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### Topic: How much output ? (Read 5783 times)previous topic - next topic

#### TempleClause

#15
##### Mar 16, 2010, 11:03 pm
Okay soo emmm to get less voltages you need at least 2 resistors ?

I think I now getting closer to understanding :-) I just had a huuuuge mistake in my basic thought which was that a resistor is reducing voltage.

Cheers

#### TempleClause

#16
##### Mar 16, 2010, 11:55 pmLast Edit: Mar 16, 2010, 11:56 pm by TempleClause Reason: 1
Okay so I tested some things out and came to the conclusion (and i really hope this one is right now ;-) )

Digital and Analog pins are reading Voltages. But a Resistor isn't reducing the Voltages .

Okay I now fully understand the basic button example and now know why they use a pull down resistor I least I think I do.

Correct me if anything of the following is wrong.

So as I learned electricity seeks for the path with the least resistance.
So when the button is open (not pressed) it would go from pin 2 through the restistance to GND . At this point the resistor doesn't bring us any important thing.

When the button is closed (pressed) it goes from pin 2 to 5V. And in this case we need this pulldown resistor because otherwise both paths would have the same resistance. And the electricity doesn't know where to go cleary. :-)

Greets

#### MarkT

#17
##### Mar 17, 2010, 12:16 am
Resistors basically enforce Ohm's law, so if you feed them a current they respond with a voltage drop, if you force a voltage across them they pass a current such that V = IR

So rather than reducing anything they maintain a (precise) relationship between voltage and current.

Its like a thin water pipe that maintains a relation between the pressure and water-flow rate.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### retrolefty

#18
##### Mar 17, 2010, 12:37 amLast Edit: Mar 17, 2010, 12:39 am by retrolefty Reason: 1
Quote
When the button is closed (pressed) it goes from pin 2 to 5V.

Actually the current flow does not go into or out of the pin 2 connection. A digital input pin has a very very high input resistance, it 'takes' almost no current, it just 'senses' the voltage level presented to it and creates a digital LOW or HIGH reading when you perform a digitalRead() function.

All the real current/no current (electron) flow in your example flows from the ground pin to the +5vdc pin. When the switch is off (open) there is no current flow, but the input pin 2 senses 0vdc (ground) through the resistor wired to ground, that's called a pull-down resistor. The input pin2 will read as a LOW at that time.  When the switch is closed there is a current flow from ground through the resistor to +5vdc, and the input pin 'senses' this +5vdc voltage level and creates a HIGH reading when you perform the digitaRead() function. Actual current flow into or out of a digital input pin is in the microamp range and for external circuitry purposes can almost be treated as a no connection when doing a ohms law calculation for sizing external components.

That makes sense? First step is to learn to really really grok ohm's law. After that things get easier to understand and learn. If not grokked, then things get much much harder to understand and learn.

Lefty

#### Grumpy_Mike

#19
##### Mar 17, 2010, 09:33 am
Quote
So as I learned electricity seeks for the path with the least resistance.

No that's one of those vast oversimplifications that is actually wrong.

Electric current always flows according to ohms law. Suppose you have a 10K resistor across a 5V line then you will get 0.5mA flowing through it. Now suppose you put a 10 ohm resistor across the same line. Ohms law says it will have 0.5A flowing down it. However the 10K resistor will still have 0.5mA flowing down it. The current doesn't ignore the paths of high resistance and seek out the path of lower resistance instead.

#### TempleClause

#20
##### Mar 17, 2010, 05:55 pmLast Edit: Mar 17, 2010, 06:04 pm by TempleClause Reason: 1
GREAT! Now I don't understand anything once again .    :-/

So could someone pls explain me the buton example (what's happening exactly) so I as a complete noob understand it ? and than maybe i'll understand the more complicated things as well.
As Lefty said I need to understand the basics first, and it looks like I don't hehe . But Lefty's description was allready pretty good for me to understand. Only thing I didn't get yet is:
Quote
When the switch is closed there is a current flow from ground through the resistor to +5vdc and the input pin 'senses' this +5vdc voltage level and creates a HIGH reading

Why is it 5 Volts even if it has a resistor between ? I've connected a Resistor to 5V and than I meassured the Voltage btw the end of the resistor and ground and I didn't get 5V. So why do I get 5Volts in this button example ?

Maybe you could give me the calculation with that ohm law.

Greetings
and thx againg for beeing so firendly to me

#### retrolefty

#21
##### Mar 17, 2010, 06:22 pm
Quote
I've connected a Resistor to 5V and than I meassured the Voltage btw the end of the resistor and ground and I didn't get 5V. So why do I get 5Volts in this button example ?

Then you were either measuring the wrong points or it wasn't wired as you are describing. Can you draw what you think you had and show your measurement points.

Lefty

#### TempleClause

#22
##### Mar 17, 2010, 06:35 pm
Sure !

The circles are the 2 measuring points.

Cheers

#### pluggy

#23
##### Mar 17, 2010, 06:55 pmLast Edit: Mar 17, 2010, 06:57 pm by stephen_t Reason: 1
If the resistor is high value and you're using an old (cheap) analogue meter, you might well get a reading less than 5v downstream of the resistor.  In most cases with a digital meter, it will read 5volts.

This is because of the higher current drawn by a cheap analogue meter.
http://pluggy.is-a-geek.com/index.html

#### TempleClause

#24
##### Mar 17, 2010, 07:03 pmLast Edit: Mar 17, 2010, 07:06 pm by TempleClause Reason: 1
well I don't know if it is a good meter but it's defently digital !
So my meassuringes should be correct shouldn't they ?

I used a 508 KOhm Resistor.
And I meassure 3.37 Volts.
And without the resistor I meassure 5 Volts...
Cheers

#### pluggy

#25
##### Mar 17, 2010, 07:36 pm
Sounds like a cheap/old digital meter, with a low internal impedance.  A reasonable meter will have an internal impedance of tens or hundreds of megaohms and the voltage drop across a 508 k resistor would be negligable.  Try a resistor of around 10k and see what voltage you get with your meter.
http://pluggy.is-a-geek.com/index.html

#### retrolefty

#26
##### Mar 17, 2010, 07:38 pm
Well your latest drawing isn't a complete circuit and no current is flowing. However you can still measure a voltage potential difference between ground and either end of the resistor, both will read +5vdc.

So what is your question again.

Lefty

#### TempleClause

#27
##### Mar 17, 2010, 07:48 pm
Emm well I can tell you what meter I have...

http://www.hotair.pl/product_info.php?products_id=3900

It's russian I guess^^ but I think you should be able to read out the values you want to know.

So I searched for a 10KOhm Resistor and I meassured 5.09 without this resistor and 5.03 with the resistor . So that's just a meassuring mistake because of the poor quality of my meter ?

Cheers

#### pluggy

#28
##### Mar 17, 2010, 07:57 pm
Theres a circuit when he sticks his meter on the test points.  If he's getting ~3.5 volts downstream of a 508k resistor and 5v upstream it suggests his meter has in internal impedance of ~1m which isn't too hot.....
http://pluggy.is-a-geek.com/index.html

#### retrolefty

#29
##### Mar 17, 2010, 08:21 pm
Quote
If he's getting ~3.5 volts downstream of a 508k resistor and 5v upstream it suggests his meter has in internal impedance of ~1m which isn't too hot.....

I agree, not bad but most DVMs have 10megs or better.

But I fear discussing meter internal resistance isn't getting to the root of TC's basic understanding of ohm's law and DC circuit analysis. He is probably not asking the right questions for us to answer.

Lefty

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