What does the volume knob do - stupid newbie

[bvbellomo]

Stupid newbie question, but I bet a lot of people (outside this forum) don't know the answer either. In a typical solid state amplifier (such as a home receiver), what does the volume knob control? Current, voltage, wattage or something else?

Speaker impedance curves are rarely flat - a speaker may have a 10 ohm impedance at 1 frequency, and 4 ohms at another, both within the range of frequencies the speaker plays well. If voltage were constant regardless of frequency and controlled by the knob, wattage would be much higher at 4 ohms than 10 ohms, and I would imagine volume would be much higher too. Similarly (but the opposite) applies to current. If the amp holds wattage steady, and is 220 into 4 ohms, what happens if I play the frequency with 4 ohm resistance, turn the volume all the way up, and then play the frequency with 10 ohms?

I figured someone on here actually builds amps and would know.

 

[scott17]

Read this:

http://www.angelfire.com/electronic/funwithtubes/Amp-Volume.html

 

[jaycee]

voltage. it functions as an attenuator on the input signal.

 

[nigelwright7557]

The volume control alters the voltage going into the amplifier.
The smaller the voltage the less the output from the amplifier.

 

[sofaspud]

The classic volume potentiometer is nothing but a variable voltage divider.

If voltage were constant regardless of frequency and controlled by the knob, wattage would be much higher at 4 ohms than 10 ohms, and I would imagine volume would be much higher too.

True enough, but in any [hi-fi, normal, realistic,... insert your own favorite adjective] circumstance voltage is definitely not constant regardless of frequency.
But let's assume it is for example's sake. The watts would be 2.5x at 4Ω than at 10Ω. That is a difference of 4dB. Enough to notice, but not a greatly significant increase.

 

[bvbellomo]

Read this:

http://www.angelfire.com/electronic/funwithtubes/Amp-Volume.html

I understand what a pot is, and I understand the voltage knob probably is a pot. However, I would think it would be terribly inefficient to run 200 watts (for example) directly through a pot before going to speakers. I understand the knob itself directly controls voltage to something inside the receiver - but does it also control the voltage of the signal that leaves the receiver? Or does it control something else? I could conceivably use a pot to raise and lower an amperage in a device that automatically adjusts the voltage to maintain the current I set (my fish tank lights work this way).

If the knob only regulates voltage, won't the current double at frequencies where the impedance is halved?

 

[bvbellomo]

The classic volume potentiometer is nothing but a variable voltage divider.

True enough, but in any [hi-fi, normal, realistic,... insert your own favorite adjective] circumstance voltage is definitely not constant regardless of frequency.
But let's assume it is for example's sake. The watts would be 2.5x at 4Ω than at 10Ω. That is a difference of 4dB. Enough to notice, but not a greatly significant increase.

I would disagree. 4dB is greatly significant. I understand that if this is the intent of the driver or speaker designer, it could be compensated for or relied on. But I disagree if you say 4dB is small enough not to care about.

 

[bvbellomo]

voltage. it functions as an attenuator on the input signal.

This could work, and I am not disagreeing with you. This might be the best way to do it. But if I attenuate my (analog) signal, won't I lose some clarity? I would think my speakers would sound better played at 100% volume on a 10 watt amplifier than at 1% volume on a 1000 watt amplifier. My experience is generally the reverse is true - high powered amps sound better even at lower volume. Of course, this could be because higher quality components are used and more time is put into design in a 1000 watt amp.

 

[scott17]

I understand what a pot is, and I understand the voltage knob probably is a pot. However, I would think it would be terribly inefficient to run 200 watts (for example) directly through a pot before going to speakers. I understand the knob itself directly controls voltage to something inside the receiver - but does it also control the voltage of the signal that leaves the receiver? Or does it control something else

Please re-read the information presented.

The "volume control" is an input signal attenuator. It is not connected to the speaker output signal.

Also, what does a "receiver" have to do with this? An amplifier has a certain amount of set gain based on the design. The input signal level, set by the input attenuator, determines the output level of the amplifier.

By the way, why is it true that better components and more time is spent designing a 1000W amplifier?

 

[AndrewT]

Bv,
forget about the speaker impedance variation with frequency for the moment.
Concentrate on understanding the input to the amplifier.
When you have that sorted in your head, then you can decide if you want to look at speaker design.

 

[chris661]

If the knob only regulates voltage, won't the current double at frequencies where the impedance is halved?

Yep.

So what?

Speakers generally are designed to work from a voltage source, and will demand current according to their impedance curve (and phase angle, but don't worry about that for the time being).
When the impedance dips down, more current flows, because the job of the output stage is to allow that current to flow.

The output stage (which here we'll consider to be anything after the volume control) has a fixed gain, lets say 10dB.
So if you turn the attenuator all the way up, so there's no attenuation, there will be a voltage gain of 10 decibels.
Wind the attenuator down to -10dB and your amplifier gain is now zero decibels. ie, the 2v signal coming in passes through to your speakers as 2v.
Turn the attenuator down to -20dB, the signal hitting your speakers is now 10dB down from 2v.

HTH
Chris

 

[bvbellomo]

Also, what does a "receiver" have to do with this?

Most people on this board probably use a home theater "receiver" bought in the past 10 years, or separates that work similarly. I have no idea if tube amps work similarly. Older, more expensive or more specialized equipment might work very differently.

By the way, why is it true that better components and more time is spent designing a 1000W amplifier?

Because either consumers are willing to pay more for larger amps, or manufactures believe so. There may be practical reasons too - a 10 watt amp can use thinner wires.

 

[bvbellomo]

Yep.

So what?

Speakers generally are designed to work from a voltage source, and will demand current according to their impedance curve (and phase angle, but don't worry about that for the time being).
When the impedance dips down, more current flows, because the job of the output stage is to allow that current to flow.

The output stage (which here we'll consider to be anything after the volume control) has a fixed gain, lets say 10dB.
So if you turn the attenuator all the way up, so there's no attenuation, there will be a voltage gain of 10 decibels.
Wind the attenuator down to -10dB and your amplifier gain is now zero decibels. ie, the 2v signal coming in passes through to your speakers as 2v.
Turn the attenuator down to -20dB, the signal hitting your speakers is now 10dB down from 2v.

HTH
Chris

So if I have a 4 ohm woofer that is 90db efficiency and an 8 ohm tweeter that is 93db efficiency, I can put them together and get flat response, even though my current is double at frequencies going to the woofer.

I realize impedance curves and real drivers are more complicated, but is this the general idea?


Suppose I have 2v input, -50db of attenuation and a fixed gain of 50db, I would have 2v of output. But I would think attenuating a signal -50db and then amplifying it by 50db would introduce a lot of noise.

 

[AndrewT]

So if I have a 4 ohm woofer that is 90db efficiency and an 8 ohm tweeter that is 93db efficiency,

Yes, if you mean 90dB/W @ 1m for the 4ohms driver and 93dB/W @ 1m for the 8ohms driver, then the two will require the same voltage for the same SPL.
Now that the two drivers need the same voltage you can design each half of the crossover. Select the crossover frequency, determine the impedance of the driver at that frequency. Design the filter to work with that impedance.
The Crossover Design Cookbook Introduction

Suppose I have 2v input, -50db of attenuation and a fixed gain of 50db, I would have 2v of output. But I would think attenuating a signal -50db and then amplifying it by 50db would introduce a lot of noise.

You are right, that much attenuation followed by that much gain can result in too much noise.
Power amplifiers generally have a gain between +20dB and +30dB.
Attenuators generally have -0dB to -50dB depending on the SPL required and the excess gain built into the system
http://www.diyaudio.com/forums/diyaudio-com-articles/186018-what-gain-structure.html