Simple bridge rect. circuit works in simulation but in real life? - Page 1
Forum section: electronics / beginners

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 20, 2017, 01:31:47 am
content: I so don't get this. This is off an 15VAC power supply. The meter reads 14.81 volts. And te LED is not burnt out because I tested it on 3.3V. Are the diodes acting like resistors?

author: IanB
author_rank: Super Contributor
timestamp: February 20, 2017, 01:42:02 am
content: Have you checked for continuity on the breadboard power rails across the gap where the "W" is? There may not be a connection between the output from the rectifier and the input to the diode and resistor.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 20, 2017, 02:13:12 am
content: Quote from: IanB on February 20, 2017, 01:42:02 amFirst thing I checked actually. But if that was the problem I wouldn't get the 14v reading on the meter.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 20, 2017, 02:34:51 am
content: Quote from: raspberrypi on February 20, 2017, 02:13:12 amNo, that's not true because the Red DMM lead is connected on the Right Hand side of the "W" and the LED resistor is connected on the Left Hand side of the "W". Check for continuity between the places I have indicated on the drawing below. Or simply move the resistor over so that it connects to the power rail on the Right Hand side of the "W".Also check the polarity of the LED (try it reversed from the way you have it.)Also, with a 14.8 VDC supply and a 2200 R (2.2k, red red red) resistor you will have only about 5 or 6 mA current through the LED. This should be enough to see it glowing, but you might try a 1k resistor here instead (brown black red) to boost the current up to over 10 mA.

author: rstofer
author_rank: Super Contributor
timestamp: February 20, 2017, 02:55:22 am
content: Electronics is a math game... In the end, it's all math!Assuming a 15V source and 2.2V drop across the LED and 20 mA forward current, you need (15-2.2)/0.20 or about 630 Ohms. But I don't know what LED you are using so I don't know Vf. You could play around with various resistors until you get 20 mA and then you could measure Vf (at 20 mA). Since 20 mA seems a more common and universal number than any value of Vf, I would try that if I didn't have any way of determining the LED specs.This is all based on Ohm's Law and it is a law, not a suggestion.

author: RJFreeman
author_rank: Regular Contributor
timestamp: February 20, 2017, 03:12:35 am
content: As well as checking continuity as IanB and alsetalokin4017 suggested, Is your LED around the right way?The flat is usually on the cathode (negative) side, and the negative side in your picture looks rounded, the Cathode is also usually the shorter leg, but of course this presumes you haven't already cut the legs.Also worth mentioning; typical peak inverse voltage of an LED is around 5 Volts, so if it is in reverse, this may have damaged the LED as the peak voltage in this circuit will be something like 21 Volts.

author: raspberrypi
author_rank: Frequent Contributor
timestamp: February 20, 2017, 03:56:21 am
content: So it was not connected over the "W" on the board. So now how are we running an LED off 15 volts? I thought they would blow up at that voltage. Also the reason I'm trying to figure ths out is beside the color of the ledsi know nothing about them. And the amps setting on my meter doesn't work so I'm trying to learn the math. Are LEDs more likely to blow up with too much voltage hooked up correctly or backwards? Or does it just die at a certain voltage no matter the polarity. How do you use resistors to lower the voltage but not the current? Say it wanted to run the LEDs at three volts but I wanted 20 mA?

author: Brumby
author_rank: Unknown
timestamp: February 20, 2017, 04:33:08 am
content: Quote from: raspberrypi on February 20, 2017, 03:56:21 amThat 2.2k resistor is how. It will limit the current through the LED.Imagine a brick (the LED) and a spring (the resistor) laying flat on a table.Lets attach a lug on either end of the brick and pull (apply a voltage). As the pull (voltage) increases, there will come a point when the brick (LED) will just break. It doesn't "give".Now let's put the spring in between one of our pulling lugs and the brick. When the pull (voltage) increases, the spring will stretch, allowing much more pulling, with nowhere near the stress on the brick as it experienced before. Obviously, the spring (resistor) could fail - and so could the brick (LED) - but these events will require much greater pull (voltage) than just with the brick alone.* For all the analogy critics out there, I know there are holes with this - stress vs strain for starters - but it is just to try and illustrate how LEDs don't 'give' and resistors do. *

author: mariush
author_rank: Super Contributor
timestamp: February 20, 2017, 04:37:48 am
content: LEDs are diodes.They turn on (start producing light) once there's a big enough voltage difference between the terminals (when the voltage goes above a threshold, or in other words the forward voltage of the LED).They resist current going backwards as long as the voltage is below some other threshold (usually around 5v), just like a regular diode..One of the reasons they fail is because of heat, and the amount of heat varies with the amount of current going through the led. The power dissipated in the LED is equal to the forward voltage x current .For example, if the voltage on the LED is 3v and the current is 20mA , then there's 3v x 0.02 A = 0.06 watts or 60 mWHere's some more details about your circuit.Your bridge rectifier (the four diodes) transform the AC voltage into a DC voltage, a voltage that goes from 0v to a peak voltage, instead of going 60 times a second down to a negative voltage and then up to a positive voltage.However, the bridge rectifier only "flips" those negative voltages to positive voltages, so you still have waves of voltage, about 120 times a second ( two times the previous AC frequency), the voltage goes from 0 v slowly up to a peak voltage and then slowly down to 0v againThe peak voltage after the bridge rectifier is equal to sqrt(2) x Vac minus 2 x Vdiode (the forward voltage of the diodes used in the rectifier), because at any point, the electricity goes through two diodes in the rectifier.So if for example you have a 12v AC transformer, your peak DC voltage would be about 12v x 1.414 - 2 x 0.7v = about 15.5vBut you don't have a steady 15.5v DC ... remember, 120 times a second voltage goes down and up. So that's where you bring in a capacitor, to charge up with electricity and give back that electricity when the voltage goes down slowly towards 0v.You can approximate how much capacitance should use with this formula C = Current / [ 2 x AC Frequency x (Vdc peak - Vdc min)]so for example, if you want at least 12v all the time up to around 100mA of current (0.1A) , and you know the peak DC voltage is 15.5v, then C = 0.1A / [2x60x(15.5-12)] = 0.1 / 120*3.5 = 0.1 / 420 = 0.0002380 Farads or 238uF , so you could probably use the next standard value (270uF or 330uF) and you'll know for sure at any time, the voltage will be at least 12v (As long as your circuit doesn't use more than 100mA).So now let's say have 12v DC minimum all the time, and you have in circuit resistor and led in series. Two things are important to understand :When two things are connected in series in a circuit, both consume the same amount of currentYou have the formula V = I x R again ...So if you want to limit the current going through the led, you add the resistor in series with the led, and pick a resistor value that would result in a certain amount of voltage being "lost" on the resistor, so that the led will "see" a voltage only slightly bigger than it's forward voltage (so that it would turn on and create light)So if your voltage is at least 12v , and let's say the led needs 3v to turn on, you must make 12v - 3v be lost in the resistor... So V = I x R .... 12v - 3v = I x R ... 9v = I x RWith your 2200 ohm resistor, that means I = 9v / 2200 = 0.0040 A or 4mA.It's a low amount of current, but the led should turn on.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 20, 2017, 11:06:17 am
content: Quote from: raspberrypi on February 20, 2017, 03:56:21 amSo now you've got it working, by carefully following the direction I and others have given you. You're welcome!(This is a trap for new players and has caught many "newbies" on this forum before; many breadboards have split power rails like this. This is done so you can easily put several, or at least two anyway, circuits on the same breadboard using different supplies. If you want, to avoid confusion, you can just install little jumpers across the "W" gap so that your power rails are continuous from side to side across the whole breadboard. See attached image. )QuoteYou are not "running an LED off 15 volts". You are running the series combination of the LED and the CURRENT LIMITING RESISTOR off of 15 volts. We covered this in another post in another thread and in this thread. Once you are over the Vfwdof the LED it essentially acts like a "wire" having negligible resistance on its own, and we use Ohm's Law and the resistor value to _SET_ the current through the system.QuoteWhy doesn't the current setting of your DMM work? Have you in fact checked the DMM's fuses for continuity and the correct type and value? It is quite common for the lower-range fuse to blow from misuse. And we have talked about how to use the Vdropacross an inline resistor to determine current flow, using Ohm's Law. The only way to "learn the math" is to follow the instructions I and others have given you, work practice problems and check your work on the actual circuit using your DMM and Ohm's Law. As far as "knowing nothing about LEDs"... entire books can be and have been written about LEDs and how they work. You don't need all this stuff at your stage of experience! I know nothing about LEDs either, except what has been written here. I just apply the math, Kirchhoff's circuit rules and Ohm's Law, and get the results I need.QuoteDon't overthink this. The LED Data Sheet will give you several parameters of the LED, including the forward voltage drop Vfwdand the maximum reverse voltage the LED can withstand in normal usage. You can get the approximate Vfwdof the various colors of LED from the graph that has been presented to you. Since you may not have the actual Data Sheet for the LED you are using, you can "guesstimate" based on the color and then fine tune the exact resistor value for your desired current and supply voltage using Ohm's Law, using the basic LED Equation which has been given to you several times. Yes, you can blow out your LED by having it try to block too much voltage in the reverse direction, or putting too much current through it in the forward direction. The exact V and I values where this happens depends on the particular LED and for this you will need either the Data Sheet or you can find it by experimentation actually blowing some LEDs.The basic LED equation (Ohm's Law) : V=IR or I=V/R or R=V/I where V = (Vsupply-VfwdLED) in Volts and I = desired current in Amps and R = current limiting resistor in Ohms.(Vsupply- VfwdLED)/(desired current) = Resistor ValueSO, as an exercise, what is the value of the current limiting resistor if you want 20 mA, your supply voltage is 3.0 volts and you are using an LED with a forward voltage of 1.8 volts?Solution: R = V/I so R = (3.0 - 1.8)/0.020 = 60 OhmsNote that you are not "running the LED at 3 volts", you are supplying 3 volts to the _entire series string_ of LED+ resistor. The circuit rules say that you are dropping the entire supply voltage over the entire circuit. The LED drops its "fwd voltage" (roughly), or 1.8 V, and the resistor drops the other 1.2 volts, if you want to look at it that way. Get out your DMM, some resistors and LEDs, and do some simple measurements. The measured voltage drop across the resistor is converted to a current reading by Ohm's Law I = Vdrop/Rvalue.Now do a little math until you are comfortable with it:1. You have a 5 volt supply and an LED with a Vfwd of 2.1 volts. What is the needed resistor value for a current of 20 mA through the circuit? How about 10 mA? 7 mA?2. You have an LED with a forward voltage of 2.5 volts, and you have a series resistor of 120 ohms. What is the needed supply voltage for a current of 10 mA through the circuit? 20 mA? 5 mA?3. You have an LED with a maximum current limitation of 40 mA, and it has a forward voltage drop according to the Data Sheet of 2.3 volts. The power supply is a 9V battery. What is the lowest resistor value you can use without risking blowing the LED?Please actually work these problems !