2 or 3 LED string
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DIWdiver
Flashlight Enthusiast
You have to get a picture of the circuit, put it on a hosting site, then get a link to the picture. While typing your post, click the "insert image" icon, and paste your link in the box that pops up. If the hosting site you are using is DropBox, you need to edit the URL. If you need instructions, I'll look them up.
Note that forum rules state a maximum of 800x800 pixels, so you may have to edit the image if it's more than that.
DIWdiver
Flashlight Enthusiast
I just installed a lamp that uses four AL8860 drivers, with strings of 3 white LEDs on each. The only difference I see between the '60 and the '61 is that the '61 has current limit on the internal FET, and costs a bit more. It's a nice little chip!
Are you planning on disconnecting the LED string from the chip? Seems to me that disconnecting the switch pin should work. I would leave the LED-inductor-diode loop intact.
You'd get lower power consumption in 'off' mode if you grounded the control pin instead.
I have also thought about shorting one or two LEDs in the string with a switch. If you are planning on switching between two or more LEDs, you can put them in series, and short whichever one(s) you want off.
Are you planning on disconnecting the LED string from the chip? Seems to me that disconnecting the switch pin should work. I would leave the LED-inductor-diode loop intact.
You'd get lower power consumption in 'off' mode if you grounded the control pin instead.
I have also thought about shorting one or two LEDs in the string with a switch. If you are planning on switching between two or more LEDs, you can put them in series, and short whichever one(s) you want off.
I'm limited to one switch. I use the SOT89 package for smaller size and still get 1.5A rating.
I was going to hook common on switch to pin 4. Connect 1 side of switch to first LED in string, connect other side of switch to second LED in string. I'm using this design on another light with 2 LED string and it has been working great with no failures. With the new design, I will have to power the chip hot all the time and break the circuit with a single switch to turn light off. Don't really like being powered all the time and it will require 5 wires. I wish there was a simpler design. Another problem is overhead voltage for 3 LEDs. I'm limited to 12 volt power source. How much does the chip use? What do you think?
I'll try to figure out how to post a picture of circuit. Shouldn't be so complicated for an old man.
and THANKS again.
https://www.dropbox.com/s/uyfxwil72d8oc31/8861 (800x484).jpg?dl=0
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Dave_H
Flashlight Enthusiast
This driver is similar though not identical to common ones like PT4115. You should have no trouble driving three LEDs
from 12v supply. Quite a few 12/24v "auxilliary/offroad" automotive lights use 3 series white LEDs with no problems,
typically 1.5-2W per LED (600mA). I find dropout starting ~9-10v with decreasing brightness below that, and eventual shutoff
well below that.
Your switching arrangement should work OK, in fact as noted above you could short one of the LEDs, with SPST switch
without a problem. The dropout voltage will be even lower.
Dave
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DIWdiver
Flashlight Enthusiast
I would prefer to short the LED, rather than switch between sides of it. As you have it, each time you move the switch, the circuit is momentarily open, and the inductor current has nowhere to go. This will cause high voltage arcing across the switch contacts. Depending on your switch and how often you change it, this may eventually damage the switch. Not only that, jumping the LED has one less wire to deal with.
The 8861 doesn't have a dropout voltage per se, but the circuit will come out of regulation when you no longer have enough voltage to push the set current through the string of components: sense resistor, leds, inductor, internal switch FET.
Vf of the LEDs should be around 3V each, the sense resistor should drop 0.1V at design current, the FET resistance is around 0.2 ohms, so that gives 0.3V. That gives
Vdropout = Vf*3 + 0.1 + 0.3 + Rinduc*1.5A.
The 8861 doesn't have a dropout voltage per se, but the circuit will come out of regulation when you no longer have enough voltage to push the set current through the string of components: sense resistor, leds, inductor, internal switch FET.
Vf of the LEDs should be around 3V each, the sense resistor should drop 0.1V at design current, the FET resistance is around 0.2 ohms, so that gives 0.3V. That gives
Vdropout = Vf*3 + 0.1 + 0.3 + Rinduc*1.5A.
DIWdiver
Flashlight Enthusiast
If you change 'dropbox' to 'dl.dropboxusercontent' in the URL, you can paste it in the 'insert image' box, and the image will show up directly in your post.
how can i use 1 switch to turn off light completely and short 1 LED?
i've been using this design for 2 years with no failures but now you've got me worried...
it says 8861 has INHERENT open circuit LED protection...whatever that means???
thanks
i've been using this design for 2 years with no failures but now you've got me worried...
it says 8861 has INHERENT open circuit LED protection...whatever that means???
thanks
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DIWdiver
Flashlight Enthusiast
If you don't switch between LEDs often, it's probably not a problem, especially if you have a big beefy switch. If you have a tiny switch with barely enough current rating, and you switch often under power, the tiny little arc there is each time could eventually erode the contacts.
The buck topology gives open LED protection because once an LED opens, there's no way to increase the current in the inductor. The inductor current rapidly drops to zero, and the output voltage equals the input voltage.
With boost topology, this isn't so. When an LED opens, the inductor current dumps into the output cap, then charges back up, then dumps into the output cap. The output voltage continues to rise until something breaks down. You have to add something to the basic topology if you want to ensure this can't happen.
The buck topology gives open LED protection because once an LED opens, there's no way to increase the current in the inductor. The inductor current rapidly drops to zero, and the output voltage equals the input voltage.
With boost topology, this isn't so. When an LED opens, the inductor current dumps into the output cap, then charges back up, then dumps into the output cap. The output voltage continues to rise until something breaks down. You have to add something to the basic topology if you want to ensure this can't happen.
very little switching and the switch is rated for 10 amps. thanks for the explanation. i feel better now.
as i've said before... YOU ARE THE MAN!!! THANKS AGAIN.
DIWdiver
Flashlight Enthusiast
If you use a single pole, double throw switch with center off position, you can wire it to get 2 leds, 3 leds, or off. It wouldn't disconnect the chip from power, just disconnect the LEDs from the drive. With a double pole, double pole, with center off, you can disconnect the power too.
Say you want to switch off the bottom LED in the string. Break the connection at its anode. Connect the switch poles to the anode and cathode of this LED, and the switch common to the cathode of the next LED up. Now, in one position the string is normal, in one position the lower LED is bypassed, and in the center position the string is broken.
This has the problem of breaking the string. I've been thinking about this. If you put an RC snubber across the switch (from common to the cathode pole), you can keep the inductive spike voltage to any arbitrary voltage across the switch. Pick a voltage well within the switch's rating, and viola! You've solved the arcing problem. Or just use a switch that can handle it and forget it. It's a pretty small inductor.
A simple way to estimate the snubber values doesn't give precise results, but I'm too lazy to calculate the exact values (which I think is pretty complicated). The estimate is close enough if you leave a little room for error. The resistor is chosen to have approximately the peak voltage across it at the peak inductor current. The capacitance is calculated by using the equations for the energy stored in an inductor or a capacitor, setting them equal, then solving for C.
R = V/I
Ec = 1/2 x C x V^2 and El = 1/2 x L x I^2
Setting Ec=El and solving for C gives
C = L x I^2/V^2
Say you want to switch off the bottom LED in the string. Break the connection at its anode. Connect the switch poles to the anode and cathode of this LED, and the switch common to the cathode of the next LED up. Now, in one position the string is normal, in one position the lower LED is bypassed, and in the center position the string is broken.
This has the problem of breaking the string. I've been thinking about this. If you put an RC snubber across the switch (from common to the cathode pole), you can keep the inductive spike voltage to any arbitrary voltage across the switch. Pick a voltage well within the switch's rating, and viola! You've solved the arcing problem. Or just use a switch that can handle it and forget it. It's a pretty small inductor.
A simple way to estimate the snubber values doesn't give precise results, but I'm too lazy to calculate the exact values (which I think is pretty complicated). The estimate is close enough if you leave a little room for error. The resistor is chosen to have approximately the peak voltage across it at the peak inductor current. The capacitance is calculated by using the equations for the energy stored in an inductor or a capacitor, setting them equal, then solving for C.
R = V/I
Ec = 1/2 x C x V^2 and El = 1/2 x L x I^2
Setting Ec=El and solving for C gives
C = L x I^2/V^2
