200mA is lethal (actually much much less is lethal) and can kill, but only it if gets inside you. For it to get into you, it needs enough voltage behind it to push into you. If there is enough voltage to push a lethal level of current into you, and enough of that current is actually available to push then it is lethal.
Think about a bullet. It's a little ...
I would like to know how to calculate battery life when using LED strip. The LED strip is powered by 8 AA batteries. LED Current Draw = 260 mA.
Right. The batteries will be discharged at 260 mA. This is close enough to the 250 mA of the third bar on the capacity chart so we can assume 2000 mAh capacity.
This is my calculation, but I'm not sure if it's ...
These LEDs are not a single radial die, they are made with a transparent substrate with many LED dies in series (probably 25) placed on it. The whole thing is then coated in phosphor. The light isn't completely uniform but it's good enough.
There's a spec sheet here: http://www.runlite.cn/en/product-detail-145.html
The resistor is there to limit the current into the input pin. The input likely has a very high DC resistance (more than 1 megohm) so negligible current flows (on the order uA) and a negligible voltage drop is produced (on the order uV or mV). The resistor is likely used to slow the slew rate of the connection (the input pin will have some capacitance, so ...
You don't need two power supplies. One is enough. Just connect the positive wire to one side of the strip and the negative to the other. Then you will have equal voltage at each LED group and this will ensure equal brightness of all LEDs.
simulate this circuit – Schematic created using CircuitLab
Using two separate power supplies has some drawbacks:
To make it easy:
1- Car battery 12V (hundreds of Amps) will not kill you. Laptop charger 20V 7A will not kill you. Therefore, the 5V 2A will not kill you also.
Most of power sources you will see are voltage source. Therefore, the current will follow the voltage (by other way the current is dependent on the voltage ) and the impedance (your impedance here) ...
I purchased one of these lamps in Italy: 3W/300lumens/9,00euros
I took some pictures with very low exposition to higlights multiple "dot leds":
I can count 28 dot-leds per each strip and 4 strips. This gives 112 dot-leds.
Assuming half the cost is for the electronics, we could ...
The power source you are looking into is a voltage source (like 99% of the power sources you interact with in your daily life). This means that 2A is the maximum it is designed to supply, but it will try to keep a steady 5V. Thus, the current is determined by the electrical resistance between the terminals.
It can't kill you because your body electrical ...
After receiving IRF540N from a reputable seller I can definitely confirm that the ones I was originally using are counterfeits.
After replacing fake one with a genuine one I got Vds=85mV on the red channel. What I was not expecting though is that the genuine FET got hot after a minute or so. And then I realized that those FETs are not generating much heat ...
The 94V marking you see on the PCB is not a supply voltage rating.
94V-0 is a UL marking indicating the flammability rating of the PCB material.
As others have said, you ideally need a constant-current LED driver or failing that a series resistor in order to drive that LED board safely.
If you know that they are white LEDs then expect each one to drop in the ...
I used Firefox Web Developer Eyedropper to grab the RGB from a photo of an Edison Light Bulb.
A lighter pixel
This is very close to the color when a 2700K 97 CRI LED was reflected off bright white paper: #F4D4AB (244,212,171) just a little less blue.
(245, 212, 131)
I do not believe the LED actually illuminates ...
This is more interesting than it first appears. WS2815 appears to have been made in more than one revision but the latest design uses a clever constant current series trick.
The lower half is a constant current (say 16ma) regulator. The top rail is at 12v.
If you want no leds, you turn off the regulator.
If you want 0xFF0000 (red) you turn PWM3 off and ...
The FET you link to is not suitable for your application. Taking a look at the "typical characteristics" table, we see this:
Vgs(th) is listed as between 2 and 4 volts - already it could be higher than our supply voltage, but not especially the "conditions" column on the right: the specification is given with only 250 microamps of current. Your LEDs, ...
The IRU1015-33 has absolute maximum input voltage of 7 V. You are operating the IC way outside its design parameters. Anything can happen, and 12V on your communication module is a real possibility. Change the regulator to whichever can handle 12V input. Or switch the input of IRU1015 to +5V rail.
The wire between your power supply isn't just a wire – it has a resistance and an inductivity. People tend to use thin wires, much too thin, and then wonder why their stuff fails in mysterious ways.
simulate this circuit – Schematic created using CircuitLab
If you switch on many of your NeoPixels at the same time, they demand a change in current ...
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 ...
What are the best materials to use for this?
These panels are built in the same way as LCD monitors with LED backlight. Well, monitors have a LCD in front of the backlight panel, but you get the idea. Here's a monitor teardown.
The important part is a thick transparent acrylic plate which acts as a light guide via total internal reflection. It is lit ...
The datasheet for the WS2811 is where you need to look.
It is a constant-current three-channel driver that operates on 4.5 ~ 5.5 V, with absolute maximum ratings of 6 ~ 7 V. The output voltage is specified as 12 V maximum. Output current is maintained at 18.5 mA on each of the three output channels.
Therefore, the maximum number of LEDs you can drive with ...
Red LEDs have a much lower voltage drop than other colors so in order to compensate, the mosfet is having to drop a significantly higher amount of voltage/power.
Red LEDs usually have a drop of 1.8 V while Blue and green are up into the 2.5-3.3 range generally. This means that the Red mosfet will have to drop about an extra volt at whatever current you're ...
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.
What is wrong with your circuit is irrelevant from the point of view of getting a working unit.
There is no need for a MOSFET driver in this case. Use the IRLML2502 MOSFET and drive its gate directly from the digital output of the microcontroller. Check the IRLML2502 datasheet and you will see that it has a maximum guaranteed on resistance of only 45 m&...
Be careful about what you mean by "in series". If you mean that you're connecting the two inputs of one strip to the two outputs of another, you're really connecting them in parallel electrically, and yes, your concern about the maximum current that the first strip in the sequence (the one closest to the power supply) can handle is valid.
On the other hand, ...
Maybe not lowest component count, but chances are that you have all parts readily available. Any general purpose low power transistor will do and resistor values aren't too critical either.
Possible disadvantage is that it acts like an open collector with R2 being the pull up resistor. This means the rising edge is determined by R2. The 3V3 output driver ...
The PIR sensor output is a signal. It is not intended to drive LEDs directly. Fortunately we have such things as transistors that can switch LEDs on and off in response to a signal like what the PIR sensor puts out.
This simplest is to use a N channel MOSFET as a low side switch. At the low voltage of 12 V, these are available that can be turned on/off ...
They probably will, in my opinion, but there is no guaranteed with such a MOSFET. The key number is the guarantee of Rds(on) at the lowest gate voltage shown in the "Specifications" section. Vgs(th) is the guarantee that the MOSFET will be almost entirely off (250uA). There is a region between Vgs(th) and your operating point and you want the voltage across ...
I have a L7805CV transistor
No, you don't have one of those, as there is no such thing.
You have a 7805 voltage regulator (which is not a transistor).
and its looks same to TIP120
That is just the package (TO-220). Having devices in a similar package, tells you almost nothing about their functionality. (For more advanced readers, the package type can ...
First you have to determine what the maximum collector current is that the transistor needs to support when on.
Divide this collector current by the gain of the transistor to get the minimum required base current. The gain of a BJT varies with operating point, so you have to look at the datasheet carefully to see what gain value you can assume at your ...
12V might be the "quasi" standard because of the 12V car batteries/electrics that were there years before the LED strips.
Of course no one is driving a single 3V led with 12V and a resistor directly. That would be too much energy loss.
Since there exist DC/DC converters which can transform voltage with a good efficiency from one voltage to another, it ...
Short answer No,
Long answer, The confusion here comes from a few common misconceptions/misunderstandings.
First being, "Current kills, not voltage" or similar.
Yes, it is the amount of current that determines how much damage is caused. However it is the voltage that determines how much current can flow.
Ohms law states V=IR and therefore I=V/R.
A human ...