? \$\endgroup\$ – masiton Jul 16 '16 at 21:26
  • \$\begingroup\$ See if the update helps. \$\endgroup\$ – Transistor Jul 16 '16 at 22:14
  • \$\begingroup\$ @masiton On a slightly different note, don't connect LEDs in parallel. Each LED string (even if it consists of a single LED) should have it's own current limiting resistor. More on that in this thread: Why exactly can't a single resistor be used for many parallel LEDs? \$\endgroup\$ – Nick Alexeev Jul 17 '16 at 1:05
  • \$\begingroup\$ The Arduino Starter Kit textbook does the same - each LED connected in parallel has it's own resistor. For the updated answer - thanks, this makes it easier to understand! \$\endgroup\$ – masiton Jul 17 '16 at 11:17
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It has to do with the logarithmic nature of the current vs. voltage in a forward biased diode.

Since they are non-linear elements, the voltage increases logarithmically with increasing current (and the current increases exponentially with increasing voltage).

You are underdriving the LED's, in both situations. So the diodes are barely turned on. They probably need more like 5-15mA each.

Let's take a look at each case.

Series case:  (9-3.3)V/(10Kohm) = .57mA through both LED's
Parallel case: (9-1.65)V/(10Kohm) = .7mA total, or .35mA per LED

In both cases the current is very low.

To prove this to yourself, start with basic diode equation:

$$ I_d = I_s*(e^{qV_d/kT}-1) $$ Back solving the ideal diode equation for voltage in terms of current: $$ V_d = ln(I_d/I_s)*kT/q $$

You can see that it takes a large increase in current to increase the voltage much. Also check out this plot for a random LED.

Let's just plug in our numbers from before (hard to see difference):

Series - .57mA => 1.0423V
Parallel - .35mA => 1.02V

This is a very low change in voltage. Since it is a random LED (I just googled 'logarithmic LED current voltage' it obviously will not match your numbers. But the trend will be similar among all diodes.

v-i plot for random LED

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You are looking at the wrong element in the circuit as to why the voltage is different at that node.

The LEDs themselves are the ones with "adjustable resistance", the resistor is constant but your LED configuration is changing, and LEDs are non-linear devices (meaning they can change depending on various factors).

The 3.3V after the resistor is the two LEDs forward-voltage drops in a row, when the LEDs are in series, and the 1.65V is a single LED forward drop. I assume these are red. One more thing - 10k Ohms as the ballast resistor is very high, so they should be very dim indeed. I don't know why you chose that, but if you wanted to see brighter LEDs, use a resistor like 1k or 1.5k Ohms.

The resistor remains the same resistance but the way you configure the LEDs changes the load, and therefore the current through the resistor changes which is how you see the voltage difference. When LEDs are in series, they share the same current and their forward voltages combine. If you were to calculate what resistor to use for a set current, you use the remaining voltage between the supply and all of the series LED voltages (9v - 3.3v = 5.7v) and use Ohms law to solve for a required current. Lets say 5mA is the target current:

R = V/I, R = 5.7/0.005 = 1140 Ohms.

If you had the two LEDs in parallel, you know their forward voltages in parallel are "the same"(this is not actually true, it's based on manfacturing tolerance, temperature, etc), so you say the voltage at the node is 1.65V. If we want 5mA going though each LED, you need 10mA through the resistor, and then hopefully LEDs will equally share 5mA each. The resistor in this case, is found like so:

R = V/I, R = (9-1.65)/0.01 = 7.35V/0.01 = 735 Ohms.

See how rather than expect the original resistor to make the LEDs act equal, we work backwards once we know our LED configuration? This is because the LEDs are NOT completely static/unchanging devices. In fact, as forward current increases, their forward voltage does too. See the datasheets for the forward voltage drop of the LED for a given forward current.

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When a LED is forward biased, it has a roughly constantly forward voltage drop. In your case, that's roughly 1.65 Volts.

When the two LEDs are in parallel, there is a 1.65 volt drop across either or both LEDs since they're connected to the same two points. With a 1.65 volt drop across the LEDs, that leaves about 7.35 volts across the resistor. Using Ohm's law, we can compute that about 0.7 mA will flow through that resistor. It will roughly split with half going through each LED. That's pretty dim.

With the two LEDs in series, there is a 1.65 volt drop across each one. That leaves 5.7 volts across the resistor. Using Ohm's law, we compute that about .5mA will flow through the two LEDs (all the current flows through both LEDs.) Sadly. .5mA is still barely enough to light an LED.

The resistor is acting exactly the same in both cases. The ratio of current to voltage is identical in both cases.

To get more current to flow through the LEDs, you need to use a smaller resistor.

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