Difference between Arduino 5V power and output of 7805 5V power

Dear Friends,

I try blink example of Pin 13 of Arduino Uno R3 and it run fine as well with other pins for LED. When i tryed to give power through simple power regulator of LM7805 with 10K Pot in VDD adn 220ohm to GND to LED, led burns !!!. I think as per ohm's law, V= IR where V= 5 as output of 7805, I = 20mA for LED to bright so only 220 or 270 ohm register to ground must work with this too to bright LED.

Friends, why same 5V output from arduino uno r3 works and output of 7805 5V is not working ? even simple 5V from arduino say pin 13 and ground near to pin 3 works directly with led without any register attached !!!

I know i'm missing some of points of electronics but stuck up what is missing please help.

Regards,

VBN

You are going to have to post a diagram of how you wired everything up.

Also, when calculating the resistor for an LED, Vr=Vs-Vf where Vf=forward voltage of an LED, Vr = voltage across resistor, Vs=supply voltage.
A standard Red LED has If=20mA (forward current), Vf = 2.1v(forward voltage).

So how much voltage is across the resistor? (Hint: think of it like a staircase, the supply lets you walk up 5 steps, then the LED forces you down 2.1 steps, so how high are you?)
The resistor needs to have 2.9 volts across it to get you back down to the bottom of the staircase.

Its a series circuit, so all current that flows through the LED must flow through the resistor, meaning that there must be 20mA flowing through the resistor if you want 20mA to flow through the LED.

So using V=IR, you get 2.9 = 20mA * R => R = 2.9/0.02 = 145R. You will notice that there are some standard values for resistors, so the minimum standard value you would want is 150R.
You will find that an LED can run quite happily at a lower current, so you can use a larger resistor if you want, the LED just won't be as bright, but its working life will be extended.

For example, at 270R, you would have I = V/R = 2.9/270 = 10.7mA flowing through the LED.

Although not good at fritzing, tried to make it. What i done is, take 7.5 V in from wall wart, given input to LM7805 linear voltage regulator, got 5V out pin from it, that pin to Anode (Longer pin of LED) and ground to 220 Register.

Wall Wart 7.5V --> LM7805 Regulator --> 5V out to Longer leg of LED, Shofter Leg + 220 Ohm Registor to ground.

You probably want to check your voltage regulator to see which leg is which.

I see no reason why that circuit shouldn't work.

I can only suggest going back and double checking your connections. I have attached a schematic to help.

Note that the + terminal of the supply goes to the Input pin of the 7805, and the - terminal goes to the GND pin.

7805-LED.png

Yes Tom, I also see no reason to fry my two LEDs.... my wallwart give true result as i checked with multimeter too with Center +ve it is DC1.5-12V 10W max and 500mA Max.

I tried one more think, i gave power of 7805 to 10K pot and output of 10K pot to LED and it worked at max, 10K or upto 5K but if i go less than that, led fired again !!!

Thanks Tom for very good answer....

Thanks all for answer and help me out..

Be very careful using a potentiometer to control the LED, if you turn it too far you can end up having much less resistance in series with the LED as you think = BANG.

Also know that the 10k pot may not be linear. If for example it is a log pot, then half way is not necessarily 5k, in fact it is usually only around 1k.

Very much true .... that's new-bee like me !!! and this even can't helped me due to new to electronics Breadboard Power Supply Kit 5V/3.3V Quickstart Guide - SparkFun Electronics and proper understanding is too low... i accept this.

VBN, you might find this site useful

vbnewcomer:
Although not good at fritzing, tried to make it. What i done is, take 7.5 V in from wall wart, given input to LM7805 linear voltage regulator, got 5V out pin from it, that pin to Anode (Longer pin of LED) and ground to 220 Register.

Wall Wart 7.5V --> LM7805 Regulator --> 5V out to Longer leg of LED, Shofter Leg + 220 Ohm Registor to ground.

You connected the power in-wrong way around and connected the LED backwards, so essentially the regulator is doing nothing at all.

A 78xx regulator is a Positive regulator, making the center pin Ground, and the other two pins positive in regard to Ground. You connected it up backwards per that Fritzing diagram.

// Per.

Zepro i think i connected is right, 1st pin is in of wall wart 7.5V 2nd is ground and last or 3rd is output of 5V. But i'm not sure that i done this one right : 5V > +ve or bigger leg of LED and ground leg or shorter led to 220 ohm register to ground.

5V > +ve of Led, -ve or 0V -> Register -> ground.

I'm not sure of above line do i need to connect Register like this one ?

5V> Register > Longer Leg of LED, shorter leg directly to ground ?

Did anybody try measuring current (amps) instead of voltages...?

It shows 0.25A @ 20m setting and 0.2 at 200m .... i'm not sure about Ampere setting on multimeter !!! there is 2000u, 20m and 200m really sorry for that but reading site suggested by JimboZA but due to non-electronics field, it takes much time than required....

I just made this, look at it and connect stuff like that and tell me what happens.

// Per.

Yes perfectly works well ... (I think i was giving resistor when grounding while in this case, it is to +ve) Thanks a lot but i will be in more better way for my further proceeding if you can let me understand why this just 2 cap addition worked fine ? or just logic behing this so that i can proceed to work with http://www.instructables.com/id/Perfboard-Hackduino-Arduino-compatible-circuit/?ALLSTEPS where i think that i will attach PIR motion sensor to get activate my door bell or passage light with LDR + PIR + Relay with perf board version to get more in-depth understanding for my first project on my own, without arduino board. and what other precautions should be taken care in that type of work.

Thanks to tom, jimboza, Zapro, fungus for answering my quety... i feel much better now and you may help me in further understanding and extending knowledge in future too please.

vbnewcomer:
It shows 0.25A @ 20m setting and 0.2 at 200m .... i'm not sure about Ampere setting on multimeter !!! there is 2000u, 20m and 200m really sorry for that but reading site suggested by JimboZA but due to non-electronics field, it takes much time than required....

If the current is unknown you usually you start at 10A and work downwards. The reason is that many meters have 250mA fuses in them which blow and leave you wondering why the meter isn't working. Make sure you've got less than 200mA before you use any scale other than 10A.

So...put the meter on the 10A scale and insert the meter between the resistor and the LED. What does it say?

nb.
a) You usually have to swap the 'positive' (red) multimeter lead to another connector on the meter to use the 10A scale (this bypasses the internal fuse).
b) You have to measure amps with the meter in series with the components, eg. between the resistor and LED. Break the connection between them and connect them through the meter.

If the meter says less than 0.2A, swap the red lead back to normal position, switch to 200m scale and read it again. If it reads less than 0.2, switch to 20mA scale, etc.

The capacitors on the Voltage regulator is to stabilise it. The capacitor acts like a reservoir of power, so it takes out small fluctuations in the power. Said shortly it helps the V-reg hold the voltage steady and accurate.

If You want more explanation, read a datasheet for a 78xx regulator.

// Per.