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31

LED forward voltages can vary some - enough that it would be difficult to set a precise current with them merely tied in series across a fixed voltage. Total Vf for the string can vary by several volts. You could very easily blow up the string just tying it to the supply as you’ve shown. So, yes, you should still have a dropping resistor. Even better, ...


16

Here's an excerpt from some Yuji LED datasheet: So yes, if you order in quantity you can get LEDs with good Vf match. Wiring them in series strings means the Vf mismatches add and substract which increases the probability that total Vf will be closer to (number of LEDs x average Vf). If they are mounted on an aluminium PCB which keeps them all at the same ...


12

You mention an IO pin; this is how you could do it with one IO pin if you are using a microcontroller (assuming one that uses 5V logic): simulate this circuit – Schematic created using CircuitLab When the IO pin is 5V, D2 will light up; when the IO pin is 0V, D1 will light up. Note that you will have to calculate the actual resistor values; different ...


11

is the number of LED enough to limit the current to around the 350 mA range? Absolutely not. do I still need the resistor added in the series? Yes, or something that acts like a resistor. A MOSFET as you have would work fine as long as it has feedback to effectively work as a constant-current source, which you don't currently have. There are many examples ...


10

I see two problems here. First, you don't have any pull-up resistors, so your switches can't actually provide a high input. 74 series chips frequently have inputs that float high, but I think the CMOS varieties (like 74HC) need a pull-up or pull-down resistor to function properly. Second, you have a lot of unconnected inputs, which are picking up noise and ...


9

The problem is that you have HC type CMOS logic chips, and in the video LS TTL type chips are used. CMOS chips don't work in a circuit that is deliberately made to utilize the characteristics of LS TTL type chips. So your inputs float unconnected when the DIP switch is off, and the CMOS chips in your circuit do not work properly with unconnected inputs. LS ...


9

First of all, diodes always need a current limiting resistor. In this case apparently at least (5V - 0.8) / 200mA = some 20-25 ohm at the very least. Otherwise you risk burning the diode. Second, the output current of your MCU is not 40mA per pin! 40mA/pin and 200mA total are absolute maximum ratings. When reading any electronics datasheet, you should design ...


9

If you want a reliable LED string, then you don't want a string of LED's without a resistor. In this case you don't want to match the 24V just by LED voltage drops. It'd be much better to use 6 LED's and a say 18 ohm series resistor to limit the current. LED's are very dependent on temperature and you don't want thermal runaway causing the LED's to burn out. ...


8

CMOS floating inputs are very receptive to stray electric fields. You need pulldown resistors on the switched_nodes. Use 100Kohm. Fresh out of university, I assisted a very senior guy debug a 3_rack cabinet logic system: the central_station telemetry receiver for South Florida Flood Control District, monitoring dozens of remote sites out in swamps that would ...


8

The absolute maximum total current from all the IO pins together on an Arduino Uno is 200 mA, so I wouldn't try this. Also, if you don't switch the five outputs simultaneously, you run the risk of blowing them up one by one. If it would work, which it won't, then yes, you would need a resistor, as the Arduino Uno has a 5V microcontroller sitting on it and ...


5

simulate this circuit – Schematic created using CircuitLab You could also try this. simulate this circuit This would also work. simulate this circuit


5

A collection from another thread, posted here for the middle schematic. Pin 6 would be the GPIO pin. For better brightness matching, swap the red and green LED positions.


4

You do have floating inputs. You simply replaced the VCC wires to DIP switches with resistors. So again, when DIP switch is closed, the chip input is connected to VCC via resistor, which is fine. When DIP switch is open, the chip input is connectes to nowhere and floats.


4

You need to use a distinct resistor for each LED. The lower forward voltage of the green LED is basically clamping the joined anodes at a voltage lower than that where any meaningful current will flow through the higher forward voltage of the blue LED's junction. In fact not only do you need individual resistors, you probably want unique resistor values to ...


4

From a different answer to the question you referenced: The above is written with assumption that you will chose the resistor in such way that is sets the current so that the current is n times the current you want in each diode where n is the number of diodes and that the current is actually larger than the current which a single diode can safely conduct. ...


4

On editing your question, I see that what you really want to do is to switch the LEDs from a microprocessor output. The easiest way to do that is in your program. Use two GPIOs, and always switch one off when the other is on. If you can't change the program, then you could do something like this: simulate this circuit – Schematic created using ...


4

Reducing lamp voltage will dim the lamp. I suspect that any LED lamp will draw so little current that you won't get much of a voltage drop. If, for example, you have a 24 V, 6 W lamp you will draw 6/24 = 0.25 A. The 50 Ω resistor will drop approximately 50 x 0.25 = 12.5 V and that should be quite enough. (You can recalculate for any other lamp.) The effect ...


4

The ac portion of the led string is not polarized. Typically a full wave diode bridge, which is polarity independent, will be part of the string to convert ac to dc and a regulator circuit after that to regulate it down to a stable dc voltage. Leds by their nature are directional which is why their are polarized as you say. Depending on the number of lights, ...


3

To block current through Q1, at least emitter voltage should be on it's base. To provide that, R5 is used as pull-up. R4 will drop voltage on Q1 base when Q3 starts to conduct, thus closing Q1. It's not cheapest solution, but easy to build using off-the-shelf components. R4 can be smaller, when more Q1 collector current is nedded. See: simulate this circuit ...


3

Notice that the top picture shows just the LEDs, but the bottom picture shows little circuit boards (the LEDs are attached on the other side, not shown). If you just have the LEDs (not attached to the little circuit boards), then you have four pins. GND, 5V, data in, data out. No problem. The circuit boards have two 5V pads and two GND pads for easy daisy-...


3

After a bit of digging, I found this: Equation 9.17 in this chapter You can manipulate that equation as - $$n=\sqrt{\frac{J}{edB}}$$ Given: \$J=1000A/cm^2\$ \$B=10^{-10}cm^3/s\$ \$d=10^{-4}cm\$ \$e=1.6\times10^{-19}C\$ Therefore - $$n=\sqrt{\frac{10^{36}}{1.6}}\approx\mathbf{0.791\times10^{18}/cm^3}$$ Calculate carrier lifetime - $$\tau=\frac{1}{B \Delta n}\...


3

You're treating the LEDs as if they're (2.5 V - 3 V) lightbulbs. Lightbulbs do not mind if the voltage is a bit lower or higher but LEDs do. LEDs respond very strongly to the applied voltage. Apply a voltage that is too low (5 V for 2 LEDs in series) then almost no current flows and the LEDs burn very dimly. Apply too much voltage ( 5 V for 1 LED or all LEDs ...


3

Coin cell batteries have a high internal series resistance. If you measure the battery voltage when the LEDs are on, then you'll see that the voltage is no longer 6V. Direct from energizer: For example, the starting IR of a 2032 battery is near 10 ohms IR = internal resistance https://data.energizer.com/pdfs/lithiumcoin_appman.pdf As for your safety ...


3

I "vote" for the code error in first place. Did you forgot to set pinMode to OUTPUT? Seems like you are doing digitalWrite when pin are configured as INPUT resulting in switching between HIGH state via pull-up enable and Hi-Z state which causes a dim light and/or unstable state when not touched by probe. Also you better add a resistor in series ...


3

... and that they have the same characteristics concerning the forward voltage. They don't. Figure 1. Variations in Vf for a typical white LED. Image source: Variations in Vf and binning. I wonder if the two schematics above are equal. Nope. The first one is a classic schedule solution, while the second is an alternative one in order to avoid non-...


3

I cannot figure out where to put a buzzer so that it goes off when the LED shuts down, and the buzzer to stop when the LED lights up. There isn't anywhere on this circuit that you can connect your buzzer that will do it. You will have to add some components or (better) use a completely different circuit. If you want to keep what you have then first you need ...


3

Go to a 3W LED and a 5V supply, put a 47 ohm resistor parallel with the LED, and 22 ohms in series. simulate this circuit – Schematic created using CircuitLab R3 can be external to the lamp assembly, R1 and D1 can be built into a lamp base, leaving the antique part unmodified, You can salvage the base of a lamp with a BA15D base, EG Sylvania 1076


3

First, is it possible? A 1000 lumen LED light usually draws about 10 W. So you would draw a total of 19 * 10 = 190 W. If you are in a 220-240 V country, that's about 1 A. This is definitely within the allowed power on a single socket (usually 16 A). Consider that you hair drier probably draws 10 times as much. Next, do you get a total luminous flux equal to ...


3

simulate this circuit – Schematic created using CircuitLab D2 - I couldn't find a Schottky diode symbol. Some folks would use a silicon diode here (i.e. 1N4001). I prefer a Schottky because they're faster. D1 can literally be any 3 or 5mm LED you find, they'll pretty much all have the same 20mA maximum. Using a 1k resistor will keep the current ...


2

If these "3 volt" batteries are little coin cells, they have high internal resistance (at least 10 ohms, perhaps much more): https://data.energizer.com/pdfs/lithiumcoin_appman.pdf That high internal resistance acts as a current limiting resistor. If you have no other resistor on your LEDs, then you absolutely must add one to use with a 5v supply ...


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