I designed my amplifier such that it can handle full-power square waves without slew-rate limiting. It's of no practical use at all, but it assures you that whatever may happen, your amplifier won't go into slew rate limiting.
If I should ever design an amplifier for driving electrostatic headphones, I would design the output current limiting for a power bandwidth in the 3 kHz to 8 kHz range. High enough for almost all music, and the lower the current limit, the smaller the shock you would get if the insulation should fail. (European FM power bandwidth: 3183 Hz, that 8 kHz comes from a very old article of Otala's research group.)
If I should ever design an amplifier for driving electrostatic headphones, I would design the output current limiting for a power bandwidth in the 3 kHz to 8 kHz range. High enough for almost all music, and the lower the current limit, the smaller the shock you would get if the insulation should fail. (European FM power bandwidth: 3183 Hz, that 8 kHz comes from a very old article of Otala's research group.)
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I think that an amplifier ideally should have as much margin on its slew rate as the amplifier has loop gain. For example, an amplifier with a loop gain of 10 should have a slew rate 10x higher than 2*pi*f*v. Nearly all amplifiers have much more loop gain than slew rate.
Ed
Ed
A good article and a long list of useful references.
https://hifisonix.com/wp-content/uploads/2018/03/SID_and_TIM_W_Jung_77-79.pdf
https://hifisonix.com/wp-content/uploads/2018/03/SID_and_TIM_W_Jung_77-79.pdf
I don't intend to use tubes, or a massive output transformer > because I want to have quite high power output to avoid the high distortion of clipping.
PS.
In the 70's/80's (?) Sanyo marketed an amplifier with a SR switch with enough SR reduction to help suppress the 'clicks & pops' of vinyl records 🙂
https://www.google.com/search?q=san...jIwoAfo3wGyBwcwLjE3LjIwuAeJPQ&sclient=gws-wiz
Loop gain is a small signal characteristic, whereas slew rate is concerned with large signals. They should not be mixed together.
That SR switch changed the Miller compensation capacitor (see the “worked examples” in my post #4 above). That changed not only the SR, but also the loop gain and thus the distortion level.
@alexcp - My reasoning is as follows: non-linearity in the output stage can cause the integrator to slew on normal audio signals. The class B output stage is a worst-case. The crossover notch causes the amplifier to run open-loop momentarily at every zero crossing.
Ideally, an amplifier should be designed not to slew under those conditions. The slew rate must be adequate to cover the amplifier's open-loop gain on a normal audio signal that would not overdrive the amplifier running in closed-loop.
A ratio x can be defined for how much of the differential pair's tail current is used:
x = Vin_peak * gm_differential_pair / I_tail
where
Vin_peak = Vout_peak / closed_loop_voltage_gain
An amplifier with x<1 can tolerate any non-linearity without slewing. x>1 is permissible as long as the output stage never becomes too non-linear. Optimal class AB biasing does that.
My amplifier has x=2. The Wolverine has x=13.
Ed
Ideally, an amplifier should be designed not to slew under those conditions. The slew rate must be adequate to cover the amplifier's open-loop gain on a normal audio signal that would not overdrive the amplifier running in closed-loop.
A ratio x can be defined for how much of the differential pair's tail current is used:
x = Vin_peak * gm_differential_pair / I_tail
where
Vin_peak = Vout_peak / closed_loop_voltage_gain
An amplifier with x<1 can tolerate any non-linearity without slewing. x>1 is permissible as long as the output stage never becomes too non-linear. Optimal class AB biasing does that.
My amplifier has x=2. The Wolverine has x=13.
Ed
Not if its correctly biased, the loop is fully closed, but there is residual distortion as the feedback can only reduce, not eliminate it entirely. If the loop weren't closed at the crossover point the amp would be unstable with a 0V input signal! Severe underbiasing will do just that. Poor thermal control of biasing can lead to this too.
If the loop ever goes open-loop, you have perhaps a class C output stage, not class B. Class B is when each device conducts 180 degrees, i.e. with no gap.
A class B conduction angle of 180 degrees is achieved with zero bias.
From the approximation to Ebers-Moll:
A transistor will conduct for any Vbe>0.
Differentiate the equation for a pair of transistors at zero bias. The slope of the transfer function will be zero at the origin.
Zero slope means that the loop is open. Even though the transfer function has no discontinuities, no amount of loop gain or slew rate can linearize it.
Ed
From the approximation to Ebers-Moll:
Code:
Ie = Ies * (exp (Vbe/Vt) - 1)A transistor will conduct for any Vbe>0.
Differentiate the equation for a pair of transistors at zero bias. The slope of the transfer function will be zero at the origin.
Zero slope means that the loop is open. Even though the transfer function has no discontinuities, no amount of loop gain or slew rate can linearize it.
Ed
My computer just told me that I had a sign reversed. Change "zero slope" to "near zero slope" and "no amount of feedback" to "no practical amount of feedback" in my previous post.
Ed
Ed
Strictly speaking, the edges of a square wave on the output of an amplifier are showing the slew rate. It will look very square at some low frequency input and less square at a high frequency input. The slew rate limit is an indicator of the high frequency bandwidth. When the dv/dt slew rate limit is reached then there is no error current left to correct or to complete the feedback loop. The slew rate limit is tied to the Gain Bandwidth Product but I don't recall the exact equation.
The edges of a square wave show the slew rate only if the amplifier is slew rate limiting. If not, the square wave shows the rise/fall time.
You can see the difference: if the edges are straight, it is slew rate limiting. If the edges form a a nice rounded shape, with or without overshoot, there is no slew rate limiting and the edges show the rise/fall time.
Slew rate limiting is a large signal phenomenon and typically only occurs at high or maximum output levels.
That is why you typically see wide bandwidth at low levels, and much smaller full power bandwidth (and that is because of slew rate limiting).
So it is often the case that at low signal levels, the edges show rise/fall time, and with rising level transit to straight-edge slew rate limit shape.
And yes, there's the frquency parameter - the higher the frequency, the earlier the onset of slew rate limiting, but only with high signal levels.
Jan
You can see the difference: if the edges are straight, it is slew rate limiting. If the edges form a a nice rounded shape, with or without overshoot, there is no slew rate limiting and the edges show the rise/fall time.
Slew rate limiting is a large signal phenomenon and typically only occurs at high or maximum output levels.
That is why you typically see wide bandwidth at low levels, and much smaller full power bandwidth (and that is because of slew rate limiting).
So it is often the case that at low signal levels, the edges show rise/fall time, and with rising level transit to straight-edge slew rate limit shape.
And yes, there's the frquency parameter - the higher the frequency, the earlier the onset of slew rate limiting, but only with high signal levels.
Jan
Your last sentence is true for sine waves, but a 10 Hz square wave with very steep edges is quite capable of causing slew rate limiting in amplifiers that are not designed to handle square waves.
Agreed, if you keep on increasing the input signal slew rate, you can slew rate limit any amplifier even at low frequencies.
I would think that is not normally the case in audio.
Although I have seen audio amps with astronomical slew rates designed for the sole purpose to impress with the square wave response.
Another way to distinguish yourself from the competition.
Jan
I would think that is not normally the case in audio.
Although I have seen audio amps with astronomical slew rates designed for the sole purpose to impress with the square wave response.
Another way to distinguish yourself from the competition.
Jan
Yes, even at zero bias a be junction will conduct a minuscule amount of electrons.
But I accept 'zero' for practical considerations 😎
Jan