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Amplifier Voltage Output - I should know this

12Many

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Here is another questions I should know the answer too, but don't. When we turn up the volume, it is the voltage level across the speaker (across the amp output) that is increasing correct? And, the current draw from the amp is dependent on the impedance of the speaker and the voltage across the speaker/amp output. V=IR, so for a fixed resistance (at a particular signal frequency which will correspondences to a particular speaker impedance), as we turn up the volume knob, we are increasing the voltage and this will in turn draw more current. Where amps get into trouble is when the impedance drops down low, and the volume is turned up, the amp must attempt to output a higher and higher current, which it may not be able to do. How would that affect the higher frequencies of the audio signal from the speakers?

One other question, is the slew rate of an amplifier related to its maximum frequency? It seems like they should be because both relate to the rate of change of the signal over time. .
What is considered a good or great slew rate for an amp. Finally, I think slew rate is change in voltage over time, but I would think an amplifiers ability to effect a change in current over time would be just as important.

Kinda all over the place with this post. Thanks.
 

DVDdoug

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as we turn up the volume knob, we are increasing the voltage and this will in turn draw more current. Where amps get into trouble is when the impedance drops down low, and the volume is turned up, the amp must attempt to output a higher and higher current, which it may not be able to do.
Correct. And since Ohm's Law is a law of nature it means that the voltage is also limited/clipping.

How would that affect the higher frequencies of the audio signal from the speakers?
Clipping is harmonic distortion so more "higher" frequencies. But, usually it's the low frequencies that demand the most power so that's what usually clips and the lower harmonics are the strongest, so most if the distortion will be in the lower & mid frequencies. ...If you clip regular program material very-badly you tend lose some highs.

One other question, is the slew rate of an amplifier related to its maximum frequency? It seems like they should be because both relate to the rate of change of the signal over time. .
I've never heard of it being a problem with audio amplifiers. I've never heard of it being audible.*

Finally, I think slew rate is change in voltage over time, but I would think an amplifiers ability to effect a change in current over time would be just as important.
Per Ohm's Law, voltage & current are directly proportional. If the voltage is slew-rate limited, so is the current and vice-versa.



* P.S.
I've never heard ANY distortion from an amplifier unless it was broken or over-driven. ;)
 

Doodski

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is the slew rate of an amplifier related to its maximum frequency?
The angle/slope of the waveform can be used to calculate the frequency if it is a sine wave. This formula is rarely used but it can be done. For square waves things can be more complicated.
 

RayDunzl

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One other question, is the slew rate of an amplifier related to its maximum frequency? It seems like they should be because both relate to the rate of change of the signal over time. .
What is considered a good or great slew rate for an amp. Finally, I think slew rate is change in voltage over time, but I would think an amplifiers ability to effect a change in current over time would be just as important.

Yes, related to frequency and also to the voltage level.

A good slew rate is probalbly one that is "good enough".

If the amplifier can change its voltage (over time) into a load, then the necessary current is flowing. If not, the voltage won't be there.

--

Here's an old amp spec:

Description: Power output: 100W into 8 ohms (20dBW), doubles with each halving of load impedance down to 1 ohm. Frequency response: 20Hz–20kHz, +0.0/– .1dB; 1Hz–150kHz, +0.0/–3dB. Distortion: <0.1% at 1kHz; <0.5%, 20kHz, full power. Slew rate: 100V/µs. Input sensitivity: 1.4V RMS. Gain: 26dB. Damping factor: >60. Input impedance: 47k ohms.

Slew Rate - 100V per microsecond.

---

Asking ChatGPT what slew rate would be required to output 20khz into 8 ohms at 100 watts:

(long calculation deleted for 28.28Vrms output)

"The amplifier would need a slew rate of approximately 2.513 V/µs or greater to output a signal of 20 kHz into an 8-ohm load at 100 watts power without distortion."
 
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egellings

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100V/uSec is way faster than needed for home audio. Something in the range of 2 to 5V/uSec would be adequate. Having a higher slew rate won't hurt; it just won't add anything useful, and it can be more expensive to get an unneeded higher slew rate. It's like asking for 100kHz bandwidth in an audio amplifier connected to a CD player with a maximum upper frequency output of a bit over a totally adequate 20kHz. The amplifier's higher frequency capability would never be used. Even if it were, any such output would be inaudible to human ears anyway.
 

radix

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You could also take a look at this thread. An amp could run out of voltage or current or limit itself (protection). And some power supplies handle bursts differently than others. IIRC, I once asked a similar question and a response was it's a bit more complicated than V=IR, but that's close. I couldn't find that post....

 

Doodski

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Yes, related to frequency and also to the voltage level.

A good slew rate is probalbly one that is "good enough".

If the amplifier can change its voltage (over time) into a load, then the necessary current is flowing. If not, the voltage won't be there.

--

Here's an old amp spec:

Description: Power output: 100W into 8 ohms (20dBW), doubles with each halving of load impedance down to 1 ohm. Frequency response: 20Hz–20kHz, +0.0/– .1dB; 1Hz–150kHz, +0.0/–3dB. Distortion: <0.1% at 1kHz; <0.5%, 20kHz, full power. Slew rate: 100V/µs. Input sensitivity: 1.4V RMS. Gain: 26dB. Damping factor: >60. Input impedance: 47k ohms.

Slew Rate - 100V per microsecond.

---

Asking ChatGPT what slew rate would be required to output 20khz into 8 ohms at 100 watts:

(long calculation deleted for 28.28Vrms output)

"The amplifier would need a slew rate of approximately 2.513 V/µs or greater to output a signal of 20 kHz into an 8-ohm load at 100 watts power without distortion."
That's great stuff @RayDunzl. Can you punch in figures for a 150kHz bandwidth audio amp to see what the slew rate is there? Pleaseee. :D This would be representative of many audio amps.
 

RayDunzl

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Can you punch in figures for a 150kHz bandwidth audio amp

BonziBuddyReincarnate2024 needs to know the wattage and ohms to give its guess.
 

NTK

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Well... Pretend it is a Harmon Kardon 70W/ch @ 8 Ohms. They have very wide bandwidth.:D

Slew Rate.jpg
 

radix

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I believe the main use of high slew rate in amps is for negative feedback to correct distortion. You want the amp to be very fast compared to the input. I'm not a control theory guy, but I believe most of the time op amps have such high BW (slew rate) that one ignores the propagation time for feedback calculations. There are some works on modeling slew-rate limited amps and feedback networks.

I'm just guessing, but I don't think any modern (since 1980s?) op amps are BW limited (slew-rate limited) for audio frequencies, unless they are being used in very high gain situations. I tried finding a table of opamps by year, but couldn't stretch my searchbar-kungfu that far.
 

RayDunzl

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Cbdb2

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Thats the slew rate needed for full power at 150khz. Audio amps never go past 20khz full power so the actual slew rate needed is less, but the rule of thumb is to multiply the amps max rate of change (which is what is calculated above) by 5 or 10 to give you the slew rate needed.
 

Cbdb2

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Slewing happens when the Amp can't keep up to the change in input. Usually because of the compensation cap in the voltage gain stage. It drives the Amp out of its linear region and into overload (the feedback dosnt work). Thats why the 5 to 10 times safety margin. From "Analysis and Design of Analog Integrated Circuits" by Grey and Meyer.
 

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egellings

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Slewing happens when the Amp can't keep up to the change in input. Usually because of the compensation cap in the voltage gain stage. It drives the Amp out of its linear region and into overload (the feedback dosnt work). Thats why the 5 to 10 times safety margin. From "Analysis and Design of Analog Integrated Circuits" by Grey and Meyer.
If the output voltage is ramping up to a curved input signal rather than following it, then we have an amp that has run out of steam.
 

DonH56

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Note power is usually expressed as average power, so the voltage is RMS, thus you need to find the peak voltage from the RMS value by multiplying by ~1.414 (sqrt(2)). ChatGPT still has a few things to learn. ;)

Slew rate is related to frequency and amplitude; the required slew rate is higher as you go up in frequency, amplitude, or both.

@Cbd2 I think you meant to say slew limiting; slewing always happens unless the output is DC.

Old article on slew rate: https://www.audiosciencereview.com/forum/index.php?threads/slew-rate.25438/
 

egellings

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Note power is usually expressed as average power, so the voltage is RMS, thus you need to find the peak voltage from the RMS value by multiplying by ~1.414 (sqrt(2)). ChatGPT still has a few things to learn. ;)

Slew rate is related to frequency and amplitude; the required slew rate is higher as you go up in frequency, amplitude, or both.

@Cbd2 I think you meant to say slew limiting; slewing always happens unless the output is DC.

Old article on slew rate: https://www.audiosciencereview.com/forum/index.php?threads/slew-rate.25438/
That's when the signal is a sine wave. Other waveforms will have different numbers.
 

DonH56

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That's when the signal is a sine wave. Other waveforms will have different numbers.
Good grief, we are debating stuff nobody really cares about. And yes that old article was just for pure sine waves. I'll leave it to you more intelligent types to figure it out for everybody else; been there, done that.
 
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