hi there, I am trying to make a video light for my camcorder powered by its internal battery. No problem, I thought. So I charged the battery and measured 8.4 volts. I strung up about 50 LEDs in a combination series and parallel. I used the right resistors so each series circuit would draw 50mA when the battery is fully charged. Worked great except as the battery drained through using LEDs and camera the LEDs grew very dim quickly. I realized that minor fluctuations in battery voltage resulted in large changes in current (brightness) through the LEDs.
No problem I thought so I made a constant current supply via an LM317. Worked great but problem is that the LM317 eats alot of power (drops
3V) so the batteries get drained very quickly. I am at a loss! Is there any circuit that gives me a constant current but without dropping 3 precious volts? Goal is to keep the brightness largely independant of the battery condition and to be very efficient.
A depletion mode MOSFET like the Supertex DN3545 would drop about
2.6V in the constant current region and act like a resistor from zero volts to that point; this is a bit better. Putting a bunch of Fairchild J202 JFETS in parallel can get you about
1.5V at IDss of about 2.2mA each; this may be more useful but you will have to pick and test them due to the variations between lot codes (and some between units). One might get close to 0.8V by using two resistors and two bipolar transistors:
C E
0-+--- --+-[R2}-+--> load | \_/ | | | | _ | +[R1]+--/ \------ C E
R1 is selected to almost saturate the pass transistor, and R2 is selected for the current limit; this ain't exactly "constant current" but it is better than a kick in the head for low drop.
Assuming you want to stay with a linear circuit: LED strings ! ! ! !/ !/ !/ Vbat ---/\\/\\---+-----!-----!-----!---- etc ! !\\ !\\ !\\ N * NPN transistor D1 V ! ! ! --- [R] [R] [R] ! ! ! ! GND---+-------+-----+-----+------
D1 is an LED. It sets the base voltage on several NPNs to be equal to its forward drop. The resistors all end up with about one volt on them and this is about all the battery voltage you lose in the circuit.
Except, for a constant-current supply, you don't want the filter capacitor on the output.
The main problem with using a constant current regulator is that you really want one for each series string of LEDs. So a few long strings is better than many short strings, but requires higher voltage. For running from batteries, you might want to put a boost converter in front of it.
Moosfet. I can't quite make out how the circuit works due to the keyboard characters you are using. Could I get any help reading these please? Or maybe the schematics are somewhere online?
Nobody: Thanks. I looked up the boost converter but it is a step-down converter. Even the boost converters are a bit too difficult for me to understand. Which one would you recommend for an output of about 5 amps? Also, using this converter, will it eliminate the constant current circuit?
God damn it, there's just gotta be some schematics out there for people who wanna do the same thing like I want.
The camera's battery has obviously not got the needed capacity for both loads.
If you use a separate battery for the camera light - using NiMH cells of at least sub C size - the voltage will then be far more stable, the variation in light level far less and the camera will be well able to cope with it.
You are probably better off using a few high wattage LEDs rather than a big string. Check out the CREE leds.
The boost converter is about as simple as it gets in switch mode. V=3DL*di/dt will get you very far in analysis. I'm not one for ASCII schematics, but I can talk you through a boost converter. The goal is to add a voltage to the power supply you already have. Hey, no use starting from zero. Slap an inductor across the supply and current builds up in the inductor. The "slapping" is done by having one side of the inductor connected to the power supply, and the other side pulled to ground via an electronic switch, usually a mosfet. Now turn that mosfet off. The current in the inductor can not change, but like a guy that had too many beers, it needs to discharge. So what if at the intersection of the inductor and switch you put a diode. The other end of the diode goes to a capacitor, whose free end goes to ground. So the inductor flies upwards like a bat out of hell, forward biasing the diode, and dumping the current into the capacitor. This raises the voltage on the capacitor.
Now if you do this forever, the voltage on the cap goes to infinity, and the universe is no longer safe. So you put a voltage divider on that capacitor. Pick the values of the resistors so the divided down voltage is the same as a bandgap when the voltage on the cap is at your desired voltage. If you use a comparator, you can sense the voltage on the cap via the divider and only do the charge/discharge cycle when the cap needs some juice. [Feed me Seymour.]
The only thing that can get ugly is if the inductor isn't fully discharged. For simplicity, you want to run in discontinuous mode, that is, make sure the inductor is fully discharged on every cycle. This can be controlled by picking your component values correctly if you are using a simple oscillator to drive the DC/DC. Or you can monitor the flying end of the inductor and wait for it to drop beneath the supply voltage. Once it is done juicing up the cap, it goes limp.
You can get fancier in that you can make the whole circuit self timed by detecting the current limit in the inductor and stop charging it when it reaches a peak value. Combine that with the comparator on the flying end of the inductor and you have a pretty fool proof DC/DC. Not a great boost mind you since it is operating in what is often called a pulse skipper, so the noise it generates is all over the map as the load varies.
Now to drive a LED string, they often do a little trick. You put a sense resistor in series with your LED string. Pick the value of this resistor so the voltage across it is equal to a bandgap when the desired current is flowing. The circuit will now current regulate instead of voltage regulate.
This sounds good on paper, but there are two gotchas in that design. If something goes wrong with the external components, the DC/DC could create too high a voltage self destruct. So you need an over voltage safety circuit if you want to make a rugged design. [When you make a chip, you make it so the dumbest customer can not easily blow it up, especially when prototyping a design, which often has flaky connection.]] The other gotcha is the sense resistor gets pretty toasty if you have high current and a full bandgap voltage. So often the bandgap is divided down.
There are a few chips that do all this. I think Micrel makes a very simple LED driver like this. For one home brew design, it's not exactly a crime to use a couple of chips, ie. have each chip drive the longest string it can handle. Or you can try paralleling strings of the proper length, though the current might not be exactly where you want it if the LEDs don't match.
I am ready to give up. Can anyone tell me where I would buy such a ready made driver. Looked at Ebay but they are all either for AC input or for an input voltage higher than my 7.8-8.5V battery. Thanks everyone.
this is probably better - I haven't looked at all types. You would use these with all the LED in parallel and each resistor with a small series resistor to balance the current through each - about 12 to 22 ohms should do the job.
I posted a reply to your post in s.e.b. that shows the results of a circuit I built with simple components that drives a string of seven Cree white LEDs at their rated 350 mA, but at an efficiency of only 70%. I intended the circuit to work at nominal 12 VDC but it will work as low as 6 VDC. If your battery stays within the range of 7.8 to 8.5 VDC, you could drive two of the white LEDs in series with a 3 ohm resistor which will provide 315 mA at 7.8V and 420 mA at 8.5V. The power lost in the resistor will be 0.53 watts with total power of 8.5*0.42 = 3.57 watts or an efficiency of 85%. The very best PWM based converters can approach 98% efficiency, but you will only get about 10% longer battery life, and you probably won't see a lot of difference from 420 mA to 315 mA compared to a constant 350 mA.
If you can power more LEDs in series by using a higher voltage supply, you can get even better efficiency. For instance, a nominal 14 VDC supply can drive four LEDs (3.5V each) directly, if you don't mind the current ranging from about 500 mA at 15 VDC to 300 mA at 13.2 VDC. You can trade off current regulation for efficiency and complexity and cost using resistors or linear regulators or complex PWM drivers.
Bottom line, I think a small series resistor and (if possible) a higher voltage battery pack will be your best bet. A lot depends on the total voltage of the LED string and the range of battery voltages. Also, there are more efficient LEDs now that produce the same light at lower currents and voltages, so you can get greater overall efficiency of light output / power input just by spending a bit more on the LEDs.
It sounds like you are looking for the simplest solution and don't want to play around with complex circuits or purchase expensive ready-made equipment. KISS is good!
This will give you 12 volts out at up to 1.2 amps on an input voltage from your 8.4v battery pack. The diagram shows nominal
5V input, but the 2587-12 chip will work with an input voltage from 4 to 10 volts.
Add an LED (or recompute R for the last string), and put 17 LED strings in parallel, where each LED string is 3 LEDs and a resistor to set the current to 50 mA in series.
That will solve the brightness problem. It will NOT solve the problem of rapid discharging, but it will lengthen the time that you have bright LEDs over what you have now. The LM2587 is a chip you can work with - 5 big through hole leads. There are more efficient surface mount chips that are likely harder for you to work with, but even at 100% efficiency (impossible) you will still deplete the camcorder battery when you power the LEDs from it. Assuming 3.3V leds, each led consumes .165 watts, or 8.25 watts for
50 of them. The battery will have to provide ~1 amp for the LEDs.
For appreciably longer life, a seperate battery pack is needed.
Unfortunately, you are probably spot on. But I thought I would work out the relative merits of driving LEDs directly or with dynamic current regulation or just a resistor or a complex circuit with some nominal efficiency. All things considered, it might be that driving the LEDs directly or through a small resistor with a well-chosen battery voltage may prove the best solution, and the battery pack will be protected from harmful depetion because the LEDs will quickly grow dim and extinguish before the battery voltage goes below its critical minimum level. So my efforts are mostly directed toward those who understand and are willing to look at all aspects of achieving a desired result.
** The application does not require the light level to be very constant- as all video cameras have automatic light level compensation. The light level could vary over a range of 2:1 and hardy be noticed in the resulting images.
BTW:
If it were MY problem, I would do the following:
Buy a 12 volt CFL lamp ( spiral glass ) of about 20 watts rating - daylight white type which have way better colour rendition than white LEDs do and pour out heaps of light.
Buy a 12 volt, SLA battery of about 5 Ah capacity.
Fit the CFL into a small parabolic, mirrored reflector and attach it to the camera.
Just buy a 12V LED flood light, they are well matched at the factory. LEDs are still lower power than CFL. White light LEDs have come a long way and are being used in auto lights and flood lights. In fact, we have some 5W LED lights replacing 100W bulbs. They are very efficient, but at higher cost.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.