Marks256

Ok, i would like to know how to use resistors. I know ohm's law V=I*R. But how do i use it? I also know that V=Voltage, I=Amperage, and R=Resistance, but where do i find the variables? With a multimeter/ampmeter? Could someone also give some real life examples? What do resistors ACUALLY do? I know they limmit current, but don't they effect voltage too? I am lost.

I do know a few thing about resistors, but i think what i know might be wrong. Will someone please re-explain them to me? As simple as possable. Thanks. I have learned alot from this forum so-far, i am hopping i can learn alot more. Also, what is the difference between voltage and amperage? Sorry for the simplicity of these questions, but like i said, i think what i know about these things may be dead wrong.

chemelec

First off, it is:

E
-------
I * R

"E" is Electromotive Force.

The Variables are any Two of the above to calculate the Third value.

Now Get some resistors and a power supply.
Also get a volt meter and amp meter.
Than Play with them and see what they do.

You will learn far more doing this, than just reading about it.

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RadioRon

Voltage is the pressure that pushes the electrons through a conductor. Amperage, or current, is the volume of electrons that passes through the conductor. Resistance is a measure of the loss of pressure as the current flows through the conductor. This loss of pressure happens because the conductor is not perfect and is sort of getting in the way of the flow of current.

We often use the flow of water through a pipe to help describe this. In this comparison, pretend we have a water pump pushing water through 100 feet of half inch pipe and then back to the pump. The pump creates a difference in pressure between its output and its input. This pressure difference will cause water to flow through the pipe from the output back to the input. The amount of water that will flow through the pipe will depend on the pressure difference and on how big the pipe is. In electrical circuits, the battery is like the pump where the + terminal is the outlet, the - terminal is the input and the pressure that it generates is the voltage across these two terminals. Electrical current flows from the high pressure terminal (the + one) to the low pressure terminal (the - one). The volume of water that flows in the pipe is like the current in our electrical circuit, the wider the pipe, the more water can flow. So the diameter of the pipe is like the resistance in the electrical circuit. A half inch pipe has a higher resistance than a 2 inch pipe so not as much water can flow through it.

That's really the essence of electricity. Volts is a measure of the pressure, amps is a measure of the current flow, and ohms is the measure of the resistance. OHm's law shows in simple mathematical form how these three things depend on each other. If you know two of the variables, you can calculate the third. If you need one to be bigger or smaller, you can see how to do that by varying one or both of the other two.

To simplify design, we usually fix the voltage first at some convenient value. How you find the values for the other two depends a lot on what you are trying to do. Usually you have some purpose, some sort of output, that you need to operate with electricity, Like a light bulb, or a relay, or an LED, or a speaker, or a motor for example. So you start there. The light bulb for example, let's start by assuming its in a flashlight. The voltage comes from two D cells, so the pressure availabe is 3 Volts. The bulb has a fixed resistance that you can look up in the manufacturer's specification or you can measure with an ohmmeter. So, it is easy to calculate how much current will flow when you hook the bulb to the battery. If you want, you could measure this current with an ammeter.

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JimB

Not quite right.

Because the temperature of the filament in the bulb changes from room temperature to 1900 degC (I guess) when operating, there is a big change in resistance.

So what you measure at room temperature is not the operating resistance of the bulb.
The bulb still obeys Ohms Law, but the complication is that the resistance changes due to the temperature.

JimB

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eblc1388

Not according to the GIF image on this webpage on voltages across bulbs.

http://www.williamson-labs.com/480_ire.htm

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Marks256

Yes, i knew that tempurature will effect the resistance. How would i calculate what voltage will appear on the other side of the resistor?

Please, tell me if this is right;
I just bought an infrared LED at Radio Shack. On the back of the package it says: Forward Voltage = 1.2v Forward Current = 100mA(0.001).

So: 1.2=.001*R

1.2/R=(.001*R)/R
(1.2/R)*1.2=.001
R=.0012

So the answer is: .0012ohms across the LED, correct?

RadioRon

Not correct. If the forward voltage is 1.2 V while conducting 100 mA then its resistance under those conditions is 1.2 divided by 0.1 (not 0.001), which equals 12 ohms.

Be careful though. Sometimes the data sheet tells you several things but which are measured under different conditions. For example, I'm not sure but that 100 mA may be the Maximum current you should put through it, not the normal operating current. If this is so, then you need to find somewhere on the data sheet what the normal operating current would be when the voltage drop across it was at 1.2 V. But I may be wrong, in which case 12 ohms is right.

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Marks256

Nope. It is 100mA. The max is 1.2A. So, where did you get .1? Oops. I know where. 100mA = .1A! I must have typed in 1ma, instead of 100mA. Ok, i understand that now. But i still dont get how resistor can bring down voltage? Or is this voltage drop a direct cause of changing the amperage?

Back to the water analogy. If you had a 10ohm resistor, and a 4.7kohm resistor, which one would have the "smallest pipe diameter"?

RODALCO

and for power

P
____

E * I

or P = I²R

where:

P = power in watts
I = current in amps
E = electromotive force ( in Europe U = used for Voltage )
R = resistance in ohms

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JimB

Maybe that is wrong as well.

I have just measured a 40W 240v lamp while cold, it shows 100 ohms.
Now do the maths.
W = (V x V)/R = (240 x 240)/100 = 576 watts.

40watts vs 576watts, I think we have a resistance change somewhere.

JimB

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Nigel Goodwin

Jimb is completely correct, and that webpage is rubbish! - incandescent bulbs have a MUCH lower resistance cold than hot - which is why the usually fail when you turn them on, due to the massive current surge.

There's little point measuring the cold resistance of a lamp, power it up and measure the current through it, and the voltage across it, then use ohms law to calculate the hot resistance.

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JimB

Apart from your maths being a bit off you are on the wrong track with the LED.
LEDs (and ordinary diodes, transistors etc) do not obey Ohms Law.

What the data on the pack is telling you is that when there is 100mA flowing through the diode, there will be 1.2volts across the diode.
But if we turn the current down to 50mA, there will still be (guess) 1.1volts across the LED, not 0.6 volts as Ohms Law would predict.
When the voltage across the diode gets down to (another guess) 1.0volts there will be no current flowing through the diode.

If you want to operate that LED from a 6volt supply, you must connect a resistor in series with the LED before you connect them across the supply.
OK so what value of resistor?
The supply is 6volts, the LED wants 1.2volts, so that leaves 4.8volts across the resistor. We want 100mA to flow through the diode (and the resistor because they are in series), so lets use Ohms Law.

R = V/I = 4.8/0.1 = 48ohm.

Also consider the power dissipated by the resistor.
Various ways to calculate this, I will use

W = I x I x R =0.1 x 0.1 x 48 = 0.48 watts. So a 0.5 watt resistor will do (just).

Problem, 48ohm resistors are hard to find, 48 is not a "preferred value". I practice we would probably use a 47ohm resistor which is a preferred value.

Caution, the ratings for the LED as printed on the packet, are they "normal working" values or "absolute maximum" values? If you want equipment to have a long and happy life, you dont run it at the maximum ratings.
If 100mA is the maximum value of current through the LED, we would design for a working current of 50 to 75mA. To do this we would need a different value of resistor.
A calculation for you I think.

JimB

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Marks256

For the last time, i am 100% sure that these are the ideal/normal/recomended characteristics. Yes, i know my math was off. I didn't quite know what 100mA equals in amps. I know now that it is .1.

Is this right? I just drew it quick. Thanks.



So, what is power disipation? Is it how many watts of power the curcuit requires?

I have another question; If i had a power supply that supplyed 2watts of power, and a curcuit that used only .25watts, would the powersupply force the two watts into the curcuit, or would it just give what the curcuit needed?

Attached Images

Circuit.PNG (11.1 KB, 18 views)

audioguru

This is the only IR LED on RadioShack's website. It is rated at 29mA which might be its max continuous. 100mA would smoke it.

The power supply for your house has very many thousands of Watts of power available. When you plug in a tiny night-light, does it use all the power to burn and blind you? Look at the power ratings of electrical things. A night-light is only 4W or 7W.

Attached Images

RadioShack IR LED.PNG (20.0 KB, 7 views)

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Marks256

Here is what it says on the back of the package;

Electrical Characteristics (25°C)
->Radiant power output (100mA): 16mW min.
->Forward voltage...................: 1.2V
->Forward current...................: 100mA
->Viewing angle to ½ intensity...: 45°

Absolute maximum ratings (25°C)
->Foward voltage (20mA)..........: 1.6V
->Reverse voltage...................: 5v
->Forward current...................: 1.2A
->Reverse current....................: 10µA
->Wavelength.........................: 940nm