Understanding gain

[Clutz]

Hey everyone,

In reading about building various headphone amps, I've noticed a lot of people find it desireable for an op-amp in a headphone amplifier to be unity gain stable, but I'm not sure I understand why. If the op-amp's gain is set to unity, then it is not increasing the output voltage- and hence it is not increasing the volume at the output. I realize there are other reasons for having an amplifier, other than simply blowing out your eardrums- but they're not all apparent to me so I was hoping someone would be kind enough to explain.

Cheers,
Clutz

 

[amb]

There are many headphones out there (low impedance types, usually) that don't need much voltage gain. They need current gain. The amp could provide either or both.

 

[Clutz]

Quote:

Originally Posted by amb There are many headphones out there (low impedance types, usually) that don't need much voltage gain. They need current gain. The amp could provide either or both.

So is the amount of current gain per a particular op-amp a fixed quantity, or is it also set as a function of the resistors used?

Cheers,
Clutz

 

[amb]

Quote:

Originally Posted by Clutz So is the amount of current gain per a particular op-amp a fixed quantity, or is it also set as a function of the resistors used?

Current gain is not like voltage gain, in that it is not set with resistors in the amp. The amp delivers current into the load (headphones) as a function of voltage swing. Basically, the relationship is simple Ohm's law I = V / R, where I is the amount of current, V is the voltage, and R is the load impedance.

The sound you hear is the result of the headphone transducer converting electrical energy (power) into acoustic energy. And power is expressed as V * I. If you think about things a little you'll come to realize that low impedance phones does it work with less voltage and more current, contrasted to high impedance phones which require more voltage and less current.

The maximum amount of current that an amp could deliver is often the limiting factor on how much clean power you'll get with low impedance headphones. Since opamps typically have limited amount of output current, a cmoy-type amp is not ideally suited to drive low impedance phones directly. For these headphones an amp with a high-current output stage will give much better results.

 

[tangent]

Quote:

So is the amount of current gain per a particular op-amp a fixed quantity

If you look in the spec table of the op-amp, you will find output current specs. Usually this isn't very much different from what your average source can put out: 20-40mA, typical.

But if you add a buffer of some sort to the output, then the op-amp's output current limit isn't relevant. The buffer's output current drives the headphones; it's easy to get/make buffers capable of putting out several hundred mA, far above what most sources are capable of. We don't need all that current, but higher current means lower impedance, and that is helpful.

 

[SnoopyRocks]

Quote:

Originally Posted by Clutz In reading about building various headphone amps, I've noticed a lot of people find it desireable for an op-amp in a headphone amplifier to be unity gain stable, but I'm not sure I understand why.

Since Amb and Tangent answered the second part, I'll address the first - unity gain stability. The short answer is that a unity gain stable opamp is easier to work with because you are less likely to have to be concerned with stability problems...basicly one less thing to worry about.

Here's an oversimplified explanation that should give you the gist of what unity gain stability means.

The opamp is being used in a negative feedback system. This means that the opamp amplifies the difference of the input and a signal fed back to the input. As more and more delay (phase shift) is introduced into the loop, the difference dwindles, eventually becoming a sum. When this happens the circuit no longer works as intended and the output can no longer be properly controlled.

When designing fast opamps care must be taken to not introduce too much phase shift. The faster the opamp, the harder this is to do. One tradeoff that the designer might choose in the quest to make a faster opamp is to require a gain greater than one. This is the case for the OPA637 for example. Think of the increased gain as way to reduce the phase shift.

Does that make sense?

I'll let someone else explain why a faster opamp is desirable.

 

[tangent]

Quote:

I'll let someone else explain why a faster opamp is desirable.

Yes, it's a sticky question, isn't it? Theoretically, you only need something on the order of 1V/uS. But empiricially, we find that 145MHz op-amps often sound better than 3MHz op-amps. Odd, innit?

 

[skyskraper]

great posts guys. most informative. starting to understand the voodoo behind amps more and more

 

[Clutz]

Quote:

Originally Posted by tangent Yes, it's a sticky question, isn't it? Theoretically, you only need something on the order of 1V/uS. But empiricially, we find that 145MHz op-amps often sound better than 3MHz op-amps. Odd, innit?

Just to be clear on this, 1V/uS- the numerator is 1 volt, and the denominator is something to do with time (micro seconds?) (i.e. theoretically we only need something that can swing the voltage by 1 volt / micro-second).

I don't understand why electronics makes me feel as though I belong on the short bus.

 

[Clutz]

Quote:

Originally Posted by amb If you think about things a little you'll come to realize that low impedance phones does it work with less voltage and more current, contrasted to high impedance phones which require more voltage and less current.

I think I almost understand this- but perhaps I'm completely off

P=I*V, and I=V/R. Hence we can combine these two equations and come up with P=I^2*R or P=V^2/R. From these two equations, I would gather that low impedence cans would get more power with increasing voltage, while high impedence cans would do better with more current.

For low impedence cans (i.e 32 Ohms), this would result in I^2 * 32, which would be a smaller amount of power compared to 600Ohm headphones for a given amount of current. In contrast to this, for a given amount of voltage, low impedence cans would have relatively more power (V^2 divided by a smaller number) than high impedence cans.

So yeah, I'm confused..

 

[tangent]

Quote:

Originally Posted by Clutz i.e. theoretically we only need something that can swing the voltage by 1 volt / micro-second).

That's right. It's very rough, but something in that neighborhood is needed to make a 20kHz sine wave at the voltages we need without going into slew rate limiting.

The thing is, we don't listen to 20kHz sine waves.

Quote:

I don't understand why electronics makes me feel as though I belong on the short bus.

If it was easy, everyone would do it.

Sounds like you're ready to curl up with a copy of AoE.

Quote:

From these two equations, I would gather that low impedence cans would get more power with increasing voltage, while high impedence cans would do better with more current.

I haven't tried to follow your logic, because I know empirically that it's wrong. I suspect the problem is that you're assuming that all else besides impedance is equal, which it is not.

Your average 32 ohm phones want only about 0.5V to reach a high volume level, while typical 300 ohm phones need more like 2V to reach similar levels. But beware of making too many generalizations from this. The 64 ohm Senn HD-570s require pretty much the same voltage to sound as loud to me as the 300 ohm HD-580s.

Since low-impedance phones don't require much voltage (as a rule), their low impedance means that current is the dominating concern. High-impedance phones need more voltage, but their high impedance means current isn't a big concern.

 

[amb]

Yeah clutz, you are a little confused.

Quote:

Originally Posted by Clutz P=I*V, and I=V/R. Hence we can combine these two equations and come up with P=I^2*R or P=V^2/R.

So far so good.

Quote:

From these two equations, I would gather that low impedence cans would get more power with increasing voltage, while high impedence cans would do better with more current...

This is where the reasoning goes awry. Let's suppose we have two pairs of hypothetical headphones, one high impedance (300 ohms) and another low impedance (30 ohms). Let's also assume that the two have similar efficiency ratings (i.e., for a given amount of power delivered, each will produce the same sound pressure level). This is not always a good assumption in real life due to many variables, but let's not overcomplicate things for now.

Let's then suppose 10mW of power will result in 100dB SPL in each of these headphones. Then, using the formulas above to calculate the voltage and current needed to achieve that in each headphone:

P = V^2 / R, therefore V = sqrt(P * R)
P = I^2 * R, therefore I = sqrt(P / R)

• 300 ohm headphone
  V = sqrt(0.01 * 300) = 1.73V
  I = sqrt(0.01 / 300) = 5.8mA
• 30 ohm headphone
  V = sqrt(0.01 * 30) = 0.55V
  I = sqrt(0.01 / 30) = 18.3mA

I think it should be clear from this that the high impedance phone needs more voltage swing and less current than the low impedance phone.

For this example, the headphone amp needs less than one-third the voltage gain to achieve the same "loudness" for the 30 ohm phone than the 300 ohm phone, yet it requires over three times as much output current capability from the amp.

 

[Clutz]

Quote:

Originally Posted by tangent That's right. It's very rough, but something in that neighborhood is needed to make a 20kHz sine wave at the voltages we need without going into slew rate limiting. The thing is, we don't listen to 20kHz sine waves.

Speak for yourself! Nothing I like better than a good 20kHz sine wave..

Quote:

Originally Posted by tangent If it was easy, everyone would do it.

That's probably true- and that's probably why I like it. I just find it surprising because I spend a good part of most days doing math (I'm getting a PhD in Theoretical Biology -- it still sounds funny saying it).

Quote:

Originally Posted by tangent Sounds like you're ready to curl up with a copy of AoE.

Ahh yes- more money to spend on things.

I'll have to consider it.

Quote:

Originally Posted by tangent I haven't tried to follow your logic, because I know empirically that it's wrong. I suspect the problem is that you're assuming that all else besides impedance is equal, which it is not.

Yes, that is an implicit assumption of my logic. I was pretty sure it was wrong- but I figured it was the best way to demonstrate my ignorance.

Thanks for all your help, it is much appreciated.. Hopefully in a few years I'll be designing amps too. I'm trying to figure out how to design a cmoy like amp which uses - using OPA134 amps- but including three OPA amps - one for each channel and one for ground.

 

[Clutz]

Quote:

Originally Posted by amb Yeah clutz, you are a little confused.

Not a terribly uncommon thing to happen with me

Quote:

Originally Posted by amb Let's then suppose 10mW of power will result in 100dB SPL in each of these headphones. Then, using the formulas above to calculate the voltage and current needed to achieve that in each headphone:

Ahh yes. I figured it was something like that- but I couldn't get it into my head for some reason.

Thanks so much for your help!

Cheers,
Clutz

 

[Ob3ron]

Quote:

...a unity gain stable opamp is easier to work with because you are less likely to have to be concerned with stability problems.

Quote:

When designing fast opamps care must be taken to not introduce too much phase shift

Quote:

Think of the increased gain as way to reduce the phase shift.

SnoopyRocks, it seems that your first and last statements are contradictory

. In the first are you saying that unity gain is more stable than a higher gain? And in the last are you saying that a higher gain will reduce the phaseshift that makes an amp unstable?

Is another advantage of unity gain that the DC offset at the input doesn't get amplified by the opamp? Or is this "advantage" nulled fulther down the road by increased demand on the current stage adding extra DC offset?