Need help designing a power saving circuit for two strings of mixed LEDs. (I don't have a lot of experience designing circuits)

Thread Starter

Zack Maness

Joined Dec 23, 2017
4
I made a few Lichtenberg figures and I want to mount them in shadow boxes. I built some LEDs into the box (Three UV along the top and one red on the left under the blue tape) and I am planning on pouring about an inch of epoxy to cover the LEDs and disperse their light. For a power supply, I am using a battery box that holds 8 AA batteries providing 12 VDC. I'm planning on adding a pot to each circuit so I can independently dim each string of lights, but I think it would be more efficient to use 555 timers. The problem is, I don't have any experience with 555s so I don't know how to design a circuit that could independently dim two LED strings with different voltage and current values. Does anyone have any advice or thoughts? Any advice on using 555s or thoughts on my current design would be greatly appreciated.

LedParallel.jpglichtenbergBox.jpg
 

ci139

Joined Jul 11, 2016
1,194
just a thought : the UV-LED-s will be dim enough without dimming . . . if the red led is too bright you only need to adjust it
the "layzyman's" solution would be a 20mA constant current source with a current mirror
so 20mA to 2x UV in series and 20mA to UV and Red in series . . . then shunt off the excess current from the Red LED by a proper resistor or pot.

a complete circuit with a likely possible current source (the voltage drops on CE junctions on real transistors may be higher + the simulation does not withcount the emitter resistance . . . so it may require MOSFET current source or higher supply voltage )
/// in simulation the 27Ω limits the current and 24Ω is illustrative but in real used to balance the currents of the shoulders of the current mirror (values of these resistors may be different in real)
/// the 560Ω is selected highest possible value so that the 27Ω still functions e.g. if the current varies with the supply voltage change the lesser value for the 560Ω resistor has to be selected
 
Last edited:

peterdeco

Joined Oct 8, 2019
95
Using your schematic and Ohms Law, put in series with the 3 UV leds a 150 ohm resistor (at least 1W to keep its cool)and in series with the red led a 500 ohm resistor ( at least 1/2W to keep its cool). A more popular 470 ohm will work. I have a schematic at work of a 555 with 100% PWM control for dimming. Unfortunately I won't be going back until after New Year's. If this thread is still alive I will post it.
 
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Dodgydave

Joined Jun 22, 2012
8,844
Easier to use a LM317 for Constant Current control, you can put your red leds in series or use separate regulators, use a 120 Ohms resistor for 10mA, or use a 470 Ohms pot with a 33ohms resistor 1/8 W in series for variable brightness.



images.png
 
Last edited:

BobTPH

Joined Jun 5, 2013
2,260
Using an LM317 to dim will not save any power. You need a PWM circuit, like the 555 circuit the OP is talking about to save power.

The idea is, make the two strings add up to about 9-10 volts if running of 12V. Then caclulate the resistor needed for the highest current you want. Then use a 555 to create a variable duty cycle and and a MOSFET to switch current to the strings. One 555 and MOSFET for each string.

Bob
 

Dodgydave

Joined Jun 22, 2012
8,844
Using an LM317 to dim will not save any power. You need a PWM circuit, like the 555 circuit the OP is talking about to save power.

The idea is, make the two strings add up to about 9-10 volts if running of 12V. Then caclulate the resistor needed for the highest current you want. Then use a 555 to create a variable duty cycle and and a MOSFET to switch current to the strings. One 555 and MOSFET for each string.

Bob
It doesn't matter which circuit he uses, it's the total current drawn that matters on the leds, and and silcon devices,, less components is Less Current!!
 

ElectricSpidey

Joined Dec 2, 2017
908
Here ya go...Merry Christmas.

Use a CMOS 555 not the NE555 as shown.
Choose a MOSFET that will do the job of 100mA or more but not so high that the gate capacitance is too high.

The schematic doesn’t show it but you should add a de-coupling cap directly across the power pins of the 555.

EDIT:
Oh yea, I almost forgot…the two 5k resistors would actually be a 10k pot.

A_555_PWM.JPG
 
Last edited:

crutschow

Joined Mar 14, 2008
24,377
PWM only saves significant power if there is a inductor in the circuit in place of the series resistor, to store energy while the PWM signal is low.
Otherwise, for the same average LED current, there is little difference in efficiency between using a resistor to control the intensity versus PWM.
 

Thread Starter

Zack Maness

Joined Dec 23, 2017
4
I made a few Lichtenberg figures and I want to mount them in shadow boxes. I built some LEDs into the box (Three UV along the top and one red on the left under the blue tape) and I am planning on pouring about an inch of epoxy to cover the LEDs and disperse their light. For a power supply, I am using a battery box that holds 8 AA batteries providing 12 VDC. I'm planning on adding a pot to each circuit so I can independently dim each string of lights, but I think it would be more efficient to use 555 timers. The problem is, I don't have any experience with 555s so I don't know how to design a circuit that could independently dim two LED strings with different voltage and current values. Does anyone have any advice or thoughts? Any advice on using 555s or thoughts on my current design would be greatly appreciated.

View attachment 195337View attachment 195333
Edit: I made this a while ago and forgot that I actually ran the UV LEDs in parallel, and forgot to add the resistors in the new schematic. but these are the resistor values I found to provide the proper current.

LichtenbergLED.jpg
 

MisterBill2

Joined Jan 23, 2018
5,222
The PWM approach does save real power because when the output is off the only power is used by the control electronics. No inductor needed unless you are creating a switch-mode regulator. Use one regulator for each kind of LED and adjust until you have the desired look.
Using ANY of those linear regulator circuits will be slightly LESS efficient than using a resistor because a regulator takes power to function.
 

Thread Starter

Zack Maness

Joined Dec 23, 2017
4
Using your schematic and Ohms Law, put in series with the 3 UV leds a 150 ohm resistor (at least 1W to keep its cool)and in series with the red led a 500 ohm resistor ( at least 1/2W to keep its cool). A more popular 470 ohm will work. I have a schematic at work of a 555 with 100% PWM control for dimming. Unfortunately I won't be going back until after New Year's. If this thread is still alive I will post it.
That would be great. That schematic would be a big help
 

Thread Starter

Zack Maness

Joined Dec 23, 2017
4
Thank you everyone for your fast replies. This
just a thought : the UV-LED-s will be dim enough without dimming . . . if the red led is too bright you only need to adjust it
the "layzyman's" solution would be a 20mA constant current source with a current mirror
so 20mA to 2x UV in series and 20mA to UV and Red in series . . . then shunt off the excess current from the Red LED by a proper resistor or pot.

a complete circuit with a likely possible current source (the voltage drops on CE junctions on real transistors may be higher + the simulation does not withcount the emitter resistance . . . so it may require MOSFET current source or higher supply voltage )
/// in simulation the 27Ω limits the current and 24Ω is illustrative but in real used to balance the currents of the shoulders of the current mirror (values of these resistors may be different in real)
/// the 560Ω is selected highest possible value so that the 27Ω still functions e.g. if the current varies with the supply voltage change the lesser value for the 560Ω resistor has to be selected
The UV LEDs are surprisingly bright, and I'm not sure how they will look after the epoxy is poured. So I was thinking of adding the ability to dim them just in case.
 

BobTPH

Joined Jun 5, 2013
2,260
PWM only saves significant power if there is a inductor in the circuit in place of the series resistor, to store energy while the PWM signal is low.
Otherwise, for the same average LED current, there is little difference in efficiency between using a resistor to control the intensity versus PWM.
Yes, of course you are right. Either a linear current regulator or PWM will save the same amount of power When compared to rinning them at full blast.

I was thinking of proposing SMPS current regulator, but that is a little more complicated.

I have built such a thing using a compararator, MOSFET, inductor and current sense resistor to run a single white LED off a 9V as a book light.

Bob
 

crutschow

Joined Mar 14, 2008
24,377
The PWM approach does save real power because when the output is off the only power is used by the control electronics.
But for a given average LED current, there's no difference (assuming the average LED current is what determines the LED brightness as seen by the eye).
 
Last edited:

ci139

Joined Jul 11, 2016
1,194
i compiled 2 variants of TLC555-'s (trending to idealistic) functional analogs - for LTSpice - if anyone want's to play around
/// estimated the quiescent current to be more than actual 150µA -- the 56k thingy below in .params . . . there might be more bugs
the 20mA is way too little for the ols NE/SA/SE555 bipolar chip.

TLC555 - TEST - BFC-AR.png

TLC555S.cir
Code:
* MORE Simplified NOT-Exact
* Functional Analog of
* the TLC555 CMOS timer
* .
* /// Pseudo--MOS conductance 1:19(20)k @ 15V ToT Supply
* /// Timer Quiescent conductance 1:56(60)k @ 15V ToT Supply
* .
* GND 3GR OUT RES MOD 3SH DCH SUP
.SUBCKT TLC555S 1 2 3 4 5 6 7 8
* supply detect
R55 8 55 24m
R89 55 9 200k
R91 9 1 2G
C91 9 1 {vCdt}
B10 10 0 V=uramp(V(9,1))
B11 11 0 V=15/V(10)
R210 10 0 10k
R211 11 0 10k
* inputs - 5
R88 55  1 56k
* inputs - 5
R85 55  5 {vRpm}
R58  5 58 {vRpm}
R51 58  1 {vRpm}
C85 55  5 {vCdp}
C58  5 58 {vCdp}
C51 58  1 {vCdp}
R99  5 90 200k
R81 90  1 2G
C81 90  1 {vCdt}
B16 16  0 V=uramp(V(90,1))/2*3
R216 16 0 10k
* inputs - 2 4 6
B12 12 0 V=u(uramp(V(2,1))/V(16)-.33)
B13 13 0 V=u(uramp(V(4,1))/V(10)-.33)
B14 14 0 V=u(uramp(V(6,1))/V(16)/2-.33)
R212 12 0 10k
R213 13 0 10k
R214 14 0 2k
* outputs - 3
A1 12 74  0 0 0  0 70 0 OR Td=11n
A2 12 14  0 0 0 71  0 0 AND Td=7n
A3 70 73  0 0 0 72  0 0 AND Td=6n
A4 13 72 71 0 0 73  0 0 AND Td=7n
A5 13  0  0 0 0 74  0 0 BUF Td=5n
B15 15 0 V=u(V(73)-.5)-.5
R215 15 0 10k
R23 21  3 910
R24  3 20 91
R25  7 22 91
S12 55 21  0 15 GSW
S21 20  1 15  0 GSW
S22 22  1 15  0 GSW
.model GSW SW(Ron=.2 Roff=1G Vt=0 Vh=50m)
.param vRpm = {20k}
.param vRqs = {60k}
.param vCdt = {2p}
.param vCdp = {1p}
.ENDS
TLC555F.cir
Code:
* LESS Simplified NOT-Exact
* Functional Analog of
* the TLC555 CMOS timer
* .
* /// Pseudo--MOS conductance 1:19(20)k @ 15V ToT Supply
* /// Timer Quiescent conductance 1:56(60)k @ 15V ToT Supply
* .
* GND 3GR OUT RES MOD 3SH DCH SUP
.SUBCKT TLC555F 1 2 3 4 5 6 7 8
* supply detect
R55 8 55 24m
R89 55 9 875k
R91 9 1 125k
C91 9 1 {vCdt}
B10 10 0 V=uramp(V(9,1)*8)
B11 11 0 V=15/V(10)
R210 10 0 10k
R211 11 0 10k
* inputs - 5
R88 55 1 R={2+15/V(10)*vRqs}
* inputs - 5
R85 55  5 R=2+V(11)*vRpm
R58  5 58 R=2+V(11)*vRpm
R51 58  1 R=2+V(11)*vRpm
C85 55  5 {vCdp}
C58  5 58 {vCdp}
C51 58  1 {vCdp}
R99  5 90 875k
R81 90  1 125k
C81 90  1 {vCdt}
B16 16  0 V=uramp(V(90,1))/2*24
R216 16 0 10k
* inputs - 2 4 6
B12 12 0 V=u(uramp(V(2))/V(16)-.33)
B13 13 0 V=u(uramp(V(4))/V(10)-.33)
B14 14 0 V=u(uramp(V(6))/V(16)/2-.33)
R212 12 0 10k
R213 13 0 10k
R214 14 0 2k
* outputs - 3
A1 12 74  0 0 0  0 70 0 OR Td=11n
A2 12 14  0 0 0 71  0 0 AND Td=7n
A3 70 73  0 0 0 72  0 0 AND Td=6n
A4 13 72 71 0 0 73  0 0 AND Td=7n
A5 13  0  0 0 0 74  0 0 BUF Td=5n
B15 15 0 V=u(V(73)-.5)-.5
R215 15 0 10k
R23 21  3 910
R24  3 20 91
R25  7 22 91
S12 55 21  0 15 GSW
S21 20 1 15  0 GSW
S22 22 1 15  0 GSW
.model GSW SW(Ron=.2 Roff=1G Vt=0 Vh=-50m)
.param vRpm = 20k
.param vRqs = 60k
.param vCdt = 22p
.param vCdp = 2p
.ENDS
 
Last edited:

MisterBill2

Joined Jan 23, 2018
5,222
But for a given average LED current, there's no difference (assuming the average LED current is what determines the LED brightness as seen by the eye).
That assertion is incorrect. That is how LEDs are dimmed in most modern efficient applications. The duty cycle is varied. There are arrangements that vary the voltage, but again, that is with a switching power supply.
 
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