Can I connect a speaker to a pre-amp?

[Zeppelin]

No the current rating does not need to be exact.

If it is high, then you will have spare current not used (the amp will only take what is needed)

If it is lower then you will not be able to get full power (especially with lower impedance speakers like 4 ohm)

If you want to maximise the power, you need to have a voltage close to the maximum allowed by the amp.

Sorry but I am not sur to undrstand it completely. When you talk about current do you refer to voltage+amperage? Or just voltage? To make it clearer, I know the voltatge part, it's the amperage part the one that makes confusion to me.

Are you saying it will work anyway if the amperage it is not exact?

 

[antcollinet]

Sorry but I am not sur to undrstand it completely. When you talk about current do you refer to voltage+amperage? Or just voltage? To make it clearer, I know the voltatge part, it's the amperage part the one that makes confusion to me.

Are you saying it will work anyway if the amperage it is not exact?

Current and amperage are basically the same thing.

It is the current that flows (like the water in a river) - the units for current flow is amps.

At least where I am, the PSU has a current rating in amps. So the current rating might be 10amps.

And yes - if you have an amplifier that has a PSU rating of 20V/3A, then it needs at least 3A to reach full power but if your PSU can give more it doesn't matter.

 

[MaxwellsEq]

Sorry but I am not sur to undrstand it completely. When you talk about current do you refer to voltage+amperage? Or just voltage? To make it clearer, I know the voltatge part, it's the amperage part the one that makes confusion to me.

Are you saying it will work anyway if the amperage it is not exact?

Current is measured in Amps. It is measured with an ammeter. In laymen's terms it refers to the amount of electricity flowing in a circuit. Current and Voltage are separate entities, but are related by Ohms law (depends on the Resistance in Ohms). When someone talks about current or Amps, they mean current not voltage.

Vottage is measured in Volts. It is measured with a voltmeter. In laymen's terms it is the "pressure" caused by a difference in potential between two points. Voltage and current are separate entities, but are related by Ohms law (depends on the Resistance in Ohms). When someone talks about voltage or volts, they mean voltage, not current.

Because of this, the amount of current a PSU may need to supply is dependent on the resistance of the load and the voltage potential across it. Let's say I have 12 Ohm load. I've purchased a very expensive 12V supply that can deliver 10 Amps. When I connect the load to it, the PSU only delivers 1 Amp. Was I cheated when I purchased the PSU?

 

[Zeppelin]

So you say it doesn't matter the total amount of voltage and amperage, just the equivalency between them?

And when does it appear the possibility of overheating? When the equivalency it is to much? O it depends more on the V or the A?

Because of this, the amount of current a PSU may need to supply is dependent on the resistance of the load and the voltage potential across it. Let's say I have 12 Ohm load. I've purchased a very expensive 12V supply that can deliver 10 Amps. When I connect the load to it, the PSU only delivers 1 Amp. Was I cheated when I purchased the PSU?

I tried to discover the answer with the ohm formula, but it didn't work haha I did "I=V/R", to discover the value of V to know why at the end the PSU only delivers 1 amp. Probably it is because the speaker (12 ohms) need more voltage and takes it from amps to convert it as Voltage (If I understood everything well)

Can you shed some light over here?
Thanks

 

[MaxwellsEq]

So you say it doesn't matter the total amount of voltage and amperage, just the equivalency between them?

And when does it appear the possibility of overheating? When the equivalency it is to much? O it depends more on the V or the A?


I tried to discover the answer with the ohm formula, but it didn't work haha I did "I=V/R", to discover the value of V to know why at the end the PSU only delivers 1 amp. Probably it is because the speaker (12 ohms) need more voltage and takes it from amps to convert it as Voltage (If I understood everything well)

Can you shed some light over here?
Thanks

Have a look at this picture. It covers all the possible variations. I wish I'd had access to such a good picture when I was a kid.

A standard speaker can be quite loud when driven with 5W. So let's start with an assumption that your use case is probably met by 5W. Let's also assume the load is a resistive 8 Ohms. I'm simplifying some other things here to do with AC, but the working out should help.

Taking P = V^2 / R. Then V^2 = P * R and V = sqrt(P * R) = sqrt(5 * 8) = 6.3V. So to dissipate 5W into 8 Ohms you need a potential of 6.3V
Taking P = I^2 * R. The I^2 = P / R and I = sqrt(P / R) = sqrt(5 / 8) = 0.8A. So the dissipations of 5W into 8 Ohms draws a current of 0.8A

We can check this, since we know P = I * V = 6.3V * 0.8A = 5.0W

How hot things get is down to current. So the PSU / amplifier needs to be able to safely handle 0.8A. Nothing is 100% efficient, so a margin is always required

What if we want 10W into 8 Ohms? V = 8.9V and I = 1.1A

What if we want 5W into 4 Ohms? V = 4.5V (note this is less than with 8 Ohms) ; I = 1.1A (note this is higher than with 8 Ohms).

What if we want 5W into 16 Ohms? V = 8.9V (note this is more than with 8 Ohms) ; I = 0.6A (note this is lower than with 8 Ohms).

So we can see that as load resistance drops, the current goes up and the voltage reduces. As load resistance increases the voltage goes up and the current reduces.

High impedance loads needs more volts, low impedance loads draw more current.

 

[Zeppelin]

@MaxwellsEq I want to thank you for the effort you are giving in making things understandable for me, but you are not really answering my two questions, which are:

1- Does my PSU have to have the "correct" voltage and current values for the speaker, or if it is "more or less" the speaker will "decide" to get more volatge than current, or viceversa?

2- When is there the possibility of overheating? If I choose a PSU that has 5A, which doesn't seem to be necessary for most of speakers, would I be entering in a dangerous overheating zone?

The first one is answered by a kind user, the 2nd is the one is still a doubt.
Your explanations are very good, but what I am only asking for is those 2 questions. After that I would be able to make more decisions.

Thanks

 

[MaxwellsEq]

@MaxwellsEq I want to thank you for the effort you are giving in making things understandable for me, but you are not really answering my two questions, which are:

1- Does my PSU have to have the "correct" voltage and current values for the speaker, or if it is "more or less" the speaker will "decide" to get more volatge than current, or viceversa?

2- When is there the possibility of overheating? If I choose a PSU that has 5A, which doesn't seem to be necessary for most of speakers, would I be entering in a dangerous overheating zone?

The first one is answered by a kind user, the 2nd is the one is still a doubt.
Your explanations are very good, but what I am only asking for is those 2 questions. After that I would be able to make more decisions.

Thanks

I've answered both questions

 

[Zeppelin]

How hot things get is down to current. So the PSU / amplifier needs to be able to safely handle 0.8A. Nothing is 100% efficient, so a margin is always required

Are you tellilng me it can overheat if I give to the amp less current rather than to much?

 

[MaxwellsEq]

Are you tellilng me it can overheat if I give to the amp less current rather than to much?

You can not GIVE current to the amplifier.

The voltage generated by the amplifier across the load causes the amplifier to pass a current which it pulls from the power supply. If you have an 8 Ohm load and your amplifier creates a potential across the load of 0.0001V you will draw almost no current. The load will not get very warm. If your amplifier creates a potential of 10000V across the same load, you will draw a huge current and things will get hot very fast.

This is all very clear from the wheel I showed above and the calculations I did as examples.

Create yourself a spreadsheet with a column for each part of the formulae I used above and drop some values in to see what happens. It really is as simple as the mathematics covered in the first year of high school.