# Tag Info

1

Of course it is possible. Whether you want to do that is another thing. 9V batteries can have more than 9.5V when new and are considered empty around 5V. Blue LEDs have approximately 3V forward voltage, but it won't be exactly 3.0V and it will vary due to manufacturing tolerances etc. So assuming all the LEDs have exactly 3.0V forward voltage at their ...

2

We take current measurements by connecting an ammeter (or a multimeter with the leads and range set to mA or A) in series with the load - not across the load. simulate this circuit – Schematic created using CircuitLab Figure 1. AM1 shows the correct way to measure the current. Typically the voltage will be measured with respect to ground (usually ...

1

It is drawing less than the paper 6 amps for several reasons. voltage drop down the long strip. Those internal pathways are PCB traces, not superconductors. cheap items of foreign manufacture are not running 100% to spec. actually, LED strips are sized to work safely on 14 volts, i.e. Automobile battery float voltage. So you are not running them at ...

0

I suspect heat to be an issue with the switch, cool down your light in the freezer then try it again to see if the delays get longer.

0

Figure 1. From the datasheet it is clear that the LEDs are wired with three in series. What you have missed is that each 20 mA powers three LEDs in series. This is the most efficient arrangement for a 12 V supply and little power is lost in the series resistors. Re-running your calculation we get:  I = \frac {20 \ \text {mA} \times 300 \ LEDs \times 3 \ ...

1

If flip dots don't suit your needs, try one of the following: A remote-reset switch, which can be flipped into the off position by an electrical signal and must be manually reset. Here's one. For a bit less money, you could get a small circuit breaker, one that you can trip on command with a pulse of current. Something like one of these, maybe.

1

How about blowing a fuse. Put a fuse on the incoming power. When there's an error then short the fuse with a transistor. The fuse blows and the board is dead. If someone tries to use the board after it won't work and they'll have to start investigating.

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There are Bi-Stable displays, both electro-mechanical (A disc with red and green painted on it that can change state) and E-ink types. Mechanical flip dots: example

2

You cannot light a visible light LED with just a single AA cell. A single AA cell provides about 1.5 volts. A red LED requires about 1.8 volts to light up, and will not light at all at a lower voltage. All other colors of visible light LEDs require a higher voltage, and so will not work on a single cell either. Infrared LEDs light at around 1.5 volts, ...

2

AA batteries may be used to light LEDs. The factors to be considered are LED forward voltage, LED forward current and battery voltage. The battery voltage should be higher than the LED forward voltage. A resistor is to be used in series to limit the LED current. The resistor value is to be calculated using the formula Battery voltage = LED forward ...

3

Read about a circuit called "Joule-Thief". Single AA battery cannot light up an led. An AA battery has a nominal voltage of 1.5V which varies from 1.1 to 1.5 depending upon how much the battery has been discharged. Most LEDs require 2.5 - 3.5V minimum to turn on. Reds LED turning on at the lower voltages while Blue LEDs require higher voltages. Two AA ...

0

I'd use two LT3476 to drive eight strings in total.. These can go up to 36V, so you could use the same controller for longer strings as well. The ICs come in a 0.5mm pitch QFN package, so check that your board house is capable of putting soldermask between those pads reliably, otherwise it will be rather difficult to solder.

1

I thought of an LED that might work for your application. It features a ceramic package that makes it more environmentally robust and also conducts heat away from the LED chip more effectively than the usual clear epoxies they use. You may be able to use a conformal coating with these, because of the ceramic package. And because you said your application ...

0

As far as I can tell from your question, this circuit might help you: It is from Nuts & Volts magazine, and part of their Bipolar Transistor Cookbook, an eight part tutorial worth going through. You would feed the output of Q (or not-Q) into a circuit that turns on your LED.

1

The white circles are caused by the LEDs not being aligned correctly with the light guide or diffuser. The backlight of a TV is an array of LEDs that illuminates a diffuser/lightguide that shifts the point source of the LED into a uniform white background. The spacing between the diffuser and lights is critical for it to function. If they become ...

0

In absence of individual current regulation for each led, the total current of leds connected in parallel is unpredictable. One led may draw more current than another one, apparently identical. The difference between red and blue is even higher. I imagine these are 12V leds. But you need to know the maximum current which can be used for each led and then, ...

0

There are two types of shift registers: First: Series in Series out (SISO) Second: Series in Parallel out (SIPO) I used a button in Logisim as a manual clock .

1

Both the power supply and the LED strip are fine with that, as long as you don't draw that from the standby power pin you don't give a signal to the LED strip while it's unpowered (WS2812 are notoriously unhappy with that) you supply the LED strip in multiple places (because the copper is too thin to transport 3.5A all from one end)

1

This question has been asked here so many times it is getting boring. A properly designed load will only draw the current from the power source that it needs. The power source, if it has greater current capacity than the load requires, is not going to somehow force the load to take five or six times the current that it needs. Assuming that everything stays ...

0

Links to datasheets ALWAYS a good idea. F9540 MOSFET 74HC595 Cin gate around 1400 pF depending on your conditions. Rpullup 1 kΩ Time constant on gate is T = RC = 1000 x 1400 E-12 = 1.4 µS You are seeing a much longer gate turnoff time that that. Are you sure the 1 kΩ is not a 100 kΩ or so? Failing that, it makes no great sense. Look ...

3

since the energy must be conserved Most of the electric energy is still turned into heat, not light. About 1/2 to 3/4. The ratio between light and heat isn't a constant but gets smaller at higher power.

2

Your inputs are as follows: A B D C are A3 A2 A1 A0 0000 will show 0 0001 will show 1 1001 will show 9 See the second half of the answer by @jonatanjo

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If you have reversing voltage, you can put two leds in opposite direction, each protected against reverse voltage by a schottky diode. You can use led of different colour to see which direction the current flows to. Only one led will shine at a time. simulate this circuit – Schematic created using CircuitLab

2

... I connect rectified 9v (using a transformer and diode) ... I'm assuming you are describing a half-wave rectified voltage source without a filter capacitor (a capacitor isn't mentioned in your question). An AC signal whose RMS voltage is $X\,V_{RMS}$ delivers the same power to an ideal resistive load as a DC voltage $X\,V_{DC}$. (NB: This equality ...

0

Fuses won't have the well defined cut-off at 3.001 A that you might be hoping for. It's more likely that the lamp will blow and protect the fuse. I don't think that any great complexity is worth the trouble to protect a relatively cheap lamp. You can add parallel resistance to divert 1 A away from the bulb but this is a potential failure and a waste of ...

2

But i'm wondering is it possible to connect it to 220v mains voltage (311V DC after using bridge rectifier) if i used a proper high voltage transistors such as A42 or 13001 instead of low voltage BJT? It's possible. It'll even work. Whether it's safe or not is something that'll be hard to answer. It's also not very necessary: grid voltage is pretty ...

1

I looked up online and there are people who followed that same idea, but I could not find any well done testing on that, any application notes or professional-level schematics. it is definitely possible that many cheaper products do use that same idea, but usually cheaper products don't care about using constant current (or safety) that much. One of the ...

1

PPF is the number of photons (usually given in moles) that are emitted from a light source within the visible spectrum. For a monochromatic LED, this is simple to calculate from the photon energy and the radiant flux given in the datasheet. For a white LED, the calculation is more complex, but it can be estimated from the LED's spectrum and radiant flux. ...

-1

Here are two images of current sink and current source (e.g in a PLC device): Note that current sink is often used with micro controllers that may have problem sourcing current. Figure 1 shows a sinking digital output that is connected to a sourcing digital input. In this circuit, the load is pulled to ground because of the sinking digital input provided. ...

1

Even clear coating will have a different refractive index than air. So, that means that the lens and reflector design of your LEDs won't work. So, no, this is not an option. Also, most higher-power LED designs depend at least partially on convective cooling, and coating would drastically reduce the ability of the PCB and the LEDs to lose heat. I hope you'...

1

The most common solution to driving large numbers of LEDs at (approximately) equal brightness is to use constant current drivers. A common size driver is 250W. If you used that size, you would set up about 5 series circuits, each consisting of about 50 LEDs. You would then choose a driver with 700mA current and 360V maximum voltage. This is a common ...

0

I would suggest you try placing the R1 immediately next to the base of Q1 so that it is between its base and emitter. That way it keeps the Q1 turned off if there is a small current going through R2. I would also suggest that you increase the value of R3 to anywhere from 220Ω to 680Ω (depending on LED's forward voltage and the total voltage applied, which ...

1

The maximum, not nominal, current draw for a driver (which are usually mains powered, this doesn't apply to bus powered) is 2mA, this is over the voltage range 0 to 22.5V and the temperature range. It applies when the driver is not transmitting, because when transmitting the driver pulls the line down to the current limit of the supply, maximum 250mA. The ...

2

maybe just unse one GPIO on the teensy to control a transitor that controls all the LEDs.

0

Fundamentally no different than running multiple wires in parallel. As mentioned you need to size the wire for the combined load, but 16 or 18 gauge will be fine for the 4 Amps 12v here. The switch may be an issue. Household 120v switches are not intended to be used with DC voltage. They depend on the AC sine wave to ensure little arcing and keeping the ...

1

Is it possible to share the positive wire from the breaker box with both light switches and go back to the breaker box with a shared negative wire? Yes you can do that, as long as cabling is thick enough to handle currents from all of your lights. Will this work or is there a better alternative? Circuit looks perfectly fine.

2

All your calculations are correct. The graph is a typical I-V curve for this LED, the drop will be higher for some parts, and less for others. The table lists the maximum voltage drop at 20 mA to be 3.4 V. They did not list a minimum voltage. This is a common practice, they don't want to reject parts because they are "too good" (low voltage drop is normally ...

0

No. you can't.Because you need a threshold voltage close to relay voltage to triggering it. Thanks.

1

No. You cannot. You can, however, use a 5V relay instead of a 12V relay as long as you put the appropriate resistor in series with the relay coil (so that the relay coil sees the same voltage across it and thus current through it that it normally needs). What you are going to need to do (other than getting a 5V relay) is to re-design your circuit to ...

1

You don't need a dimmer if you don't intend to modify the luminousity frequently. What you need is to set the right voltage so that it will draw the right amount of current. If you put 9V to it you will draw the maximum 300mA. If you want 50% less, you must draw 150mA. Idealy you could use a constant current power supply giving 150mA and 1.5W (between 1 to ...

2

There is no inductor on the dimmer PCB so it is almost certainly driving the LED directly with the PWM signal. If you attempt to filter the output just with a capacitor it will probably cause the LED to run at almost full intensity even if adjusted for a lower value. The capacitor will charge during the on period and then power the LED during the off time. ...

0

If the LED is strobing when you reduce the duty cycle, it suggests that the dimmer in question is using some sort of (poorly) low-pass filtered PWM, or the LED's drawn current is higher than the dimmer's rated output current. As you already suggested, adding a capacitor will reduce the cut-off frequency and hold the output voltage a bit longer until the next ...

0

Smoothing capacitors can be calculated using the following formula: C = I⋅Δt/ΔU where C = capacity of the capacitor in μF I = Charge current in mA Δt = half-period in ms ΔU = ripple voltage in V However, in this case adding a smoothing cap might actually make the problem worse. It is important to note that the intensity of an LED is ...

1

Here is an LTSpice XVII circuit which (at least in simulation) does what you want: When the 3.3 voltage source is connected (representing a HIGH on your microcontroller pin) there is 17mA running through your 5-LED string. When the 3.3 voltage source is disconnected (by deleting the little bridge inside the circle) the current running throught D1 is ...

2

Here is what the display circuit should look like (showing 2 of 3 digits to make the schematic readable):- simulate this circuit – Schematic created using CircuitLab Each segment has a resistor to set the LED current, common to all digits. Each digit has a transistor to pull the common Cathodes of that digit down, with a resistor setting the Base ...

2

Based on my testing (see comments), the glow can be caused by transistor leakage. In my setup, the glow starts when the voltage is 26 volts (could easily vary depending on the transistor). Capacitive coupling would have a similar effect. One 10k resistor across all 5 LEDs removes the glow in my setup. This works up to 31 V, the highest that my power supply ...

2

This is likely not due to a flaw in the driver, or in the signal source, since only some flicker. Because LED's are not absolutely identical, the voltage across each in a series string will not be equal for DC due to differing leakage currents, and for AC this would be exacerbated by differing capacitance wihin each device and from device to ground. ...

5

It shouldn't be necessary to play funny games with resistors and diodes. Lighting LEDs with a microprocessor output is a normal thing to do, and shouldn't require any kind of overly clever circuitry. Make sure you are setting the common to high, and not simply "high impedance." You want it driven to a high state, not simply floating. You want to use "...

6

The main link for the logic for the individual segments is that they must share the same inputs. Optionally, they can share computation of a given expression: for example, inverted values of all the inputs occur more than once, so you can share the output of the invertors (/A, /B, /C, /D). Also, A./B occurs twice and could be done with a single AND. If ...

4

This is the truth table from the DM9368 data sheet, which should confirm the correct decoding.

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