0
\$\begingroup\$

The following circuit will cause problems because of voltage drops:

schematic

simulate this circuit – Schematic created using CircuitLab

It should be changed to look like this:

schematic

simulate this circuit

However if we used both parallel and series resistors in this circuit, there is a potential for voltage drop issues between the LEDs. As we raise the value of the series resistor and lower the values of the parallel resistors, what determines the point at which voltage drop issues start to happen (and at what values?)

In this scenario, all of the LEDs are the same brand, with a 2 volt drop +-2% and are run at 20 mA.

schematic

simulate this circuit

Edit----

Practical application: I designed my original circuit with 18 LEDs in parallel, each with a 220 ohm resister. Now, I want to change the voltage to 9 volts (or 12 volts). Will I need to switch out all of the original arrays or can I add one resistor in series?

...Then I became curious about the limits involved in doing this.

\$\endgroup\$
  • 5
    \$\begingroup\$ How many volts does a voltage drop drop if a voltage drop does drop volts? \$\endgroup\$ – CharlieHanson Apr 12 '16 at 17:48
  • 1
    \$\begingroup\$ It's easy enough - though tedious - to work through the arithmetic, so since you set up the problem why not try to find the solution yourself before wasting others' time with what seems to be frivolous nonsense? And aargh please don't dog-leg R4. \$\endgroup\$ – EM Fields Apr 12 '16 at 18:28
  • \$\begingroup\$ You might get some insight on LED forward voltage - why does series resistor take excess voltage? if you use the simple LED model in my answer. \$\endgroup\$ – Transistor Apr 13 '16 at 18:17
  • \$\begingroup\$ @EM Fields- it suffices not to respond to the question posed if you find it to be frivolous nonesense. I'd hate to discourage folks who are asking questions to gain insight, learn , or to seek guidance on how to approach problems whose solutions may not be as obvious to them... \$\endgroup\$ – jrive Apr 13 '16 at 22:20
  • \$\begingroup\$ @jrive: Then, instead of allowing yourself to be led astray by others' attitudes with which you disagree, by all means jump in there and follow Mr. Pefhany's excellent lead re. educating the querent. \$\endgroup\$ – EM Fields Apr 13 '16 at 22:35
2
\$\begingroup\$

It's not an easy calculation. You can get some idea of the worst-case situation by looking at max/min forward voltage drops, and the slope of the current/voltage curve, and the LED temperature coefficient from the LED data sheet.

Without detailed statistics on LED voltage-current-temperature characteristics it will be difficult to guess what will typically happen. Cheap consumer products often parallel LEDs directly with no resistors, but then nobody much cares if one LED is 30% brighter than the next and the whole thing only lasts a few thousand hours or less.


Edit:

Here is a simulation you can play with. I've altered the saturation current of the model for D2 to make the Vf high by 4%

40mA current splits with D1 getting 23.1mA and D2 getting 16.9mA. That's if they are all held at exactly the same temperature. If instead I assume that they are thermally independent and have a 50°C rise nominally, then the difference between the two would be 16°C and if the tempco is -1.7mV/K then that would cause another 1.7% difference between the Vf's, leading to more temperature rise etc.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • 1
    \$\begingroup\$ "Without detailed statistics on LED voltage-current-temperature characteristics it will be difficult to guess what will typically happen." Nailed. \$\endgroup\$ – EM Fields Apr 12 '16 at 19:03
  • \$\begingroup\$ How does eacd LEDs voltage-current-temperature characteristics effect the problem? \$\endgroup\$ – Hoytman Apr 13 '16 at 2:00
  • \$\begingroup\$ Missmatch between the parameters will interact and effect differing currents and therefore differing dissipations (and thus different temperatures, and different current). All of that will affect the brightness and reliability of each LED. \$\endgroup\$ – Spehro Pefhany Apr 13 '16 at 2:47
  • \$\begingroup\$ Lets say each LED has a 2 volt drop +-2% and are run at 20ma. Could you give a general ball-park estimation? Also How does temperature effect this problem? \$\endgroup\$ – Hoytman Apr 13 '16 at 16:34
  • \$\begingroup\$ See simulation in edit above and feel free to add resistors and play with it to get more insight. \$\endgroup\$ – Spehro Pefhany Apr 13 '16 at 17:12
1
\$\begingroup\$

At any given level of LED drive current, adding another milliamp will increase the voltage by some amount. The ratio of marginal voltage to marginal current may be called the marginal resistance. If two LEDs in parallel have slightly different voltage drops, the difference in current flow will be roughly equal to the distance in voltage drops divided by the marginal resistance. If the marginal resistance is small (as it is with some LEDs), the difference in current will be quite large.

Adding e.g. a 10-ohm series resistor will increase the marginal resistance of the LED+resistor combination by ten ohms. That may not sound like much, but it's huge compared to the marginal resistance of some LEDs, and might thus reduce current variation by an order of magnitude.

If one were using resistors as the only current-limiting devices, there wouldn't be much point using N+1 resistors for N LEDs (versus simply using N resistors). The approach may be advantageous, however, if one is trying to use a transistor-based circuit to control current to many parallel LEDs. If one has ten parallel LEDs driven from a supply that produces a regulated 1 amp, the LEDs will receive an average of 100mA, but some might receive 150mA and others 50mA. Adding a ten-ohm resistor in series with each LED would require that the supply be capable of producing one more volt than would otherwise be necessary, but would nearly eliminate variations in drive current.

\$\endgroup\$
  • \$\begingroup\$ Let's say I bought several resistor arrays (220 ohms, to be used with a 5 volt power supply) for my circuit (which has 18 LEDs in parallel) Now, I want to change the voltage to 9 or 12 volts. Can I simply add one resistor in series, or will I need to change all of the resistor arrays? \$\endgroup\$ – Hoytman Apr 15 '16 at 1:31
  • \$\begingroup\$ Could you describe marginal voltage / current / resistance a bit more please? I think you are giving me that answer that I am looking for. \$\endgroup\$ – Hoytman Apr 15 '16 at 1:55
0
\$\begingroup\$

The issue with the first circuit is that the three leds have a different forward current at the same forward voltage. The drop across the three leds will be the same. One led can't take 3 volts while it's parallel led only takes 2. Unequal voltages can't happen in parallel circuits.

And as each has a different current across them, one may dominate the amount of current available, and may go into thermal runaway conditions, killing it. Which kills the next one for the same reason. Etc. Highly depends on how much current can be pulled, i.e. the resistor value.

The third does not have this issue, as each is limited to the current by their individual resistor.

The voltage drop across each led and resistor will be equal. Current may vary.

schematic

simulate this circuit – Schematic created using CircuitLab

Notice that V-Total is the same as the Voltage source, 5 Volts. It will always be 5V. Notice that V-1 (R1 + D1), equals V-2 and V-3, but A-1 has a different current than A-2 and A-3. Yet V-4 and the nodes V-1/2/3 equal V-Total. And A-1 + A-2 + A-3 = A-4.

The very basics of series and parallel circuits.

Re: Your edit:

Practical application: I designed my original circuit with 18 LEDs in parallel, each with a 220 ohm resister. Now, I want to change the voltage to 9 volts (or 12 volts.) Will I need to switch out all of the original arrays or can I add one resistor in series.

Yes, you can. You need to resize R4 so that it takes up the same current at higher voltage. The formula stays the same. Basic Ohms law R = V / I.

Since you know I, and the new V, just calculate for R. Assuming the values from above, we can change R4 and the Voltage Source but keep the same current of 15.45 mA.

R = (9 VSource - 1.909 VForward) / 0.01545 A = 459Ω

Of course you need to chose the next standard sized resistor. But we'll ignore that.

schematic

simulate this circuit

If you are just adding a resistor to an existing circuit like 2, then VForward would be the old VSource, i.e. 5V.

\$\endgroup\$
  • \$\begingroup\$ your last statement is not 100% accurate. The voltage across the resistors will be similar , as will the current, but will be a function of the V_f vs. I_f curves of the LEDS. \$\endgroup\$ – jrive Apr 13 '16 at 22:29
  • \$\begingroup\$ @jrive are you saying that the voltage across R1 + D1 will be different then the voltage across R2 + D2? Where does the voltage difference go? Thin air? \$\endgroup\$ – Passerby Apr 13 '16 at 23:49
  • \$\begingroup\$ I misunderstood what you meant. The voltage across R1+D1 is definitely equal to R2+ D2. I was referring to the voltage across each -- v across R1 is not necessarily equal to the v across R2, and neither will the voltage across D1 necessarily equal the voltage across D2. \$\endgroup\$ – jrive Apr 13 '16 at 23:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.