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I wired up a simple circuit (below) which was designed using the "Every Circuit" website/app. It's a small LED circuit used inside a costume prop. When the master switch is pressed, the 3 LEDs light up. Then when a "momentary push" trigger is pressed, the other (top) green LED lights up. All works fine, but when the trigger is pressed in, the 3 lights in series seem to dim ever so slightly.

The "dim" can be seen in this video over on Instagram (hopefully you can see it) https://www.instagram.com/p/Bqh2x6YDASG/

The dim doesn't bother me too much, and I might just leave it as it adds a bit of character to the model, but I was wondering if anyone knew how to fix it so that the LEDs don't dim at all.

My electronics knowledge is extremely basic. I've heard of using a voltage regulator, but I didn't think it would be needed as the circuit seemed to run fine when I built it on "EveryCircuit.com".

Any help is appreciated.

Circuit

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  • \$\begingroup\$ Welcome to the site. Playing with LEDs, batteries and resistors is a great way to learn about DC fundamentals. \$\endgroup\$ – mike65535 Nov 27 '18 at 14:30
  • \$\begingroup\$ That toy desperately needs a well tuned buzzer to go with the light effect. LOL \$\endgroup\$ – Jeff Wahaus Nov 27 '18 at 15:07
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    \$\begingroup\$ Cheers @mike65535! @JeffWahaus - Yeah I thought about adding sound. I bought the components to do it but haven't had time to play with it yet! \$\endgroup\$ – Rovanite Nov 27 '18 at 16:56
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    \$\begingroup\$ How momentarily is the switch pushed? Is it just a blip, or a few seconds? I'm wondering if a capacitor rated at 16 V and, say, 1 μF, across the battery (after the on/off switch) would hold up the voltage long enough that the brightness change was not noticeable. \$\endgroup\$ – Andrew Morton Nov 27 '18 at 19:37
  • \$\begingroup\$ @AndrewMorton It's in a prop, so the button can be pressed either extremely quickly, or can be held in for many seconds, so no specific time. I'll try this out to see if it works, but I may nee to get some capacitors first. \$\endgroup\$ – Rovanite Nov 28 '18 at 16:06
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The LED's dim when you close the switch because you're drawing more current from the battery. The simulation you're using assumes an ideal battery with a constant 9V. In reality the voltage from the battery will depend on the load (current draw) and the voltage will drop as your circuit draws more power.

Your schematic shows that you're driving the LED's at a high current. You'll find that many LED's are just as seemingly bright at smaller currents, like around 10mA. I would recommend wiring the 3 always lit LED's in parallel each with their own resistor and use as large as possible resistor values (small enough so that the LED's are bright enough for you but as large as possible). This will reduce the current draw and will result in less dimming when you activate the 4th LED with the switch.

schematic

simulate this circuit – Schematic created using CircuitLab

Blue LED's typically need more current to look bright. Red typically look bright at lesser current.

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    \$\begingroup\$ Configuring these in parallel like this makes them less sensitive to variations in the input voltage (which is good), but they also draw three times as much current, and most of the power is lost in the resistors. That may or may not be a problem. \$\endgroup\$ – pipe Nov 27 '18 at 14:41
  • \$\begingroup\$ It doesn't draw 3 times as much current if he optimizes the resistor values for each led and drive them at the minimal acceptable brightness. If you drive each LED at around 10mA you'll save power. \$\endgroup\$ – Jeff Wahaus Nov 27 '18 at 14:53
  • \$\begingroup\$ No, in the original setup he draws 28 mA from the battery for the LED string. This 28 mA goes through every LED. In a parallel configuration, even at a lower 10 mA per LED he will draw a total of 30 mA from the battery. \$\endgroup\$ – pipe Nov 27 '18 at 14:57
  • \$\begingroup\$ LOL, okay I didn't do the math or simulation but if he draws only 9mA per LED then he'll save power at 27mA. \$\endgroup\$ – Jeff Wahaus Nov 27 '18 at 15:12
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    \$\begingroup\$ "increasing the resistor values would potentially help the "dimming" be less?" Possibly, it depends on how much current the battery can supply. In parallel the button draws 1/4 the current but in series the button draws 1/2 the current. So the dimming effect will always be greater if the LED's are wired in series. \$\endgroup\$ – Jeff Wahaus Nov 27 '18 at 17:17
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Try altering your circuit to look like this

schematic

simulate this circuit – Schematic created using CircuitLab

Now see what happens when you close the switch. Jeff Wahaus described the effect, but this will show it to you. Batteries are not simple voltage sources - they have what is called output impedance. And, if you think about it, this makes sense. If the output impedance were zero you could get infinite current when you short the battery outputs.

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  • \$\begingroup\$ Will do! I've got plenty of things to test out now :) \$\endgroup\$ – Rovanite Nov 28 '18 at 16:04
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I was wondering if anyone knew how to fix it so that the LEDs don't dim at all.

A better 9V battery would do it. Using a more efficient LED would go a long way as well.

The type of batteries makes a big difference. Three 3V coin cells would have low capacity. But they are lithium and have a flat discharge curve so the LEDs remain brighter over the lifespan of the battery.

Voltage discharge curve for a lithium battery.
enter image description here

A 9V lithium battery would do much better than a 9V alkaline.
enter image description here

With a 100 Ω resistor on the 3 LEDs, you are drawing only about 5ma when the battery is fresh. It appears the LEDs have a low luminous intensity (mcd) and the battery is weak.


When powering LEDs with a battery it is a good idea to use high intensity (brighter), high efficacy (more light per watt) LEDs. A good LED will cost about 25¢. That cost will quickly be made back in battery savings.

The LEDs you are using are not very bright to begin with. If you use brighter LEDs your eyes may not perceive any difference.

The best 5mm round blue and green LEDs are currently the Cree C503B-BAN-CX and C503B-GAS-CB

The LEDs in the video were were very dim, maybe had a luminous intensity of a hundred mcd. The Cree LEDs have up to 90500 mcd (green) and 23500 mcd (blue). Both these LEDs emit about the same number of photons (µmols/s), it's the perceived sensitivity of your eyes that change the luminous intensity.


Your simulation is not using the correct forward voltage. Blue and green LEDs have a forward voltage around 3V. Red, orange, and yellow are about 2V. This makes a big difference when the total forward voltage for the 3 series string of LEDs is greater than the 9V battery. Especially when, with use, the battery discharges down to its cutoff (dead) voltage.

I've heard of using a voltage regulator, but I didn't think it would be needed as the circuit seemed to run fine when I built it on "EveryCircuit.com".

Actually for a battery powered LED circuit a constant current (CC) controller is the typical option. Although if lighting a single white, blue, of green LED a 3.3V voltage regulator would also work well as the difference between the supply voltage and the LED's forward voltage is not that great given very good efficiency.

There are many current regulators made especially for alkaline, Lithium, NiMH, and LI-ion batteries.


If you drop the current of the Cree blue LED to 4 mA and the green to 1 mA, the LEDs will still emit over 200 mcd. By reducing the current the forward voltage drops below 3V. Then you use a supply voltage of 3V-3.3V and LED Vf of 2.9V to calculate the resistor values.

enter image description here

The problem with battery powered boost CC controllers at low current and low voltage, is the efficiency is not great. For your project it would be most efficient to use a boost step up voltage converter. Step up a the battery voltage to just over the LED's forward voltage.

Using the Cree LEDs you can run the green LEDs at 1 mA and the blue at 4 mA. Using an efficient low current and low voltage switching voltage converter like the TI TPS61261 and tweaking the output voltage to just above the LEDs forward voltage, you will have a very efficient design.

With a 3.3V regulator, you use a 400 Ω resistor for the green LEDs and a 100 Ω resistor for the blue LEDs. If you adjust the voltage to 3V then use a 25 Ω for the blue and a 100 Ω for the green you get even better (≈10%) efficiency.

enter image description here


Bottom Line

Using a high intensity LED you can lower the current draw on the battery so the voltage does not drop when you turn on an additional LED.

When you use a battery with sufficient capacity, the voltage will not drop when you turn on an additional LED.

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