Current Voltage Resistance Basic Questions

current voltage resistance

Hi all I just joined and hope u can help
,
I have a couple of begginners books and looking at some circuits I would like some clarification if you can. I have been building simple circuits following diagrams and getting things working without any real undersrtanding. I buy parts, they come with instructions which are mostly simple to follow.

Could you help with basic questions below ->

Connecting up LED's and push button switches shows that the IO pins of the microcontroller are pulled low with a 10K resistor. So then you hit the button and open a path with only 470 Ohmn resistor.

Thats all good and I get it. However then they show a diagram with LED's connected to Vdd+ and the IO Pins. No switch used. No Pull Down resistors used.

They say that Pull Down or Pull UP resistors are used to stop the Pins floating.

So you use code to tell processor PIN 1 HIGH or PIN 1 LOW. So If you are not using a switch then the PINS don't Float ?

Also with regard to I = V/R , If the voltage is 5V and the resistor is 470 Ohms
there is a LED drawing 1.4V
I = 3.6/470 = .007mA
So can you also say that the LED resistance is 244 Ohms therefore ->
Is this ok , R = 5/.007mA so then R = 714 Ohms ?

Just can't wait till I get onto voltage dividers !

calculate voltage resistance

Thats a lot of questions in one message!

The input pins of a chip are designed to place a little load as possible on the circuit that drives them. This is so the they have least possible influence on the driver which could also be feeding other devices. The drawback to this is that because the input draws such little current, it cannot discharge a voltage unintentionally present on it. For example, static build-up or just signals picked up when the pin or connections to it acts like a radio antenna. For that reason we use either pull-up or pull-down resistors, they are a source of current to pull the pin to a known high or low voltage. The value is chosen to be low enough to discharge unwanted voltages but not so low that the circuit takes too much extra driving. Some of the drive current inevitably goes down the resistor and this to some degree reduces the amount reaching the IC itself. Sometimes the resistors (or an equivalent weak current source) is built in to the IC, sometimes you have to provide it outside.

The output pins of a chip have a circuit behind them that intentionally provides signal to drive other devices. Because it can source current from the supply when high, or sink it to ground when low, there is always a discharge path for unwanted signals so pull-up and pull-down resistors are not normally needed, the chip itself provides the discharge path.

There are two exceptions to the output pin rule;
1. Tri-state pins have a third condition, the pin can be internally disconnected from the driver behind it. This is usually to allow several output to be joined together with only one being active at a time. In a tri-state condition, the pin doesn't source or sink current so a pull-up or pull-down may be advisable. It depends on the chance of all the connected pins being tri-state at once. If any is driving, it discharges the whole network and resistors are not needed.
2. Some devices are called "open xxxx" where you can substitute 'collector', 'emitter', 'source' or 'drain' for xxx, depending on the kind of device. These can either sink current to ground or supply current from the power rail but not both. Whichever way they pull, sometimes a resistor is needed to return them to the other state.

You should not normally use LEDs connected to an IC without resistors in series with them. LEDS try to draw as much current as you let them and the resistor is needed to keep it within safe limits for the LED and the pin driving it.

Your math for the LED is correct to a point but LEDs and most other semiconductor materials do not have a linear voltage to current ratio. Their resistance will appear to change as the current changes so you can't always say it equals 244 ohms.

The correct way to calculate the series resistor is:

(supply voltage - LED voltage) / LED current

Which is what you actually did. Get the LED voltage and current requirements from the manufacturers data sheet.

Brian.

led resistance formula

expelleior said:

Also with regard to I = V/R , If the voltage is 5V and the resistor is 470 Ohms
there is a LED drawing 1.4V
I = 3.6/470 = .007mA
So can you also say that the LED resistance is 244 Ohms therefore ->
Is this ok , R = 5/.007mA so then R = 714 Ohms ?

Just can't wait till I get onto voltage dividers !

Click to expand...

There's a typing mistake:
I = 3.6/470 = .007A (or 7mA) !
LED is not a linear device and its resistance is a function of voltage applied to its pins.
Similar to any diode, LED is a diode too, in forward current, LED has a drop voltage about 1.4-1.6V. This drop voltage gets value ~1.4V when I_forward~0.5mA (R=U/I ~ 2.8 K Ohm) then it changes only about 100mV while forward current changes from 5mA up to 10mA (R= U/I ~ 1.6V/10mA= 160 Ohm).
So for LED, V/I is not constant value.That why you can not use LED (or diode) as a voltage divider.
In other hand, based on this diode property, you can use LED (or diode) as a low cost reference voltage (like a zener diode)!

led voltage resistance

Many thanks to you both for a complete explanation and for correcting my typo, which may have led to further confusion in my math

cheers ,
Mike.

voltage resistance formula

One more question if i may to vhelp me understand current flow.

If I have a few 470 Ohmn resistors in parallel and last one is a 1K resistor, there would be no flow through the 1K resistor at all ?

thx, mike.

guide questions in current,resistance and voltage

If the 1K is also in parallel it will share some of the current but only a little under half of one of the 470 ohm resistors.

Ohms law still applies, the current through the resistor = voltage across it divided by its value in Ohms.

If you need to work out the effective resistance of all your resistors in parallel, the method is to add the reciprocals of each value then take the reciprocal of the result.

It's easy with a pocket calculator:
1 [divide] R1 [M+=]
1 [divide] R2 [M+=]
for each resistor R1, R2 and so on. then
1 [divide] [RM] [=] gives the result.

Brian.

basic questions resistors

I was surprised that any current would flow through the 1K resistor as i believe current onlyn flows through the path of least resistance. But this is obviously an over simplification of the laws of electricity.

But then i made a little circut with a LED and a short circuit with a small resistor and the LEDc did not even slightly light up.

Added after 17 minutes:



I need some help with calculating resistor values. Sorry the GIF is fuzzy and Lamp.png is the only picture worth looking at. thx for ANY help. Mike.

ps. Im doing a diploma in electronics and microcontrollers next year so hopefully that will help me a lot!

led voltage to current ratio

For the above Lamp222.png

I have a 5V battery powering a lamp that requires 60mA . The lamp uses 3V+

I wanted to vfind out the resistance of the lamp -> 3/.06 A = 50 Ohms



When I want to calulate the current flowing through the circuit I also did this
3V/50 Ohms = .o6 Amps

So i need to drop the Volts by 2V..... 2/.06 = 33.3 Ohms

Does this mean all I have to do is use a 33.3 Ohmn resistor or there abouts ?

Thanks.

how to work out voltage, resistance

The lamp draws 600mA (0.6A) when its filament is cold and 60mA (0.06A) when operating. Normally you don't need to worry about the cold current when there is a resistor in series. You calculate the value needed to achieve 60mA with 3V across the lamp and let the filament heat up slowly because it is starved of current. As soon as it heats up, which is normally within a fraction of a second, the steady 60mA condition will be reached.

Your math is 'sort of' correct.

The resistor has to lose the excess voltage at the running current so the value is:

(5 - 3) / 0.06 which equals 33.33 Ohms.

The schematic shows 166 Ohms. It is difficult to predict what the voltage across the lamp or current through it would be because the characteristics of the lamp are not linear. As you can see from the cold current being 600mA and the warm current being only 10% of that, the filament resistance increases dramatically as it heats up. What it's resistance would be with the current limited by 166Ohms is anybody's guess, perhaps the lamp manufacturer specifies it in their data sheet.

A lamp specification is normally given as "the current that flows with the specified voltage applied across it and at normal operating temperature".

Returning to your earlier point about taking the path of least resistance; it only applies to lazy people, not electrons.

Brian

calculating current and voltage questions

Thanks Brian. I made up that 166 Ohm using some crazy math i dont rember I think i was probably making some adjustment for the cold current?
Anyway I understood your explanation and now to request or look up data for components. Many thanks.