25

No, that's a really bad circuit. You have 100 LEDs, all in parallel. Bad idea. Since this is a assignment, I'm not going to give you a better circuit outright. However, consider that LEDs want to be driven with a fixed current, not a fixed voltage. The change in current due to a change in voltage is very large when the LED is lit. Conversely, the ...


24

There are many reasons: - Cost: This is probably the biggest reason. The lighting market is quite competitive. There are lots of Chinese players offering aggressively cheap prices. This forces the other brands (even the bigger European ones like Osram, Philips, etc.) to use "cheaper" components to keep themselves in the game. As an employee for one of those ...


20

Lose the series resistors. It's not clear what you think they do for you, but they won't do anything useful. You have the LEDs connected to constant current drivers. Series resistors only get in the way and make the constant current drivers require a larger compliance range. Also, do the math. (300 Ω)(20 mA) = 6 V. You say your are powering the ...


18

There is one other way, much less commonly seen. Good for one LED, very simple, you can throw anything from about 4v to 20v at it, and it happily gives the LED a fairly constant current. Blue is the input voltage, 20v to 4v. Green is the current to the LED, about 12mA. Red is the power dissipated by the JFET, datasheet here.


18

I'd suggest your B340A diode is avalanching since you are far exceeding its reverse voltage rating. The diode must standoff the voltage you generate, so you need a voltage rating above your capacitor voltage. I'd use something in the 75-100 volt range, and perhaps an ES07B would suffice.


16

None of the above. All you want is Deep Red (650nm) and Deep (Royal) Blue (450nm). You do NOT want Full Spectrum, UV, IR, Far Red, Only White or anything besides Deep Red and Deep Blue. Red White and Blue is always a safe bet. Red and Blue come way before white. White (aka Full Spectrum) is more like supplemental. I do some consulting work regarding ...


16

Because doubling the current paths halves the power dissipation in each path, allowing use of cheaper resistors.


15

I suspect something to do with the strong magnet (I wouldn't have suspected it, except that it must be an interesting answer or you wouldn't have posed the question). Perhaps having the magnetic field induces a current in a loop somewhere. Was the leakage higher when there was more area between the wires connecting to the battery? Or it could be a current ...


15

Here is the summary of my investigations and my findings on the mysterious case of the random leakage current. When I began dismantling the units, I also got curious and wanted to research a rational cause for that odd behavior. I had no luck until I almost inadvertently moved the magnet near a powered-up PCB. The ammeter jumped to about 2mA and stabilized ...


15

XY problem. Your best efficiency solution from a power and cost point of view to drive a 1A LED load is to use a LED driver IC. There are many to choose from; the Diodes Inc PAM2804 seems likely to meet your needs. It supports PWM dimming. These chips are very inexpensive (about 15 cents Digi-Key price, half that in volume.) You could drive up to 3 of your ...


14

Here is a collection of LED driver options you can play with. simulate this circuit – Schematic created using CircuitLab


12

I believe you are exceeding the peak inverse voltage (PRV) rating of D4, the 40 V Schottky diode. During your switching cycle when the SW pin on the 2586 goes to 0 V, D4 becomes reverse biased due to the level at the output at the top of C34. With the output set to 57 V, this exceeds the 40 V reverse rating of D4. This can only be ...


11

It's a buck-boost configuration. If it was connected in boost then the input voltage could not exceed the LED voltage or they would burn out. There's more voltage across the inductor and the switch has to withstand more voltage with buck-boost. simulate this circuit – Schematic created using CircuitLab The voltage feedback is to protect the IC if ...


11

As always, READ THE DATASHEET! The chip you use has programmable constant current outputs, so you don't need external resistors to determine the current, the chip does this for you. The current sunk by each output is ~ 10 x the current out of the Vref. In your case 10 * ( 1.2 V / (1k2 + 680 ) Ohm ) ~= 6 mA, which is OK for a normal (20 mA max) LED.


11

The fact that there's multiple operating modes indicates that HH5K1 is not in fact a simple device, but most probably a cheap microcontroller¹ or dedicated ASIC that does the blinking. I'm afraid that means you can't swap anything to do what you want. So, build a flasher from scratch that does what you want. I rarely do that, but here, using a NE555 might ...


10

Your driver will output a fixed current through a string of LEDs. Since your LEDs are rated at 700mA @ 3.6V, this driver will indeed supply adequate current for the LEDs without burning them out. How many LEDs you can put in series on a driver depends on the drivers voltage range, in this case 18 to 36V. So you can attach 18/3.6 to 36/3.6 or 5 to 10 LEDs ...


10

You mean instead of single resistor? It depends on the use case. Assume a power of 600 mW has to be dissipated across the resistor. Instead of choosing one single resistor of standard wattage, say 1 W, one will choose a resistor (double value of intended resistance) and a power wattage of 500 mW.. Effectively, you have same resistance value and you are ...


10

I read about a method of limiting current using 2 transistors (schematic attached below). All linear solutions, resistor or whatever, will have the same efficiency. If you run the LED at 1A, it will draw 1A from 5V, so that's 5W. If there's 1.6V across the LED, then there will be 3.4V across the current regulator (whatever it is) so the LED will get 1.6W ...


9

Your connecting wires are too thin, you are dropping voltage in your leads. You need to select the right size of wire and potentially use a star connection to drive the strings. If your strip is using SMD2835 LEDs these will be FULLY bright at 12V DC. The typical LED characteristics (when warmed up) will be as below: Notice that the Vf is around 3.3V ...


9

That is a pretty weird way of "driving" LEDs. OP's schematic: A few issues: essentially you have an "equivalent" resistor, shared by the LEDs (if the LEDs have any imbalance - which they have - the current draw will be uneven) power draw - when the LEDs are "off", the FET will be drawing the current (even more than the amount ...


9

This is a constant-current output off-line switching power supply. In normal operation the output voltage is limited by the LEDs to between 10 and 21VDC. You should not connect it to an LED array with a range of forward voltage that is outside that range, in other words, so a single LED or two in series, in most cases is not acceptable. If you disconnect the ...


9

A current mirror can only servo the output current if it has the voltage headroom (compliance) to do so. If D2 is lower voltage drop and tries to hog the current, then Q2 will reduce its current by current mirror action. If D1 hogs the current, Q2 will be unable to increase current into a higher drop D2, it's out of compliance voltage. There are several ...


8

There is no way for your MOSFET to turn off. Figure 1. Move R39. The problem is that when the opto turns on Q1's gate capacitance is charged up. When the opto turns off there is no discharge path. If the gate voltage gradually leaks away Q1 will gradually turn off resulting in substantial current through it while there is substantial voltage across it so ...


8

According to the datasheet, the LED's Vf is 3.05V @ 1500mA, which is close enough to the 1400mA constant current provided by your power supply. The power supply's maximum output voltage is 69V. 69V / 3.05V per LED = approximately 22 LEDs in series. Say 20 to have some margin.


8

tl, dr: you either need a buffer that has really low Rds(on) to ensure that your series resistor is the only thing (or at least, the dominant thing) setting your current, or you need to sense the current and regulate it. That's easy enough, as you'll see below. Also, to set the current in each direction you'll also need to steer the current so that each LED'...


7

This kind of symptom smells of missing pulldown (or pullup, depending on topology) resistor on a FET gate. All is fine if the system powers up without the startup transient activating the FET. If that is successful, all is well. If not, the system latches into some wierd state that manages to draw some current. I notice this is a single-sided board. ...


7

Do COB LEDs usually need electrically insulating from the heatsink? The LED chips inside the module are already isolated from the heatsink on the back. They must be because these modules contain many LEDs in series and that is why they need around 20 - 30 V to work. Each LED needs around 3 V so there must be many in series. This can only work if all LEDs ...


7

Oh, you can drive an LED with a contant voltage. It's only that this voltage has to be in a very narrow band, depends on chip temperature and is not easy to regulate. Look at the I/U chart of a small power white LED. See how the current varies between 1µA and 10mA (nominal current) in the small band between 2.25V and 2.5V? And how much it depends on ...


7

LEDs used for street lighting usually employ a DC/DC converter of some kind, with tight current control at its output. So, providing a steady current isn't reducing efficiency nor does it add unnecessary components which could fail, nor does it reduce the lifetime of the LEDs. It's the simplest and most efficient way to drive a high-power LED array. Steady ...


7

You're making a wrong assumption that the efficacy goes up with higher power level. The opposite is true, to any meaningful power levels the efficacy decreases whenever you increase the current. PWM is used because it's very easy to implement. If you set your current to the maximum you want to use, you can have linear brightness control by just adjusting ...


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