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The peak current is only 2mA if there are no voltage spikes on the AC. Voltage spikes will also arc across a little resistor that has a 250V absolute maximum voltage rating causing very high current in the led. I think you should rectify the AC, it feeds a series resistor with a capacitor to ground as a filter then another series resistor to feed
Hi, This solution doesn't use a PIR, but I can't think of any solution for what you describe... If the room only has one entry point, maybe an ir led across the doorframe could be used to latch on for entry and off for exit from the room.
Don't you think that when a transistor turns on then the current in the led attached to it is determined by the voltage across the resistor in series? The resistor is R1 and R4. You did not say your battery voltage. The kit says 3V to 15V. If you use a 9v battery then it is small so it will not last long. You need to look on the datasheet for yo
A narrow beam led is important. I sent signals from one to a photodiode across the room, several feet. I also mounted a magnifying glass in front of the photodiode. The setup worked fine.
led required very low current to glow. No it does not! The led uses as much current as it is fed. But its current is limited by the 1k resistor in series with it. If the led is a 2V red one then the 1k resistor has 5V - 2V= 3V across it then Ohm's Law calculates the current to be 3V/1k= 3mA. The Cmos IC with a supply
A circuit diagram would avoid misunderstanding regarding to the depicted above, as for instance if you are using capacitors for softening the current pulses across the leds.
The losses due to light emission should be negligible, and the power consumption of the led can be took solely on the current across its terminals against its drop voltage. Even assuming that the OP is referring to a continuous operation ( not pulsed ), the power density over this sufrace size seems feasible to operate at the standard FR4 PCB i
This is basically a continuous Buck regulator switching oscillator than regulates the pulse width for an average DC voltage across Vbe controlled by D1 zener voltage , Vbe and Ie*Re and duty cycle. Offline buck designs have improved over this and work in principle like active PFC to provide a sinusoidal current fluctuating aro
What do you mean by "My circuit works on 5V. Controller works on 3.3V"? if you mean the leds are powered from 5V, consider that even if the PIC port pins are high (~3.3V) there is still 1.7V across the led which might be enough to light it. Brian.
I think any expectations for bright, long hours of storage and fit into a small space are not realistic. Even the smallest I would consider is too big for your application and too short a runtime. A CR-123A Primary Lithium 3.0V directly across 20mA = 60mW approx with compatible white led with only 65h expected life. using[URL="www.batte
Two series power white leds will work fine directly across two 3.6v 14500 only when Charger is OFF. They will be much dimmer at 3V per led which is when the SoC of LiPo drops to 0. MOSFET switch must be much less than ESR of led's which is around 1 Ohm @ 1Watt. So choose <<100 mOhm. with logic level drive.
IS YOUR INDUCTOR CURRENT smooth? Have a look at the voltage across the sense it a regular triangle wave? If not then maybe you have instability. Sometimes, surface mount inductors cam make noise esp if the inductor current is not smooth.....also maybe the 1khz pwm may be making the inductor coils vibrate and make noise...maybe yo
You just have to check for the output current only. The voltage doesn't care, as the led will self adjust the voltage across its terminals. By using a PWM signal and using the output current as feedback signal you could accomplish your task (led dimming). Think about led as a simple diode - you don't have to set the (...)
I assume that your car is a small toy car because the leds will be VERY dim. You have 5 White(?) leds in parallel. A white led has a forward voltage of about 3.2V so the current limiting resistor will have 12V - 3.2V= 8.8V across it when the battery is 12V. Then Ohm's Law calculates the total current in the 1k resistor and (...)
As explained the voltage across a led is a crude way of controlling the current through it. Put a small resistor in series with the earthy end of the led and monitor the voltage across it, this will be proportional to your led current, which you want to stabilise. Frank
My question is related to thread . but since the thread is old i am starting a new one. I want to build a sensor using led so i need a constant brightness and temperature compensating driver (so no drift in wavelength with T) with a good precision. I have come across the CL2 series from Microchip (prev
I think a schematic would help but I'm guessing the issue here is the 12V relay and leds being mixed. For the relay to be used safely, the catch diodes in the ULN2003 should be connected to the relay supply (12V) so when connected to leds fed from 5V they forward conduct and leave 7V (12V- 5V) across the leds. The cure is (...)
Short answer: no. 5V (less the led drop) across 100K would only give you about 30 microamps. Unless you have an optocoupler that can work with this low current, you'll need to find another method. I'm not sure what you mean by "My scope is need to check the voltages". If you are using a scope, what do you need optos and rectifiers for?
It depends on whether your led shares a common anode or common cathode. The simplest scenario is you connect the leds to spare contacts on the relay. If the led is common anode, you could connect the common pin to the relay supply, one cathode through it's resistor to the collector of the transistor (so it is across the (...)
The only ways to do it are: 1. to pick up the original remote control codes on an IR receiver and use a logic analyzer to see what they are 2. open the remote control and connect the logic analyzer across the IR led. Each manufacturer and each equipment they make uses a different code system so trying to guess it when there are millions of combina
Hi there , i'm working on a pulse oximeter project using an arduino atmega168 to build a pulse oximeter . was hoping if i could get help with the code , i'm using two led'S A TSL235R AND an LCD. Appreciate the code might be complex to calculate the oxygen level and just wondering if someone has it or came across it ? Many Thanks
It would probaby work with 1K, you could drop to 470 Ohms to be absolutely sure you have enough led current. However, you have the optocouplers wired wrongly and they will not produce any output as shown. The MOSFETs probably want connecting differently as well, switching a capacitor across the supply wont achieve much. I would drop the 10uF capac
A logic level MOSFET would be more suitable in this circuit. However, to answer your question, the value for R3 is calculated as (supply voltage - led forward voltage - drop across MOSFET) / led current. If I assume you are using a normal indicator led with a Vf of around 1.6V and the MOSFET is fully conducting, the (...)
But the voltage rating of the led is used in the calculation of the resistor value. The resistor value is the voltage across it divided by the current you need. The voltage across the resistor is the supply voltage minus the voltage rating of the led. The voltage rating of the led is not one voltage, it is a (...)
The black capacitor is across the ground and supply pin, the yellow resonator is across the oscillator pins and the black transistors are between the output pin and the led. I would doubt it is possible to reprogram the device though, these kinds of IC are mask programmed with specific device address and command bits at the factory. It (...)
Hello everyone, I need a code for LPC2138 to run UART with interrupt. I have pasted my code below. I am switching on an led inside the interrupt if '1' is received. But the led constantly stays on and that too its dim with a voltage of 3.3V across it. Please help!! #include __irq void UART0_ISR(void); int main() { PI
There is mechanical version of this. You need to use voltage to turn the internal motor. This best suites for the high resistance potentiometer. You may come across some IC type digital controllable potentiometer, but that mostly applied for low power and small value resistance
Hi All, It has been long time since I used my electronic knowledge and I'm a little bit rusted :) I have a simple camera flash circuit (similar to this): 99700 I have 330v 102uf capacitor as the main capacitor, and the signal measured across the capacitor is: 99701 As you can see in the pic
It can be any reasonable voltage that the opto-coupler transistor is happy with. When AC is applied the led in the opto-coupler lights on positive half cycles and when it lights, the transistor on it's output side conducts and discharges the capacitor so the voltage across it drops very low. The resistor and supply (5V or whatever) pull it high ag
expanding on that... When one polarity is applied, current flows through the led and it lights up. In this state it will drop voltage across the resistor to maintain a steady voltage across the led. For indicator leds, this is typically around 1.5V and for high brightness types, about 3.5V. When the (...)
Basically you want to drive led with AC source... For this you can use Capacitor voltage divider instead of voltage divider. there will very less dissipation in capacitor voltage divider. One important thing is...18V across single led is too much...Thats why it is flickering and died in 3-4 turns. calculate capacitor voltage divider so that (...)
As led's are current sensitive devices, testing them with a direct variable power supply or 3.6 volt battery is never recomended. Try searching a constant current source to test led's. Connecting the X led to the tester and measuring the voltage across will give you the 99% perfect forward voltage of a unknown (...)
Dear Madfish Since leds are like batteries and nonlinear, unlike resistors, in order to use Ohms Law you must treat the leds as a fixed voltage with a series resistance. Then when voltage is applied the current will depend on the voltage difference across the series resistor. But when putting many in a string such as in letters for your (...)
I suggest you to go with Fig:2 because of the open collector configuration. The first circuit will be preferred for logical output. The first circuit will also works, but unnecessary power dissipation will happen across the resistor even if the led turn off...
No! The transistor has to meet two critreria, it must be able to witstand at least the maximum voltage across it and it must be able to dissipate the heat produced in it. 2N3055 ratings vary by manufacturer but are typically around 60 V and you should use a transistor that could witstand surges so one rated at 300V+ would be more suitable. It mu
Instead of looking for a "best" optocoupler, look to select one that you can use for your purpose. And you have loads of choices. As long as your current is limited to the led using a resistor, you can apply high voltage to the optocoupler led as the remaining voltage (besides the optocoupler forward voltage) is dropped across the (...)
To give a bulb a long life when its "flashed" on and off, they are operated at black heat, that is to say they have 70% of the working voltage put on them to keep the filaments hot(tish). I would suggest that you work out the circuit to see how this is done. It could be a resistor across the back of the switch. if you remove the resistor you may f
The measurement taken by the PIC is of one voltage relative to the other so you have to measure across the led using ground and the ADC input as the two connections. A word of warning - make sure it is safe to connect the two grounds first. Although they are both called ground, a true connection to Earth may not be present in the power (...)
That's a completely different circuit! Firstly, you will never get 80mA across a led, current flows through components, not across it. Secondly, whatever current flows through the led has to flow into the 74138 and that causes you two problems: 1, the output of the 74138 will not go to zero volts when low and 2, the 74138 (...)
They need a resistor or even better, constant current source. If you connect 12V directly across them they will almost certainly die very quickly! Brian.
Install the red led across the switch that controls the pump. It will permit a small amount of current will go through the pump. If this is not allowed then disregard my suggestion. - - - Updated - -
you can also bypass the rectify so that the circuit will be simple. also you have to follow the step above by tahmid... In that case you can not use a single resistor , you also have to add either a diode in series with the led or an anti-parallel diode across the led to protect it from reverse voltage. Alex
second thing if any switch is pressed in room means like bulb,fan, the APR circuit stops working and busy led will glow. Can any body help me? Do you have sufficient decoupling? Place bypass/decoupling capacitors across all power lines.
leds are current driven, that means that they are rated at some current, say 30mA. The volt drop across any one led might vary a bit about 2.2V, so if they are connected all in parallel as in the first figure, it is likely that one led will "hog" too much current and be very bright, the others might be very dim (or out!). (...)
Hi, Currently i am doing a Battery Charger. in this I want to have audio and visual indication when the battery is reversely connected. battery reverse indication i have done using a resistor and led like we do for negative supply indication. but my question is about audio indication. Can i directly connect a BUZZER across the battery.M
@js - you are forgetting the resistor only drops the 5V down to Vf of the diode so there will only be about 2V across it. 2/0.02 = 100. Brian.
I came across with the website some other time, and it was addressed to me by David Lee , saying I can find SX70 and ER and assembled PCB with led from Cree on this Asian site. I took a look but could not find the place where I can buy those items except from the banners where it ha
Sorry, ignore this rubbish: You also have one side of the led connected to ground, and your code is setting the pin on the other side of the led to zero (=ground) also. So it has no potential across it. Connect the grounded side of the led to VCC instead of ground. Or, keep the connections as they are and change the (...)
leds need about 2V to turn on. You can't expect to put 2V across them and expect them to work, ESPECIALLY if you've also got a resistor in series. You'll need to add some gain. - - - Updated - - - Actually, it's not clear what you are trying to do here. If you are just trying to light an led when a +/- signal is present
You can run a small high brightness led like that at about 2mA for a long time, maybe 10 hours or more before the battery gives out. If all you need is a few seconds it should last a long time. Consider that most "keyring" led lights use nothing more than a CR2032 and a white led straight across it. They actually rely on (...)