Voltage question

It is obvious to those who do know, that you do not. I'm not trying to hurt your feelings, but just being accurate.

Do you understand the basic relationships between voltage, current and power?

Have you heard of Ohm's law? It relates voltage, current and resistance.

You refer to the data sheet for each device to find out the numerical values or operating voltage and current and how the voltages changes if you pass various currents through the LED. Speakers also have data sheets that describes their power rating, resistance and frequency response.

Read their data sheets. Then work to understand everything you read, there.

Then you are ready to attempt to understand an existing design, and eventually to produce your own designs. But until you can analyze the operation of someone else's design, you can't possibly evaluate your own.

This process is more like learning a new language than it is like memorizing a few foreign swear words.

Volts are what you get out of a battery or power supply. If you connect the + and - side of a battery (a voltage) with a resistor, a current will flow. There is a law, called Ohm's Law, that says that

Voltage / Resistance = Current

So, if you have a 9 volt battery, and you connect the + and - leads with a

100 ohm resistor, you get 9 / 100 = 0.090 amperes of current. People often use milli-amperes, or mA, to mean 1/1000 of an amp. Thus, a 100 ohm resistor across 9 volts gives you 90mA.

As the current flows through the resistor, it dissipates energy. Energy per second is power. This power can be computed by the formula

Power = Voltage * Voltage / Resistance

If you substitute in Current * Resistance for Voltage, it is also

Power = Current * Current * Resistance

So, you can use either current and resistance, or voltage and resistance, to compute the power dissipated by a resistance.

Power is expressed in units of watts, or W, and the variable P is often used. Current is usually abbreviated as I. Volts are given as V. Resistance is usually R. Thus,

V = I * R P = I^2 * R P = V^2/R

Now the brightness of LEDs is controlled by the current you pass through them. For your purposes, the voltage across them is nearly constant, and depends on what kind of LED it is.

If you have a 9V battery, and a standard red LED, and want to light it up, you need to figure out how much current it'll take, and limit the current to that value. That is usually done by putting a resistor in series with the LED. This means the 'anode' of the LED (the long lead) connects to the

• terminal of the battery, the 'cathode' of the LED connects to the resistor, and the resistor then connects to the - terminal of the battery.

For a standard red LED, it will generally light up with about 5mA of current. (remember 5mA = 0.005 A), It will also have about 1.4V across it when it is lit. To figure out what value of resistor you need, you do a little math:

9V (the battery) - 1.4V (the LED) = 7.6V (across the resistor) 5mA * R = 7.6

Thus,

R = 7.6 / 0.005 = 7.6 * 200 = 1520 ohms

A standard 5% resistor value near this is 1500 ohms.

Speakers are different, in that they are alternating current devices. An alternating current device has some kind of varying voltage across it, producing a varying current. People don't like to think in these terms, and often the values go between negative an positive, so they have come up with a reasonable way to express the alternating values with a single number. This the 'root mean square' thing, abbreviated RMS. Basically, it means that you take the square of whatever is varying along the waveform (thus making it always positive) and then average it over one or more waveforms. At the end, you take the square root of the result.

For example, your AC plug has a voltage across the two plug wires that varies between 170V and -170V, in a sine wave pattern. However, if you do the RMS thing to it, you end up with a number which is about 120V. All AC voltages are expressed this way. AC currents are also expressed this way, as the root mean square of the waveform.

Happily, if you do it this way, then power becomes just the product of the voltage and the current, at least for a sine wave like your AC power. If you want to find out how much power your toaster uses when cooking bread, for example, you measure the RMS current it is drawing from the plug, and multiply that by the voltage supply. This will also be expressed in watts, abbreviated W.

Speakers are like this too, because the voltage across them is an AC varying voltage. The apparent power output for a speaker is simply the RMS voltage across it times the RMS current. Unfortunately, there is a wrinkle, which makes this 'apparent' power not equal to what a resistor would be using in the same situation. Thus, the speaker may use less power than what you would expect using the RMS calculations.

For more than that, please go to the library and check out a basic book on electronics.

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If you don\'t know "how much power to put into a circuit", then
you\'re not a designer.

Thank you all for your responses!

You're right, I don't really know how to design circuits, I meant I know how basic electronic components work together, except for the amount of power.

Actually, I do not have a specific circuit in mind, I just thought that if I trying to learn about electronics, knowing how much voltage and resistance to apply in a circuit was something I should know!

Do I need to find out how many amps each load uses?

If I understand correctly, volts, amps, and ohms are completely separate units, and you don't convert between them. You supply power in volts, resistors use ohms, and loads consume amps. To get the necessary resistance, you use

Resistance = Voltage / Current

So, if you have a 6v power source, and a load that uses 5mAs of current then you get

Resistance = 6v / 0.050A

which gives you 120 ohms. The nearest resistor would probably be 100 ohms. (This is not for a circuit, I am just using it as an example.)

Thanks for your help, David

It's more the other way around - first you design a circuit that does what you want, then you work out how much power it's going to need and choose a suitable power supply. (Which can be significantly over-rated without a problem, but not under-rated).

I bought myself a small 12 volt battery and charger for doing beginner circuits, which has the advantage that you know it's unlikely to harm you as it's not connected to the mains and 12 volts can't cause a shock. (Though you can overload the battery by shorting it out, which could start a fire if you weren't watching.)

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P.S. you would do better to think of a simple project you'd like to have a go at, make a start on it yourself, then ask any questions which come up as you go along, rather than posting general questions like this.

e.g. 'I'm trying to build a flashing LED light circuit and I've got stuck with X - how can I sort this out?'

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Again, Thank you all for the help

So, do loads consume amps or watts (or both)?

I think I'll try experimenting a little with basic circuits. You're right, that would be a good way to learn.

Thanks Again, David

a

stuck

There are usually 2 ways to determin the voltage. One is how much do you have to start with. That is if it is used in a normal 12 volt car system, you design it to work from a low of around 12 volts to a maximum of 14 to 15 volts. Often you will see a value of 13.8 volts as this is a nominal voltage as the car is running. The other way is to find the maximum voltage the major components are rated for and use a somewhat lower voltage as the supply voltage. Then you factor in the cost of the higher components. YOu have to find a cost effective medium.

After you decide on the voltage then you calculate the current the device will draw at this voltage for the power you want. After that you look at catalogs and find a device that will handle the voltage/current/ power. A simple example might be a 12 volt output power supply. You start with the standard transformer of around 18 volts. Go with a diode bridge that will handle the current you want. Usually the voltage rating of any power diode will be enough. Look at the cost and pick one. The pass transistor(s) will have to be looked at. Some will be rated for a maximum current but if you take the voltage drop and caclulate the power, the maximum power may not be high enough. YOu either use more transisitors or a differant one.

Electrical loads usually have 4 components, Operating voltage Current flow in amps Resistance in ohms Power in watts

For example, if the load is a 100 watt light bulb operating on 120 volts, it will draw 0.83 amps and have a resistance of 144 ohms.

You can try this calculator for other resistive loads. Just enter any 2 of the components to find the other 2.

But there are also non-resistive loads, such as a capacitor, that consume no power (no watts), but do have amps flowing through them, and also a voltage present.

Yep, you have designed your first circuit. Should you connect it up and measure the values of current etc. you would notice that it was drawing more current due to a mistake in your calculation (hint, we must check our decimal points but such is life :-) So to design things that process is repeated for every component in your design. You would probably fast track your learning experience with a bit of study and get a better idea of the use of terminology ie. you dont "Put power into a circuit" it consumes power. You seem to be on the right path and as problems arise you will get help from the folks on this NG as long as you keep it civil.

Best of luck

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Regards ......... Rheilly Phoull

It looks to me as if you're confusing the concepts of voltage and power. You apply voltage (V) to an effective resistance (R measured in ohms) and a current (I, measured in amps) of V/R flows. The power, P, is found by multiplying the voltage and current, and is measured in watts. This is a measure of the amount of energy (E, measured in joules) used per second - e.g. how fast your battery is running down, or how much energy is being radiated from a light bulb per second.

Try it - you'll learn more that way, and probably enjoy it more as well. Or else think over why you want to learn this stuff in the first place.

good luck,

andy.

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Current passes through loads (pushed by the voltage across them) and everything in series with them. Power is what is consumed (turned into heat ant possibly some other form of energy (mechanical if the load is a motor, for instance)

both - a current flows through the load, driven by the voltage potential you put across it. This causes power to be drawn from the supply, and used by the load - either radiated as heat or used to do something useful, like running a motor or lighting a light.

have fun.

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basically yes. You also have the option of giving them less than rated power if you wan't less light, sound, etc.

more or less you do, you atleast need a "ballpark" figure of how the transistor will perform.

Bye. Jasen

It can with wet hands, if you try real hard...

stick a 5W indicator globe in series with your circuit, if there's a short it'll light up. for small low powered circuits it won't have any other effect.

Bye. Jasen

They pass amps through them, and convert the energy represented by the watts watts into some other form of energy like sound, light, motion, but mosstly heat.

talking of loads consuming amps or watts is sloppy talk (which many people use, myself included, it basically means the above.

learning by doing is best, by discussing in next best, by listening or reading is third best.... I'm here to learn too, that's why I post so often.

Bye. Jasen