# Tag Info

4

It seems that you have got a useful answer, to the wrong question. This isn't really an answer to what you should have asked, but it's too long for a comment and it does help you (and future readers) by explaining what you should be researching. And since you now have an (accepted) answer, it's too late to radically change the question above. You asked about ...

0

There are, in fact, LEDs that have the resistor for a particular operating voltage already built-in, so you don't need an external one. For instance, here's a spec sheet for LEDS that can be directly attached to a 5V pin. I have a few of these directly connected to my Arduino Nano (5V operating voltage) - no problems so far, and they're on almost 7/24.

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Since LEDs have manufacturing tolerances, and also the power supply you are using to power up the ESP32 also has some tolerance, it is possible that connecting LED directly to an IO pin will exceed the safe current value. So the maximum forward voltage value for your LED is 2.8V, but that is the best case which causes least current to flow. The worst case is ...

1

OK, first of all there are at least three standard I2C speeds: Classic 100kHz (10µs bit time) Fast 400kHz (2.5µs bit time) High speed 3.4MHz (only the data phase is high speed, for simplicity 300ns bit time) these works as normal bidirectional I2C buses (the high speed has a negotiation trick for compatibility and the transceiver is different but the ...

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The photodector module you are using consists of an avalanche photodiode, a transimpedance amplifier and some biasing circuitry. So the optical power that is received is converted to a current by the photodiode. This current is then converted to a voltage by the transimpedance amplifier. You can measure this voltage using the BNC connector. So the meaningful ...

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For a 3.3V logic device the CMOS drivers are 25 ohms +/-25%. A 65mW White LED has an internal ESR of around 15 Ohms above 2.85V threshold approx. Thus direct drive is possible but for an indicator with today's efficacy will be incredibly bright. Even 5mA is adequate, unless you want max brightness. Although not very precise on tolerances this will work ...

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For IR input 0.5 [A/W], is a common sensitivity for silicon photodiodes as well. This refers to output current for input Power at the wavelength in datasheet. RMS = 0.707* peak for sine wave 0.5A/W = 2W/A or 2mW (in) per mA (out)

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Reflective sensors may range from UV & yellow for paper money to IR for most other things. It is possible to operate at +/- 5V with floating current limiting to drive the emitter and have a logic level drive the 0V floating to earth ground with an open collector and Zener limiter. I.e. two way Tx/Rx on one cable assuming some case ground is avail. ...

6

Yes, it's fine. Most LEDs are rated at 5V reverse voltage, which is much less than the Vf of a blue LED. Multiplexing LEDs also exposes them to similar reverse voltages, so every blue or white LED numeric display is probably operated the same way.

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Get a few 5m LEDSTRIPs with remote controls , stick it up and call it a day's work for \$25 or get creative and buy a box full. You can easily reach an outlet. They use IR remotes, so you have to aim and are practically cheap now. But get the white CCT you like and RGB for fun and W/m you need. Blocking direct light with indirect light is the trick to ...

4

You could do something like this: simulate this circuit – Schematic created using CircuitLab R9 + R10 is 7.07K, which is optimized to maximize the difference between Vcc/2 and the voltage with one switch and two swiches (about 1V nominal difference in each case). 1% resistors are suitable and cost about the same as 5% these days. Here are the voltages ...

2

A circuit like this will get you there: simulate this circuit – Schematic created using CircuitLab The idea is that you choose the R values and the zener voltage so that when the desired number of switches are closed, you get a voltage above the (-) input on the (+) input. In this case 1 switch will only get you 6V but two will get you 8V and the ...

3

This is a simple voltage adder circuit. Resistance values are just for show. In this circuit you can play with the switches and see that the output voltage changes based on how many switches are thrown. You can also see how the output voltage changes with resistance. Reading Vout will tell you exactly how many switches are thrown if you set the range of the ...

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Let's start with your ideas concerning voltage. 12 V suck. Really. Let's do some simple math. Assume you want to illuminate the whole area with at least 150 lx. A rough calculation gives the need for 6000 lm for basic brightness. In fact you will want to spend additional power on a lot of spots and you'll loose much light depending on the remittance of ...

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Some times ago, I designed a PWM controller for a LED strip. In first tests, I verified that perceived brightness was not linear with the PWM duty cycle. For example, with 50% d.c. PWM, the perceived brightness was much more than the 50% of its maximum value with 100% d.c. PWM. In fact, human eye perceives light intensity in a logarithmic way, because only ...

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Ok so I just had to deal with this. According to this link, while the emitted light is linear with the increase in PWM duty cycle, our perception of it is not linear, but logarithmic. We can transform it to be perceived as linear by applying an exponential function to it. With a python example: exp_scaler = 100/(math.exp(1)-1) def change(v): pwm....

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the dice will need some sort of wireless power receiver to receive the charge. unfortunately Qi receivers, while inexpensive, seem to be reasonably large https://www.aliexpress.com/item/32898114152.html maybe wireless tealights would be a better starting point. https://www.aliexpress.com/item/1005001965026911.html but still fairly large.

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If you connect a capacitor to a resistor such as a small LED, and then in case it in epoxy, but then place it on a inductive charger such as a wireless phone charger, will the capacitor charge and then release light? Nope. In order to receive the AC magnetic field emitted by the wireless phone charger, you need a coil. These chargers are basically high ...

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Perhaps you should reconsider your idea. Most LED lamps will come with their own driver that is 230VAC (or 110 depending on your region). LEDs are current-driven and not voltage-driven, which means that if you have a 12VDC circuit, you will still need to have a driver for each lamp, defeating the purpose. The only case where a driver is not required is if ...

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You just described something like rail lighting system used in museum and such… they did it with halogen burners, but it applies to LED lamps too. They solved the power drop simply using bars as conductors and also hanging the lamps mechanically. So it seems a fine thing if you are using an adequate wire cross section (adequate as in, depends on the distance)...

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The LED with the lowest Vf is the red one, at about 2.1V. 5V-2.1V is 2.9V, so in theory it should not be a problem for the driver, which has a Voh of 3.3V (the 3.3V rail) and therefore not forward bias the LED. In practice, what happens is that the LED has enough sub-threshold leakage current that even this minuscule amount will cause it light dimly, sinking ...

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Those pushbuttons have a LED with resistor, as those are meant to be driven at 12V. Driving it with the MAX7219 cannot drive the pushbutton LEDs very bright, because it is meant to be used with 5V supply. As per the question itself, the datasheet does not say what is the minimum resistance, but it does not matter. No matter how much you want current out of ...

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Based on the comments (later they have been moved to chat) and my subsequent work, I want to share my last position, self answer in a way. Because I finally succeeded to get what I want. And I know there will be newbies who face them exactly. No, there is not such a safe platform including all available DC motor specifications. Try not to buy/use electronic ...

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The solution was to add a 5V level shifter. As pointed out in the update, the controller shifts out data with a high voltage level of 3.7V. The 2 cases are: Full brightness: Power supply is around 5V => WS2812 expects high voltage as 0.7x5 = 3.5V Dimmed: Power supply increases to around 5.3V => WS2812 expects high voltage as 0.7x5.3 = 3.71V So, more ...

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You have about 270mA total, so a MOSFET would not be a bad idea. simulate this circuit – Schematic created using CircuitLab The MOSFET should be a logic-level N-channel type with low rated Rds(on) at 3.3V or less drive voltage, and rated for 30V or more Vds. If you are running long wires to the LEDs you should pick a higher voltage MOSFET. The AO3400 ...

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Something like this might work... simulate this circuit – Schematic created using CircuitLab The transistor selection is based on a max collector current of 100 mA or so. There are many other possibilities. The R1-R3 values were chosen to give ~20 mA of LED current. Choose R4 so that you get the required amount of base current out of your MCU.

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simulate this circuit – Schematic created using CircuitLab Figure 1. A circuit suitable for demonstration. A single-pole, double-throw (SPDT) switch is required. In a real car the headlamps are typically 55 W on filament lamps and the circuit would have to switch more that 10 A into cold filaments. For this reason the switches control relays which do ...

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Can I do a 9V supply to get the current with adjusted resistor values or will it hurt the leds? So long as the current through each LED is correct, the LEDs don't care what the supply voltage is. But if you are controlling the LEDs individually, beware that the voltage drop along the cable will depend on the current, which will depend on how many LEDs you ...

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It seems like there is no voltage limit for LEDS if you adjust the current and power dissipation accordingly. So it should work with 9V and the proper resistor. The parallel connection is not a great idea but that would have to do for now. Thanks everyone.

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Unless you investigate fluorescence, you can't get any colour out of a filter if it isn't already in the source. If the colour is off that much then you're going to need a lot of filtering to remove the deep red portion of the band that you've got and you won't be left with much light. It's not clear from your question whether you need three different LEDs ...

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This isn't any different for RGB LEDs than for any other set of 3 LEDs: You can't put them in parallel with a single series resistor. Every color LED needs its own resistor. but my 4m power cable is 3 ohm an this is the problem Yes. This means you need a better cable. I don't know which cable you're using, but even a single 0.180 mm diameter copper wire ...

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1.5mm wire is about AWG 15 which is an insignificant loss at 6A @ 6m x 10 mΩ/m = 60mV, also the WS2812B operates from +3.5~+5.3 V. But the 25 ~ 50 Ohm CMOS driver into a 10pF CMOS load with a 3m delay line of around 15 ns is going to ring enough to corrupt the signal integrity for CMOS Clock and data. The solution is simple, but look at the differences in ...

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One chip consume 60mA at full brightness.150 is 9A. Power conductors on strip can not conduct so much current. I got voltage drop over 5m strip 0.3V with current 350mA. Run power wires in parallel to strip and connect at few point to strip will improve it.

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Apart from switching current-sense resistors for adjusting the output level as suggested by Kartman and Jasen, another good option is to use the CTRL pin since its intended use is brightness adjustment. As stated in the datasheet, setting this pin's voltage between 0.125V and 1.25V will give a brightness level of 10% to 100%. The REF pin outputs a regulated ...

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as Kartman suggests switching the curren t sense resistor for a higher resistance will dim the LEDS maybe install two resistors in parallel but have the switch in series with one of them. so you get onew resistor for dim or two in parallel for full brightness. That is use bedlamp switch type of button that toggles between on and off. simulate this circuit &...

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In general you must not connect different LEDs in parallel and I wouldn't even connect different color temperatures in parallel even though they share the same datasheet (in case of LM301). The different phosphorus coatings might introduce different power dissipation, which can lead to thermal runaway (a pretty important effect you should look up). For high ...

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Your assumption is correct. If you connect both the anode and cathode of an LED to + nothing will happen. Since there is 0V potential no current will flow.

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I know this thread is getting old, but I just wanted to add a point for future readers. Diodes of all sorts often have an inverse Vf vs. temperature curve. Well, curve might be misleading since it's most often linear. In fact, diodes are sometimes used as temperature sensors for this reason. Knowing that, and that when you put current through a device, it ...

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Replace the batteries in your multimeter and check the gate-source voltage again. Set the pin low; should see 0.0 to about 0.5v. Set the pin high; should see Vdd to Vdd-0.5v. If you still get the same readings, measure from the PSoC ground to GPIO pin. If you STILL get the same readings, then the GPIO is unable to power the pin properly for some reason. The ...

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Often, when you try to come to terms with a big and unfamiliar technological stack, you find yourself in this nothing works situation. There are so many things that could be wrong that you can't even formulate a precise question. To find the solution you need to go step by step: Is the problem on micro-controller side, or is the problem on LED side? Detach ...

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Data can be boosted to 5V with something like an SN74HCT125 (this is a quad channel, so you could use four different pins from the SMT32 to drive 15 LEDs each). As mentioned a resistor in the line is also considered best practice. I've found 33 to 120 ohm work best for me. Also a good idea to put some sort of reverse current protection in as WS2812B do not ...

2

The MOSFET might be leaking a small amount of current which is enough to make a small LED glow a little. The MOSFET you're using is rated for 10 Amperes which is overkill for a small LED. If you used this MOSFET to switch a 5 A lightbulb on/off you would not notice the leakage current. One solutions is to use a MOSFET more suitable to small currents, for ...

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My guess is that you did not connect the ground of your microcontroller (Arduino?) to the ground of the MOSFET circuit. Make sure you have a solid connection between the two, and then measure the MOSFET gate voltage with respect to ground when the output is both 0 and 1.

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You are confusing the data input and the power supply to the WS2812. The WS2812 (single or a chain) draws its power from the 5V connection. At full brightness a single WS2812 can draw 60 mA IIRC, so that is 1A for every 16 WS2812's! Hence it is a good idea to bypass power (and ground) of a WS2812 (or similar) LED strip with a thick wire that jumps let's say ...

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If STM32 has Vdd max = 3.6 and WS2812 is CMOS input rated for 3.5 to 5.3 with Vih max >= 0.7 * V+ then you need a level shifter to extend past < 30% & > 70%. Due to mismatched impedance of interconnecting cable some ringing occurs so a series R <= cable impedance improves damping of ringing to Input. Twisted pairs might be 240 Ohms and driver ...

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Thanks for all the help. I will definitely check out the solution @ringk89 came up with. I had idea of my own, trying to solve this using PWM. This is the solution I came up with: Simple sim: falstad sim It is somewhat exponential, and I have tested (without the inverter so far) it on a breadboard and it seems to work great. Values are a bit different as ...

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It sounds like you might be using the power meter incorrectly. The power meter has a wavelength setting, but this only affects the calibration. It does not change the operation of the sensor. The wavelength setting is used to compensate for differences in the response of the sensor so you can get an accurate reading. For instance, most sensors have a peak ...

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If this is a fairly low-frequency wave you want to represent, you could sample your waveform with a microcontroller ADC, then convert the waveform voltage to separate green LED and red LED PWM outputs. Then you have control over the visual effect, including doing a non-linear (gamma) mapping to improve the visualization of voltage. Why non-linear mapping? ...

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Nothing stopping you from putting log scale and linear scale on it. Use the big log scale (or even the small log scale) to create a graphic in some graphic program. Group it. Then scale. Linear, 6 lines. Align edges and space evenly. If $I_F$ = 120mA, then $V_F$ = 2.94V. A = 108, say 110mA and 2.9V. B = 128, say 130mA and 2.95V.

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You are trying to get more information from the graph then is really there. Remember that this graph is for a typical LED and the device you have in your hand may be significantly different. The graph shows you the shape of the relationship between current and forward voltage, but it can't give you an exact value for any given physical LED. If you need ...

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