New answers tagged

-1

Sorry to bump such an old post but yes this is what auto electrical designers would call the CAN Bus system. It is a small amount of current used to test the system for bulb outages. Nothing is as simple as it used to be, and apparently we aren’t trusted enough to notice our own bulbs are out..


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the following will work very well. Values are chosen to provide Leds with 10-11 mA, more than enough for the Leds to be bright. With Switch ON : With Switch off: Hope this helps :-)


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the 2 ohm resistor is for the inrush current protection of the LED diodes, and the 4.7 ohm is for stability to prevent thermal runaway.


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The capacitor acts like a capacitor. It charges up like it normally does, but when the voltage potential measured at the anode reaches the LED's turn on voltage, the diode's junction internal resistance goes down, and the current flows through the diode, discharges the capacitor until the voltage potential has lowered past the junction off voltage, then the ...


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Looks like a stuck clock fuse protection resistor where the resistor burns up and opens before the LEDs burn out. The resistor if rated for 0.5W and drawing 0.5W = 1/4A*2Ohms will run at max temp <150'C . But maybe they used a bigger power resistor. Looks like a poor man's or old technology safety solution when you can get SMD CC chips instead now which ...


4

The QTLP690 LED has a maximum pulsed current of 160 mA however, the constant current driver circuit formed around Q6 and Q3 can sink a current peak of about 200 mA. This is based on Q6's base being about 2.65 volts (1.25 volts from the LT1634 and two lots of 0.7 volts from the diodes). The 2.65 volts will become about 1.85 volts at Q6's emitter and drop by a ...


2

H bridges are for driving high current coils or motors. You have no current limiting R. CMOS (74HC family ) has about 50 OHms driver R so you can include this with your current limiting R. Since Iv intensity in xxxx mcd is common now , the curent can reduced to a few mA for indicators with a single Rs = 470 to 1k.


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Here is how I drive bi-color LEDs: simulate this circuit – Schematic created using CircuitLab Use a logic family that can source or sink 20ma or whatever you need to drive the LEDs. I actually don't use inverters (that solution was closest to your circuit), but a serial to parallel shift register (like a 74xx591) to drive 4 bi-color LEDs using the 8 ...


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Don't use that IR receiver. It was meant for serial communications and therefore has application specific circuitry integrated into it which will limit it's broad applicability. Use a phototransistor or photodiode instead. These components are the fundamental optical receivers in electronics and all IR receivers and anything similar contain either one or ...


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If you are controlling each LED independently then each should have its own resistor. If you are controlling a whole line/string at a time then using a single resistor per line may work out well enough. Position does not matter as long as the resistor is in series with each LED or string. The only issue that might change this is if there is a chance you ...


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As others have noted, there's no problem as such with having both sides of an LED at +9V. It'll just be off, just as if both sides were at 0V. I do see a couple of other potential issues with your circuit, however: Your voltage notation is confusing. Assuming that the triangle symbols in your circuit represent ground, which is at 0V by definition, all ...


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No, you don't need a relay. There is never actually any reverse voltage across the "Ready to Arm" LED. It goes off because there is +9V on both ends of it, for a net voltage difference of 0V.


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ADD: The left cct cannot run in CCM (continuous conduction mode) because the switch also shorts out the LED to 0V. But besides that now you have a low ESR path with a discharged cap that draws as much current as the inductor can swing V+/DCR and causes a high Q current resonant cct and adding pulses adds fuel to the fire with burnout on the series resonant ...


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One more reason to point out is that the strips are quite resistive (being "thin" strips of copper) and are usually ran through long lengths. That causes a considerable voltage drop, so with 12V you have plenty of margin for that. Granted, you can see the voltage drop on very long runs (i.e. brightness or color shift) but its better than simply not having ...


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In a simple configuration like this, there is a tradeoff between efficiency and accuracy. The forward voltage of each LED is slightly different, and it changes with temperature. If you are to run the LED strip at 3V, all LEDs must be in parallel, with no individual current limiting for each LED. That would lead to uneven lighting, and probably thermal ...


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There are lots of issues with using 3V directly. For starters, the forward voltage of an LED is related to the wavelength of light it produces, ranging from about 2V or less for red to about 3.6V or more for blue. So no single voltage is going to work for all three colours and devices that run from a single supply would have to be given enough voltage for ...


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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 the 3V led with 12V and a resistor directly. That would be too much loss. Since there exist DC/DC converters which can transform voltage with a good efficiency from one voltage to another, it does not ...


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If it really is capable of switching 30A DC, then you can connect both LED strips in parallel to the 'LOAD' connection. Looks like you want the TM630A-4 which is the 12V version, not the one in your photo which is the 250V AC version. Looking at the datasheet for the power supply mentioned in the comments, it looks it has two separate outputs, so you won't ...


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Thank you all again. I found the answer. I am not that familiar with solid geometry. What I was looking for is the area of a SPHERICAL CAP. The formula is pretty simple. A = 2 x Pi x R^2 x (1 - cos(a)). I will close this question.


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Remember that the "discharge" pin (pin 7) is essentially a copy of the output pin, except that it is open-collector — it can only sink current, not source it. But you could use it to control your LED without affecting the rest of your circuit: simulate this circuit – Schematic created using CircuitLab


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For a quick and simple solution, (if you don't mind an inversion of the LED function), you could just add a pull up resistor to +5V then connect the LED between that resistor and the 555's output pin. For a more modest LED current use a resistor of 270 ohms or more.


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From the NE555 datasheet: we can see that with a 5 V supply we can expect around 3.3 V at the output when loaded with 100 mA. Your LED of course consumes less current but that will not increase the voltage by much. So the behavior you see is to be expected. I expect that even without the LED the voltage at the output will not reach 5 V! That's a ...


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If you are performing this repair as an exercise in engineering, then you could definitely build a replacement circuit using op-amps. After all, that is what the original IC used in it's design. It would be a somewhat complex circuit, however, when compared to the compactness from integration of the original chip. If you simply want to "future proof" the ...


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If you wanted to replicate this with monolitic IC's then the datasheet would be a good place to start. Instead of a regular op amp the first amplifier is probably a log amp, and then followed by comparators. The resistors set a range of voltages for the comparators from lowest to highest, the external resistor between 6 & 7 sets the range for all. The ...


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No, a comparator won't do (also, how's that more "modern"?). You can try to approximate the logarithmic response using discrete semiconductors and a lot of hand-tuning, then use a parallel output DAC plus LED-driving transistors. There's a reason these ICs were popular even back when ICs were expensive! The alternative was pretty complex to implement. For ...


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Am I missing any options? I'd measure the voltage drop across the strings, if it's vastly different (like more than say 5%) then use different CC drivers. If not then it's fine to parallel them. Separating out the strings will allow for better control of the current, but if your not calibrating the LED's or have any requirements, then the configuration ...


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Just Googling "555 astable circuit" will list plenty of web sites. This one: 555 Oscillator Tutorial gives the equations for the on time and off time.


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Your suggestion could work. I have done this with a single white LED where space and cost and simplicity were paramount. The boost diode wastes voltage and hence power. Your 33V string has about ten times the voltage that I had on yesteryears' birdsnest. The prospective power savings will not be great in percentage terms. You will waste more power in the ...


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These look like DIN loudspeaker plugs. Odds are you could find some extension cords for those too! I'm not sure how would you still source them, these new connectors aren't very new at all, and seem to be quite rare these days. One can still apparently find these listed on our favourite auction sites and the like as B&O connectors, as the brand still ...


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It's almost certainly down to the filter. It's hard to make a filter that has a very sharp cut-off between transmission and absorption, and make it cheap enough for use in consumer electronics. So the IR-blocking filter in your typical phone camera will still let a substantial amount of IR through, and the visible blocking filter on a typical IR remote ...


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The human eye response stops at around 700nm to 750nm, you might be picking up a small amount of red because there are still a few photons in the red range produced and the eye is really sensitive to them in the dark. (its been theorized that the eye can be used as an individual photon detector.) Source: http://hyperphysics.phy-astr.gsu.edu/hbase/vision/...


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Using 33uH inductor in the circuit, LED can draw 110mA current. This is the not full power required to get full brightness of 1W LED but 110mA gave good light. Thus, 1W LED works with QX5252F in the solar garden light circuit. This thread can be closed now.


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if you look closely at the matrixes you will see that they are not regular square crosspoint matrixes, but instead charlieplexed. because of this each logical row and column has only 8 positions which limits the number of RGB leds that can be fitted. thus there are only 16 places in each matrix where common-cathode or common-anode RGB leds will fit, ...


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An infra-red LED can detect the raw infra-red signal from a remote. An oscilloscope can view the voltage across the LED (diode). A standard oscilloscope appears as a 1MEGohm resistance, in parallel with some capacitance (perhaps 25pf). The diode itself adds some capacitance as well. If you use a 1X oscilloscope probe, its additional capacitance will cause ...


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I do not know if you are connecting this to 120V or 240V ac mains, but this source voltage (U) and the resistor (R) is what defines the current through the optocoupler input LED, not the other way around! So for example you need a current of I=20mA with a given U=120V ac mains, then you can calculate R = U/I = 120V/20mA = 6kOhms. The resistor dissipative ...


4

It would be best if you provided current limiting for the GPIO port. It depends on the current gain of the NPN transistor, but at minimum a 200Ω series resistor from the GPIO to the NPN needs to be used to limit the current to less than 16mA. (and also use less than a total limit of 50mA on all ports)


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You need to create a bridge rectifier using the common-cathode LED and two other diodes: simulate this circuit – Schematic created using CircuitLab The red LED is lit when A is positive with respect to B. The yellow LED is lit in the opposite case. Build 6 of these; hook three of them up one way at one end of the locomotive, and hook the other three ...


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Figure 1. OP's thoughts. In your first circuit the LED will light while the PLC output is off. If the opto-isolator turns on then it will try to short-circuit the 5 V supply to ground. That would destroy the opto-isolator if enough current flows through it. As it turns R1 will limit the current through the opto-LED to \$ I = \frac {V}{R} = \frac {24}{10k} = ...


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Really all you need is a single series resistor. simulate this circuit – Schematic created using CircuitLab


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The operating current range is between 3 and 50 mA, so a typical small LED current of 10-20 mA will drive both components. Either schematic will work. I would go with the upper one; if either diode fails, the external indicator will indicate it.


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Optical geometry. Emitter: 5mm D LED say 65mW ~ 100mW Assuming +/- 30 deg Lambertian response below half power Target: 0.01m² = 100 mm x 100 mm square or some equiv round area. Calc: LED Emittance= input (0.1W) Pwr * ζefficiency(?) / Area/ wavelength[nm] = 0.1Wζ/385 nm/π{2.5 mm}²/ = 13.25 * ζ W/m²/nm Irradiance ratio of spreading light from π*{2.5 mm}...


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Using a battery that is not brand new so its loaded voltage is 8.0V and using a 3.2V blue LED, then the current is (8V - 3.2V)/220 ohms= 21.8mA. No problem when most 5mm blue LEDs have a maximum continuous current of 30mA.


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Here is how this works ----- simulate this circuit – Schematic created using CircuitLab I picked a 180 ohm resistor, because at 20 ohms/volt we are guaranteed a 50 milliamp intersection at the left axis. Thus we scale 20 ohms/volt by 9 volts, finding a 180 ohm resistor provides that current. Note other LEDs will have lower forward voltage, moving ...


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Lumens DO add directly. If the illuminated area is common to two sources then Lux (luminous intensity) = lumen per square metre also adds linearly. If you has say 3 LEDs (each of say 1200 lumen output), each exactly and uniformly illuminating a 1m x 0.5 m area and you placed the areas side by side to make a single 1m x 1.5m area, then Each area would ...


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After contacting the website about the amperes of the ring LED. 25mA So therefore if my supply can handle 1500mA, 4 x 25mA = 100mA which is well within the rating of the supply. Many thanks to @Hearth and @peter-bennett


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Put the reed switch from the input of a microcontroller (like Atmega, PIC, STM32, ...) to Vcc and put a pull-down resistor from the input the GND. You can now read the switching of the input like with any other button (don't forget to debounce if the reed switch contacts aren't covered with mercury). Now you can change the modes in software and enable ...


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Your circuit is underpowered - you need 2.5v-3.5 v depending on LED color. You'll learn the voltages needed for differing LEDs pretty quickly. When I started I used this as a cheat sheet. While a multimeter can help you determine if the LED is good, I prefer to use a Cr2032 (3V) coin cell battery instead. It's just faster than messing with my MM. And ...


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Most colors other than red and IR require more than 1.5V to light. You can try a CR2032 or CR2025 lithium button cell across the LED. That kind of battery has more voltage (about 3V) and enough internal resistance that it won't likely damage the LED. You could also use 2 or 3 AA cells in series, but you must add some series resistance (try a 1K\$\Omega\$ ...


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A reed switch and magnet is the same electrically as a pushbutton switch. The rest is simple interfacing and programming a micro (or designing a logic circuit with debouncing and some flip-flops or whatever to retain the states, but a micro would be, by far, the most appropriate solution in the real world). Speaking of the latter, if this is a homework ...


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You do need connections between the negative (grounded) terminals of the lights and the negative terminal of the battery. If these lights are on a vehicle, the metal frame and body of the vehicle can serve as the negative/ground connection, as the negative terminal of the vehicle battery will be connected to the vehicle frame.


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