10
\$\begingroup\$

I'm researching into using one resistor to control a 7 segment display. Not because I want to do so (could save some time), but just to understand concepts.

Some say that slight differences between each LED can cause very unequal lighting, which make sense, but at the same time they say that this could cause damage.

I don't understand this: On an LED Voltage-Current graph, does the LED "pull" a certain current like it has resistance?

If I have 2 LEDs in a parallel circuit, both parallels have to drop 2V, but one LED is 2V at 5mA, and the other is 2V at 10mA, this does NOT make sense to me. I have a resistor before the circuit, lets say from an Arduino 5V, that drops 3V (to give 2V to the LEDs) at 15mA, so 200 Ohms. This is what I dont understand: Wouldn't the current be shared between the two, giving 7.5mA to each? What is the usage of the 'voltage versus current' graph? My guess is to show what equal lighting takes.

Hopefully someone could explain where my understanding went wrong, am I not factoring in LED resistance, or do they follow strange rules?

\$\endgroup\$
  • 5
    \$\begingroup\$ I STRONGLY advise you to do it and see for yourself what happens. This will be much better than anyone trying to explain it. The main problem is that the brightness of the segments will depend on how many segments are lit. Don't be afraid to try things on the bench. \$\endgroup\$ – mkeith Oct 15 at 2:28
  • \$\begingroup\$ dont assume those datasheets are written in stone , hell you could have 2 40ohms resistors and will be 38 and the other 43 , nothing is perfect. the led they used to draw that graph is not a perfect identical led to the one you have. this is something you must get used to in electronics. diodes, transistors etc..all those devices have "characteristics" values like gain and drops .... do not assume the datasheet is giving some absolute value that is true all the time for every copy of that device ever made \$\endgroup\$ – Edwin Fairchild Oct 15 at 2:38
  • 1
    \$\begingroup\$ When you do it yourself and see what happens... it works. Lots of people do this and you don't see their LEDs randomly going up in smoke. \$\endgroup\$ – user253751 Oct 16 at 10:07
7
\$\begingroup\$

I don't understand this: On an LED Voltage-Current graph, does the LED "pull" a certain current like it has resistance?

Yes, to understand intuitively this phenomenon known as "current steering", you can think of the LED of as a resistor... but having nonlinear resistance with the property to keep up the voltage across it constant. It is interesting to see how it does this "magic". Here is a simple but very intuitive explanation.

If the LED was an ordinary (ohmic) resistor, the voltage across it would be V = I.R. But it behaves as a "dynamic" resistor that decreases its resistance R when the current I increases and v.v. so their product - the voltage V, does not change.

When you connect two such voltage stabilizing elements in parallel, they form a dynamic current divider. Each of them tries to set its voltage across this network... and if there is at least a small difference between their voltage thresholds, they vigorously change their resistances to divert the current ("to 'pull' a certain current").

The advantage of this simple explanation is that you can emulate this arrangement by two variable resistors (rheostats) in parallel.


EDIT - answering the OP's questions in the comments below

Not because I want to do so (could save some time), but just to understand concepts.

Exactly... all we need to understand concepts... "to see the forest for the trees". In this way, we can make a connection between seemingly different circuit phenomena.

Do you think you could explain another situation? If I had 5V with a 300 ohm resistor, and after it split into parallel, each side had an LED. One's datasheet showed 1.9V at 12mA, the other 1.9V at 30mA (a large difference). This seems like it would "pull" 42mA, but the resistor would only allow 10.3 mA to come through.

First of all, we have to create a notion about the non-linear resistance. Most people do not feel a need for this; they simply accept that it is what its IV curve represents... they take it for granted. But we are both different from them and we need an even deeper explanation. We want to know not only that the curve is such (nonlinear) but also to somehow imagine how such a shape is obtained in principle.

1. Presenting the red LED1 as a "dynamic resistor". The most natural and intuitive way to answer this question is to imagine LED as a varying resistor (rheostat) that changes its own resistance. This is illustrated by the graphical interpretation in Fig. 1 where the supply voltage varies in the range 0 - 5 V. The resistor R can be considered as internal resistance of this real source. As its voltage increases, its IV curve moves (translates) to the right. At the same time, the IV curve of the static resistance R1 of the nonlinear "resistor" (LED1) does not stay immovable but rotates counterclockwise. As a result, the intersection (operating) point OP1 moves upwards along and pictures the LED1 nonlinear characteristic. So, the idea is to dynamically change the current resistance of the element.

LEDs as dynamic resistors

Fig. 1. LEDs as dynamic resistors (graphical representation: top - red LED1; bottom - green LED2

2. Presenting the green LED2 as a "dynamic resistor".

In a similar way, we can explain how the non-linear IV curve of the second "resistor" (the green LED2) is obtained. The only difference is that it is more sloping.

Could it be that both of the LED's I could count as resistance, then solving for the exact current per LED?

Exactly... This is what I will show with the following drawing...

3. Presenting the two LEDs in parallel by an equivalent "dynamic resistor". I fully accept your idea to replace the combination of two nonlinear resistors (red and green LEDs) in parallel with only one "dynamic resistor" (represented by a blue IV curve).

LEDs in parallel as equivalent resistor

Fig. 2. LEDs in parallel as equivalent resistor: top - nominal R (74 ohm); bottom - higher R (300 ohm)

If we supply the LED network through a resistor R with a nominal value of 74 ohm (3.1V/42 mA) - the upper Fig. 2a, everything will be fine and will correspond to the datasheet. But if you supply it through R = 300 ohm - the lower Fig. 2b, the IV curve of the real voltage source (aka "load line"), will tilt significantly... and both voltage and common current will decrease.

Is there some resistance that I am not thinking about?

No, there are no other resistance; only the LEDs have increased their static resistances.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ @Blake, I understand your desire to understand (at the lowest intuitive level) what is happening because I am too... Circuits with nonlinear elements can be solved graphically. I will illustrate it but later, in a few hours... \$\endgroup\$ – Circuit fantasist Oct 15 at 5:39
  • 1
    \$\begingroup\$ @Blake, I uploaded two pictures to make you start thinking about this (your) idea. I wonder what is your first impression... Do you get the idea behind the drawings before I started explaining them? \$\endgroup\$ – Circuit fantasist Oct 15 at 15:11
  • 1
    \$\begingroup\$ When the switch is open, each LED will drop 1.8V while passing 10mA, having an effective resistance of 180 ohms; the total current through the circuit will be 20mA. When the switch is closed, the voltage across each LED will be only 1.5 volts, making their effective resistance much higher, so the circuit behavior will be dominated by two 120-ohm resistors in series, passing only 12.5mA. \$\endgroup\$ – supercat Oct 15 at 19:58
  • 1
    \$\begingroup\$ I was inspired by an article about automobile traffic flows; I'm not sure that the circuit is especially useful, but I think it helps to illustrate that things like diodes are "active" circuit elements rather than purely passive. \$\endgroup\$ – supercat Oct 15 at 22:28
  • 1
    \$\begingroup\$ @Circuitfantasist: I would write 12.5 or 12½ rather than rounding, since for purposes of discussion all values are infinitely precise. \$\endgroup\$ – supercat Oct 15 at 22:44
20
\$\begingroup\$

LEDs’ forward voltage drop varies from device to device and by more than you might think. This causes a few issues for the user.

enter image description here

Figure 1. A Lite-On LTST-C170TBKT. Image source: Lite-On.

The Lite-On LTST-C170TBKT InGaN blue LED datasheet, for example, shows the forward voltage varies between 2.8 V and 3.8 V at 20 mA. This is quite a variation. The lower value would, just about, allow operation with a low value series resistor at 3.3 V whereas the 3.8 V sample might not turn on very brightly. On higher voltages control of the current by series resistor would vary for different LEDs.

enter image description here

Figure 2. LTST-C170TBKT-Vf-variation is rather wide. Image source: Variations in Vf and binning.

Here notice that if we applied 3 V to a group of these LEDs some would draw one milliampere and others could draw 50 mA.

I don't understand this: On an LED Voltage-Current graph, does the LED "pull" a certain current like it has resistance?

Resistors don't "pull" current. They allow a certain amount to pass and the amount (I) is directly proportional to the voltage (V).

enter image description here

Figure 3. You can consider any diode as being a little like a non-return valve with a spring. Image source: What is an LED?.

You may be able to imagine that if you paralleled a set of many of these non-return valves with assorted spring pressures that as you raise the water pressure the one with the lowest resistance would open first and hog all the current. The ones with higher spring pressure would only open a crack. The first to open will determine the pressure drop across the whole lot. In the same way the diode with the lowest forward voltage, Vf will conduct first and it will determine the voltage applied across the parallel group.

If I have 2 LEDs in a parallel circuit, both parallels have to drop 2V, but one LED is 2V at 5mA, and the other is 2V at 10mA, this does NOT make sense to me.

Hopefully it does now.

I have a resistor before the circuit, lets say from an Arduino 5V, that drops 3V (to give 2V to the LEDs) at 15mA, so 200 ohms [small 'o']. This is what I don't understand: Wouldn't the current be shared between the two, giving 7.5mA to each?

No. Due to the differences in Vf one will conduct more than the other.

What is the usage of the 'voltage versus current' graph? My guess is to show what equal lighting takes.

It shows the relationship between current and voltage for a typical LED of that type. Figure 2 may be more useful.

Have a read of my linked articles. They may help.


From the comments:

"others could draw 50 mA", I thought it was the resistor, not the LED, that allowed the current through, though?

In my 3 V LED discussion I am applying 3 V directly to the LEDs, not through a resistor.

Also, having the different forward voltages doesn't match my understanding of how each parallel branch drops the same voltage.

As Figure 2's blue line nominal curve shows the Vf is not a constant. Once it gets into the linear region it is slightly off vertical so small changes in the applied voltage result in large changes in current. Now imagine a bunch of blue lines all slightly offset to the left and right of the one drawn and you should be able to see that a vertical line drawn from any point on the V-axis (representing the common voltage applied to the parallel LEDs) will result in different currents in each of the parallel LEDs.

Does the current through one branch increase because the wires have to drop the voltage?

Generally not. It's because each LED's curve will be slightly different. There will be tiny differences due to the internal connecting wires but these are metallic, not semi-conductors.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ "others could draw 50 mA", I thought it was the resistor, not the LED, that allowed the current through, though? Also, having the different forward voltages doesn't match my understanding of how each parallel branch drops the same voltage. Does the current through one branch increase because the wires have to drop the voltage? \$\endgroup\$ – Blake Oct 14 at 21:00
  • 2
    \$\begingroup\$ See response to comments. \$\endgroup\$ – Transistor Oct 14 at 21:09
13
\$\begingroup\$

If you open almost any LED datasheet which gives its voltage and current specification, you will find out that the LEDs are not perfect devices, they have tolerances.

So if you buy a bag of hypotethical red LEDs, and test them by driving exactly 20mA through each of them from a constant current source, you will see some variation what the LED voltage is.

Even if the typical LED voltage at 20mA is said to be 2.0V, you might have lowest voltage of 1.8V and highest voltage of 2.2V.

If you connect two LEDs that have very different voltage characteristics in parallel, like the the 1.8V and the 2.2V LEDs, with a single resistor, and try to provide 40mA in total for the two LEDs, most of the current will be flowing via the 1.8V LED and it may get damaged from the excess current, and the 2.2V LED might not even light up at all.

So the assumption that all LEDs are identical, and can share the available current identically, is only valid when the LEDs are guaranteed to be identical enough in the datasheet from the manufacturer.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ If some LEDs drop different amount in parallel, how does that work? If there is 2V over one parallel line, and the LED drops 1.8, the 0.2V is left, does the wire itself have to drop the rest of the voltage by pulling a lot of current through? How does that work? \$\endgroup\$ – Blake Oct 14 at 20:53
  • 3
    \$\begingroup\$ It was explained in the post. Imagine that LEDs are either ON or OFF only in this example. The LED that requires only 1.8V to light up will start conducting current at 1.8V and will limit the voltage over both LEDs to 1.8V. The LED that requires 2.2V to light up will not light up, as it only has 1.8V over it. You should experiment this by paralleling a red and a blue led to see it in action. \$\endgroup\$ – Justme Oct 14 at 21:18
  • 1
    \$\begingroup\$ which is why you buy a bag of leds. You select the ones that happen to have similar characteristics. You can buy multimeters that can test it (or you can set up your own little testing rig @ desired voltages) \$\endgroup\$ – Stian Yttervik Oct 16 at 8:15
  • 2
    \$\begingroup\$ @StianYttervik No, you really don't want to do that when making a real-world product. You would buy binned LEDs directly from manufacturer, or design a circuit in such a way that tolerance of LEDs is not a problem. If you are making a single one-off device for hobby use, and want to spend your time measuring individual LEDs to get down to one resistor, then you are free to do that, but the time could be better spent to other things if getting down to one resistor does not really matter. \$\endgroup\$ – Justme Oct 16 at 8:46
  • 1
    \$\begingroup\$ @Justme I am certainly making that comment from a hobbyists perspective, aye. I fully accept the hardware designers' view might be different. \$\endgroup\$ – Stian Yttervik Oct 16 at 9:03
5
\$\begingroup\$

If you use one resistor for a 7-segment (say common anode) LED display, the display segments (all other things being equal) will be radically dimmer when all 7 are illuminated vs. when one is illuminated. That's regardless of how well they share current, which is also a problem, particularly at relatively low LED current.

You could drive a single seven segment display with a single resistor (say in the anode circuit) by splitting time into 7 or 8 slices and energizing one segment at a time. Of course you would do it so rapidly (hundreds of Hz typically) that the segments appear to be continuously illuminated. The resistor would have to be around 1/7 or 1/8 of what you would need to drive the segments statically to the same brightness. The peak segment current would also be 7 or 8 times higher .

This approach quickly runs out of steam with larger numbers of segments but it's do-able for one or two digits.

| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

It's the same issue as your earlier question. Finding current through LEDs in series-parallel circuit

Wiring multiple LEDs in parallel to a common load resistor will divide the current between the 'on' segments because of our friend, forward voltage (Vf.) Each lit segment has a fixed forward voltage, however the resistor current will be the same regardless of how many segments are lit because Vf doesn't change (much).

With a common load resistor on the digit:

  • The numeral '1' will light just 2 segments, so each segment will get 1/2 the resistor current. That's the best case.
  • The numeral '8' lights all 7 segments, so each segment will get just 1/7 the resistor current. It'll be quite dim.

So instead, typical 7-segment LED display systems will have one load resistor per segment, not per digit. Fancier LED drivers will use constant-current per segment, and so will have no resistors at all other than the current-setting resistor on the IC. Here's an example: https://www.ti.com/product/TLC5927

This goes for multiplexed displays as well. However, with multiplexed displays you can share the per-segment resistors between digits since only one digit is activated at a time.

If you were to only light one segment at a time, you could use just one common resistor. However, you would be faced with running the segments always at 1/7 duty cycle, so your display will be 1/7 as bright. That might be acceptable, so long as the refresh rate is high enough (40Hz or more for the entire digit.) Add more digits and the problem gets much worse in a hurry.

| improve this answer | |
\$\endgroup\$

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.