New answers tagged

0

Taking a wild stab at this, without knowing your battery longevity constraints: Tie the battery output to an AP1603 to step up to 3.3V from as low as 1.1V Use a voltage reference well below 2.7V such as the LM4041 1.225V that has quescient current below 100uA, between ground and the non-inverting input of an op-amp; with bias resistor tied between 3.3V and ...


1

As you have guessed, such a low resistor value means that any variation in LED Vf, which will happen as it heats up, will translate into a substantial change in current. Since LED forward voltage decreases as they get hotter, more current leads to more current. It might smoke, or it might not, but you'll have to fiddle with it, measure current when hot, try ...


1

R1 and C1 to taste. I have no idea what the general consensus is on which opamp is suitable for breadboard/protoboard casual use, but it seems like the consensus is that the 741 belongs in a museum. The MCP601 is a readily-available single-supply rail-to-rail (more on that in a second) opamp that gives 22mA Isc at 5V supply, comes in a DIP-8 package, and ...


0

700mA through an LED will generate a lot of heat. Make sure the heat sinks on the LEDs are adequate. If not, they will overheat and fail rather quickly.


1

A typical string of LED holiday lights uses either resistors or capacitors to limit the current, as well as a reverse-blocking diode, as shown in the below schematic. Capacitors tend to be less common for some reason. Perhaps it's cost-related, but I wouldn't expect a capacitor to cost a whole lot more. Anyway, if you can isolate one of the canisters and ...


0

If 120 volt AC hit both the LEDs in parallel and the output of the power supply, it's very likely that everything has been damaged including the power supply. Your initial set up is good but you have to replace everything. You can verify the output voltage of the power supply with new leds. But if it's defective you risk to blow up leds again.


0

Me, I would wire an ammeter in series with the LED. Then measure the amps. Then, I'd get a constant-current driver that runs that exact current (at that general range of voltage). But then, I'm presuming I have ordinary LEDs which are not ohmic (i.e. do not obey Ohm's Law or random websites in fancy fonts). If I had super-magic ohmic LEDs that mysteriously ...


1

3V on the LED is what you'd expect from a white LED. Forget about the 9V supply. Instead use a USB "charger" ie a 5V 1A power supply. This should be much more useful as you'll be able to plug the light on a powerbank, on a cellphone charger in the car, etc. Also the resistor power dissipation will be manageable. Cut a USB cable and connect to the ...


0

It is not a good idea to connect two voltage-type elements in parallel (here, a charged capacitor and a diode) since the current will vigorously distribute (steer) between them. In your case, this means a vigorous fade-in. If you still want to insert a resistor, do it in series with the LED to "soften" its "hard" IV curve. This will make ...


10

There is no danger to the LED from removing the voltage source V1 or shorting it to ground. The LED itself limits the voltage across the capacitor, i.e. if it was safe while the voltage source was still connected, it's also at least as safe after it has been disconnected. However, if you allow C1 to charge up to greater than the LED forward voltage and then ...


1

As others have mentioned, photoelectric effect (solar cells) and photoemission (LEDs) are pretty much inverse processes. My high-school-physics mental model is like this: Electrons orbit a nucleus in shells. The physical size of those shells dictate how much energy the electrons have (and somewhat depend on the type of atom). Thus, to move from one shell ...


1

Assuming the motors are on the same power supply and not on separate power switches, LEDs in parallel will only tell you the power supply is active. You need to detect some current flow (motor active) but not too much current flow (motor stalled or bogged down) - and you want to do it without losing very much voltage in the detector ( 2V for a LED would only ...


-1

You can try to connect the led in parallel with the motor use a current limiting resistor for the led else you will burn it out and whenever there is power to the motors the leds will turn on and when there is no supply the leds will turn off. Ideally you can use a 220 ohm resistor it would be fine and a supply of 12V


2

12 Mbd over 45 km with a single wavelength with multi-mode fiber will just not happen, due to the physics of light propagation. Really, really good multimode (OM4/OM5) fiber (that's mainly about consistent geometry) has some 3 GHz·km bandwidth-distance-product at 850 nm, and something like 500 MHz·km at 1300 nm. That means that after 45 km, you get 500/45 ...


0

The distance the data will be required to travel is 45km and the light source needs to transmit data at a bandwidth of 12MHz, and at a data rate of 12Mbps. Ideally, I want to use a light source with a central wavelength of >1300nm. Given multimode and low bandwidth, an LED is probably a good option. Notably absent here was a specification for TX power, ...


2

The chosen opamp MCP6241 is quite suitable for this application as it is rail-to-rail input/ouput, and Vcc as low as 1.8v. You have two options to handle it: 1) add a resistor between Vcc and inverting-pin of opamp. Because when the + pin is zero, the - pin tries to reach zero as well but because of input-offset, an error appears on the output. so you have ...


1

Clearly that circuit not work well with near zero reference voltage. Maybe input voltage offset of the opamp. The resistant value is very sensitive if you choose to low then the offset voltage can cause more current at zero voltage ref. as I = V/R. if you select too high value then the current will satterate at low value. the maximum voltage across resistor ...


0

it is an interesting circuit and it works correctly although you do not get what you expect. It blinks and in less than one second turns off: The circuit, how it is made will work only once, only one time, the shift register circuits, the 4015, are never reset so you need to add some logic to that circuit to reset the 4015 after all LEDs are turned off It ...


0

To have a repetition cycle of descending LEDs illumination, you need negative feed back. Best bet is use an inverter on last LED to drive Reset instead of Vdd. the inverter gives negative feedback. using a couple XOR Gates and 1 inverter makes it random and there is 1 combination case for a maximal length random sequence (MLS PRSG) Look it up. 2^8-1 The ...


0

Presumably the 555's threshold pin is connected to the inverting input of the op amp. Swap the inputs over on the op amp. 555's threshold pin to the op amp's non-inverting input and the pot to the op amp's inverting input. And do put a current limiting resistor in series with the LED.


1

You can calculate the capacitance between your traces using the formula in this answer. It will be very small, less than 1nF for the entire board probably, so negligible compared to a ceramic decoupling cap costing a few cents. It would be important if you were interested in crosstalk between one trace carrying a high frequency signal and an adjacent traces, ...


2

I know that if I decrease the clearance between GND and VCC is the same as adding decoupling capacitors that will help in case of power supply voltage fluctuations. If the GND and VCC are on the same layer the capacitance will be too small to matter, the plate area is small. If the GND and VCC are on adjacent layers, then the plate area can be large enough ...


0

The easiest way would be to use a transistor to pull the PWM-Dim pin down to PWM-. The PWM-Dim is also the on/off signal for the power supply. Open (no connected, floating) is on. Connected to PWM- is off. That means it has a pull up connected internally. All you have to do is pull down PWM-Dim. Your PWM controller provides up to 5A of current, and is ...


2

Theoretically you could use the transistor's current gain to calculate the required Base resistor value. However this is a poor method of determining Collector current because (due to process variation) the current gain may vary widely between individual parts, as well as with temperature and (to a lesser extent at medium values) Collector current and ...


2

It's the same issue as your earlier question. Finding current through LEDs in series-parallel circuit Wiring multiple LEDs in parallel to a common load resistor will divide the current between the 'on' segments because of our friend, forward voltage (Vf.) Each lit segment has a fixed forward voltage, however the resistor current will be the same regardless ...


5

If you use one resistor for a 7-segment (say common anode) LED display, the display segments (all other things being equal) will be radically dimmer when all 7 are illuminated vs. when one is illuminated. That's regardless of how well they share current, which is also a problem, particularly at relatively low LED current. You could drive a single seven ...


6

I don't understand this: On an LED Voltage-Current graph, does the LED "pull" a certain current like it has resistance? Yes, to understand intuitively this phenomenon known as "current steering", you can think of the LED of as a resistor... but having nonlinear resistance with the property to keep up the voltage across it constant. It is ...


20

LEDs’ forward voltage drop varies from device to device and by more than you might think. This causes a few issues for the user. Figure 1. A Lite-On LTST-C170TBKT. Image source: Lite-On. The Lite-On LTST-C170TBKT InGaN blue LED datasheet, for example, shows the forward voltage varies between 2.8 V and 3.8 V at 20 mA. This is quite a variation. The lower ...


13

If you open almost any LED datasheet which gives its voltage and current specification, you will find out that the LEDs are not perfect devices, they have tolerances. So if you buy a bag of hypotethical red LEDs, and test them by driving exactly 20mA through each of them from a constant current source, you will see some variation what the LED voltage is. ...


3

The typical output behavior is clearly shown on the datasheet: If you keep the output current to < 10mA you'll typically see less than 400mV drop (sourcing, the way you have it) or about 350mV (sinking).


0

Old question but I wanted to close this with an answer. I wound up placing a relay in series with the driver output. It was acceptable to wait for a bit in my application. I therefore turned off the LED driver and opened the relay contacts. I then proceeded to measure the voltage on the output of the LED driver and waited until it discharged down to a ...


3

350 mW/sr means \$(4\pi)(350\ {\rm mW})\$ into the complete sphere around the source (if the source were to be emitting equally in all directions, which of course it isn't). So at a distance \$r\$ from the source, the peak irradiance is \$\frac{(4\pi)350\ {\rm mW}}{(4\pi)r^2}=\frac{350\ {\rm mW}}{r^2}\$. As shown in figure 4 of the datasheet, you can indeed ...


0

A quick google search suggests it's the electrons jumping into holes that give off photons. Looks like there's plenty of info out there on it. At a shallow level, it maybe the exact opposite of the solar cell effect. In fact, LEDs also work as solar cells.


2

You don't wire a voltmeter in series with the load. Instead you wire it in parallel with the load because you're trying to measure the voltage across the load. simulate this circuit – Schematic created using CircuitLab Figure 1. A voltmeter is wired across the load. An ammeter is wired in series with the load.


0

For assured safety you need to use a DOUBLE INSULATED transformer, as you have no safety ground on the secondary circuit which possibly may be subject to the effect of weather (especially rain, damp) and human contact. There is much that could be improved regarding the circuit, as noted in the other replies, notably regarding flicker at 2 X line frequency. ...


0

If you dial down the voltage to 50 V then monitor the current rise with temperature as Vf reduces -0.1%/‘C or ~-0.5V then you should be able to adjust the current and temperature so that they do not overheat with a 700W supply to have margin, as it is a bad idea to load a supply to >=100% for reliability reasons. You might then be able to Make a dimmer ...


4

UPDATED - MAJOR CHANGE due to new information: WARNING The question has probably been misinterpreted by everyone (including me) due to a lack of complete information. I/We thought there were 20 series LEDs per string and that the OP's diagram showed the WHOLE LED setup and that each LED shown was a 1.2W unit. With the added (clear and nicely lit and focused)...


1

Without any reservoir capacitor on the output of the bridge rectifier, the LEDs will flicker strongly at 100Hz. To convert it to 50V AC, you will need to re-calculate C1 and R2 to get the correct current through the LEDs. Also white LEDs require about 3.0 to 3.3V to run. A string of 20 thus requires about 60 to 66V. The peak voltage of 50V AC (about 70V) ...


1

You can solve all of this difficulty, complexity and heat management by using a centre-zero moving coil meter. (And, yes, I realise that this is not a direct answer to the question at the top of the page. It is an alternative.) Figure 1. A μA moving coil meter movement. Image source: [ESR}(http://esr.co.uk/electronics/test-panel5.htm). Many of these are ...


1

Depletion FETs like DN2540 are ON with Vgs=0V. Similar to a JFET, adding a resistor in series with the source pin makes a constant current source. Since this MOSFET has a body diode, we need two in series so it acts like a current source in both directions. Only one resistor is needed. Unlike a JFET, this one withstands pretty high voltage. This makes a nice ...


1

It's very Soviet looking. Those large LED displays were frequently made with series LED dice so if there is an open connection the whole segment will go out. If the display digit or display is potted there's not a heck of a lot you can do other than replace the entire bad part. Maybe swap the MS digit? You might be able to rotate it 180° to get working ...


0

IIRC, analog write on Arduino uses PWM (pulse width modulation). The whining noise you hear from the power supply might be because of the PWM. The voltage you measured between common positive and red negative of the LED strip (I assume that is what you measured) also sounds reasonable. If you measure DC voltage with a multimeter, it will measure the average ...


0

OK, let me add another fantasy suggestion about the nature of the resistor:) As you need only a polarity detector, why not replace the resistor with a capacitor... or better by a network of a capacitor and (protecting) resistor in series? So, add only a capacitor to the Spehro's circuit. I think 1-10 nF (non-electrolytic) and 1 k resistor will do the job. ...


12

Figure 1. Image source: Adafruit. The data is modulated on the 38 kHz carrier as shown in the image above. The 38 kHz is transmitted in bursts and it's the bursts your camera is detecting. You are correct that your camera's sensor will integrate many 38 kHz pulses in one video exposure "frame". Having the carrier frequency makes the system much ...


5

No, because the 38kHz is turned on and off to send data, it won't be continuous 38kHz signal. And the code is repeated few times per second. But yes, if it was a continuous 38kHz modulated light, the camera would see it just being lit.


4

Remotes don't transmit continuously. The output signal looks like bits modulating a 38kHz signal, then a pause like a few tenths of a second, then another transmission. So it'll look like it blinks, but that's probably not an effect of the 38kHz modulation, rather the pauses between retransmissions.


1

Thanks for all feedback, ideas and support! They were really helpful! I manage to solve the problem by: Lowering the values of the pull-ups on the SDA and SLC close to the arduino (470 ohms each) And I used a thin Cat6A cable - less than 4mm diameter - (with 28 awg twisted pairs inside and no shielding), instead of my previous 6 core wire with shielding. I ...


0

I think the simplest answer is to use an 'off the shelf' MR16 LED lampbulb such as any of these .... https://www.ebay.co.uk/sch/i.html?_from=R40&_trksid=p2334524.m570.l1313&_nkw=mr+16+12v+led+bulbs+-halogen&_sacat=0&LH_TitleDesc=0&_osacat=0&_odkw=mr+16+12v+led+bulbs And step up the battery voltage a bit using something like this ... ...


1

Instead of through hole LED, use an SMD LED. They can shine with much less current than THT ones and give enough visual effect over a broad range of voltages. However, 3.5 to 60 V is still too broad. To reduce the range, use a 10V Zener diode. This will reduce the range to 3.5V - 10V. Chose R2 to fit with 10V, bearing in mind that there is already R1 ...


1

Here's one way of doing it, using an LM358 dual op-amp and a few resistors:


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