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I know that Leds in parallel with one resistor is bad.

enter image description here

I was wondering if this is still considered bad to do since all power is run through 1 resistor then split and run through other resistors before going to the led.

Only reason I ask / want to do this is i have a lot of leds that need power but don't have enough of the right resistors. I am looking to see if this is a safe alternative. Also the power going through R1 in this case will be well within the breakdown limit.

Thanks for any help.

Edit: In the example above D3 is pin 0 of one 7 segment display and D2 is pin 0 of another 7 segment display. The other pins will be the same setup as above.

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closed as unclear what you're asking by brhans, Marcus Müller, Finbarr, laptop2d, Sparky256 Dec 23 '18 at 1:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ The current regulation won't be as good as if you just used two resistors, since each LED is allowed now to affect the other LED through \$R_1\$'s voltage drop. Do you really need to go to three? \$\endgroup\$ – jonk Dec 19 '18 at 19:18
  • \$\begingroup\$ I do not know what you mean by go to three. I have tons of 100 ohm resistors and that is why i was hoping to do it this way. In what way would one led affect the other in this config ? Thanks \$\endgroup\$ – deathismyfriend Dec 19 '18 at 19:22
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    \$\begingroup\$ You have a 5 V source, why not put two LEDs in series, sharing a resistor. You need fewer resistors and waste less power. \$\endgroup\$ – Colin Dec 19 '18 at 19:23
  • \$\begingroup\$ @Colin They need to be in this config. The leds above are actually pin 0 on a 7 segment display. 2 different 7 segment displays are being used in this circuit. I only did the circuit in the example above with minimal parts. \$\endgroup\$ – deathismyfriend Dec 19 '18 at 19:26
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    \$\begingroup\$ @deathismyfriend The mathematics is pretty easy to show. But the picture itself that you show makes this obvious without math. \$R_1\$ is in a common path. Differences in LED 1's current will impact LED 2's current because of it. Isn't this obvious at a glance? If not, perhaps the mathematical approach would clarify? Do you know how to work out the question/answer process for computing the impact of one upon the other? (This is especially true in many applications for 7-segment displays where not all segments are ON.) Could you spend more time describing exactly what you are trying to achieve? \$\endgroup\$ – jonk Dec 19 '18 at 19:33
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Yes this will work. The concern with parallel leds is overblown (pun intended), but having one main resistor with smaller parallel resistors for each led string will work, as long as you calculate the right resistance and power ratings for it. Basic Ohms law applies. As the current draw of your LEDs plus small resistance changes, the current and thus the voltage across your main resistor will change and then the current and voltage across your small resistors and LEDs change (imagine a see saw trying to balance itself with moving children on it). You may need some trial and error to find the right value.

The resistor in essence is a poor voltage regulator for your led strings. You would be better off with using an actual voltage regulator and each of your small resistors. With the voltage regulator you get correction of the output instead of a variable drop based on the current draw.

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  • \$\begingroup\$ I didn't think of the resistors as in parallel. That is a good point. I need 200ohm drop from the 5v source to each led. To give the proper current to the led. Would my design above be incorrect then ? \$\endgroup\$ – deathismyfriend Dec 19 '18 at 19:29
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    \$\begingroup\$ The OP has indicated in a comment that this setup is to be used in a 7-seg display application. Considering that the current through the one 'common' resistor will now change depending on the digit being displayed, does your answer still hold? \$\endgroup\$ – brhans Dec 19 '18 at 19:54
  • \$\begingroup\$ @brhans so the question changed after the fact and people downvote. Even when my answer specifically address the variable issue and suggests a voltage regulator instead, meaning it's still a valid answer for the new information? Nice. \$\endgroup\$ – Passerby Dec 19 '18 at 20:52
  • \$\begingroup\$ @brhans I never said that only one input will have the resistors. So when different digits are displayed nothing will change as each segment input will have resistors. In the example above D3 is pin 0 of one 7 segment display and D2 is pin 0 of another 7 segment display. \$\endgroup\$ – deathismyfriend Dec 19 '18 at 20:53
  • \$\begingroup\$ @deathismyfriend - you should edit your question to clarify what you're actually planning to do here. It's entirely possible that you could share a single resistor between 2 LEDs (or between segments on different 7-seg displays) as long as you're never going to be driving both LEDs at the same time. \$\endgroup\$ – brhans Dec 19 '18 at 21:21
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As you have a 5 V source, LEDs with 2 V forward voltage, and plenty of 100 ohm resistors a more sensible approach would be to put two LEDs in series, and use one resistor per series pair. This will require fewer parts, and waste less energy in the resistors.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ This will not work as I can't do it in this instance. \$\endgroup\$ – deathismyfriend Dec 19 '18 at 19:37
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If each LED should have 200 ohms in series you can either put two 100 ohm resistors in series for each or two in parallel to give 50 and then 100 to each LED.

Four resistors either way, the first way is better.

schematic

simulate this circuit – Schematic created using CircuitLab

Shorting the section above R3 degrades the current sharing a bit, but the nominal current is the same if the LEDs are identical.

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Only reason I ask / want to do this is i have a lot of leds that need power but don't have enough of the right resistors.

There may be a simple solution. As I understand, you have a 5V supply and need 200 Ω per LED. That means you will drive the LEDs with 15 mA. The problem is you have 100 Ω resistors so have to double the number of resistors.

If the above is true, then all you need to do is add a 3.5V LDO and use 100 Ω resistors. The all LEDs will get 15 mA using a single 100 Ω resistor supplied with 3.5V rather than 5V.

What I am not sure about in your circuit is whether D2 and D3 will ever or never both be on at the same time. So I do not know your desired current.

The above 15 mA may not be exactly what you want but you can drop the 5V to the required voltage to get the desired current with a single 100 Ω resistor for each LED.



I know that Leds in parallel with one resistor is bad.

This is not really applicable to your situation. When the LEDs are in parallel it does not matter much when each resistor is the same value. The concern with parallel LEDs applies more to parallel strings of multiple LEDs. It's the mismatch in the LEDs forward voltage that causes an imbalance. With single LEDs in parallel the imbalance will not very likely be significant enough to be perceivable to the human eye.

If you truly wanted each LED to get the precise same current you would have to choose each resistor value based on the forward voltage of each individual LED. This is a moot point becasue the human eye is not very good at distinguishing differences in luminous intensity between two sources of light.

I have experimented with strips of LEDs in parallel driven with a 2 A constant current driver. One strip draws 800 mA and the other 1200 mA even though the forward voltage were very close e.g. 42.2V and 41.95 when powered individually with 1 A.

The point being 800 vs 1200 is a significant imbalance but it was not perceptible and had to be measured.

In your case the single resistor, R1, should be eliminated if you must use a 5V supply. If you have two 7 segment digits side by side one displaying a one and the other displaying an eight, there may be a perceptible difference in luminous intensity between the two digits if R1 remained in place. This could be true if your R1 was limiting the current for all seven segments.

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