Once again, a capacitor does not have a slew rate limit. Try not to use the term "slew rate" at all. Then you will free yourself to grasp what an exponential response is.
An inductor cannot change current instantly. This would require infinite voltage. So you are asking the wrong question. Eg: ask about what V would result in a certain change of I to within 1% within t seconds. If you ask what speed an inductor will change from 1A to 2A then there is no answer as it depends on what V is applied. Eg: you can ask what V drop will occur across the inductor if it is subjected to a dI/dt = 1A/us.
An amplifier's output cannot change current instantly nor would it even if its output voltage was a perfect step. It sees inductance in the speaker cable and output inductor (if you use one) and resistance to ground through the Zobel and it sees non-zero impedance in the psu rails. Try estimating how much dI/dt an amplifier needs to supply and then work out how much voltage drop will result across the psu caps.
T-bam, you still do't grasp that slew rate can be applied to voltage, current, AA guns, whatever. Audio types have such a narrow view!
Slew rate as such is an undefined thing. Slewing of what? Voltage, current, guns?
Just as voltage cannot change instantaneous, neither can current. It takes time to slew the voltage on a cap, just as it takes time to slew the current into a an inductor. Completely exchangeable.
I think I made it clear, I will not continue this discussion. Those who want to learn already have done so. Too bad you're refusing.
Jan
Slew rate as such is an undefined thing. Slewing of what? Voltage, current, guns?
Just as voltage cannot change instantaneous, neither can current. It takes time to slew the voltage on a cap, just as it takes time to slew the current into a an inductor. Completely exchangeable.
I think I made it clear, I will not continue this discussion. Those who want to learn already have done so. Too bad you're refusing.
Jan
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You get a linear change in voltage/current in an ideal capacitor/inductor when the applied current/voltage is constant. If the applied current/voltage reduces according to the voltage/current already achieved in the component (e.g. a CR/LR circuit) then the change is exponentiallly decaying.jan.didden said:
If the circuit is LCR (and all real circuits are!) then you get some portion of a damped oscillatory/transient response, but a short enough segment of this will look like a straight line.
I agree 100% to Jan didden. Slew is about change and rate is per time hence its applicable to any thing.
As far as the situation is that consider if we have an amplifier ckt with 400V/uSec so how much should the supply cap ESL should be?
consider Im running the amplifier at +/-100V and into 4ohm load so consider its clipping considerations with healthy 80v/4ohms about 20A so if the psu capacitors are able to deliver 20A in 1 micro seconds then the slew of 400V/usec will make the justification.
So in that case what should be the ESL of the capacitor? Any practical ESL values measured lets do a case study here.
As far as the situation is that consider if we have an amplifier ckt with 400V/uSec so how much should the supply cap ESL should be?
consider Im running the amplifier at +/-100V and into 4ohm load so consider its clipping considerations with healthy 80v/4ohms about 20A so if the psu capacitors are able to deliver 20A in 1 micro seconds then the slew of 400V/usec will make the justification.
So in that case what should be the ESL of the capacitor? Any practical ESL values measured lets do a case study here.
The current rise will resemble this graph, starting at 1A and ending at 2A, with a time constant of L/25.
The slope (rate of change of current) varies throughout the current rise.
The maximum rate of change of current, di/dt (max), is just after the load switch at t=0+ and is 25/L.
https://upload.wikimedia.org/wikipe...svg/230px-Series_RC_capacitor_voltage.svg.png
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There is no direct relationship between output voltage slew rate and your supplying bulk cap.
You cannot assume that a 4Ohm load soaks 100A/us, because this applies to an ideal ohmic resistor - in reality you are far away from that. Due to some inductance, output current slew rate will be ways slower than this expectation value.
On the other hand, assume a pure capacitive load of 1nF. During transition the current under ideal conditions is constant according to
I = C * dV/dt = 1n * 400 meg = 0,4A.
Assume a poor bulk cap with an ESR of 0,1 Ohm. The supply voltage dip during transition computes to
dV = di * ESR = 0,4 * 0,1 = 0,04V
These simple examples might illustrate that there is quite little relationship between output voltage slew rate and the properties of the supplying bulk cap.
Is this an established equation? why is it 1A there? is it the delta for the current change? Since i (t)= {2 - e^(-25t/L)} x 1A
can you give me some clarity on the following example:
Lets consider the change in load is about 0A to 20A in 4ohms with an inductance of 200nH so how much does it takes to raise to 20A?
The equation was derived from the circuit discussed,
and the 1A factor adds the proper units for current.
You can also write the equation as i(t) = 2A - 1A x e^(-25t/L)
Note that at t=0 the exponential e^0 is 1, giving i(0)=1A,
and at large times the exponential is 0 due to the minus sign, giving i(limit) =2A.
The time constant for your case is L/R = 200nH/4R = 50nS.
A single time constant exponential curve reaches within 1%
of the final value in 5 time constants, since e^(-5) = 0.0067 = 0.67%.
Then the total change (within 1%) takes 5 x 50nS = 250nS.
The initial or final current values don't affect the shape of the time response,
or the time constant. They serve as scaling factors to the exponential curve.
Only the L and R affect the time constant, so 250nS is the time for
any current change whatsoever to complete within 1%.
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This is important. It means that whatever supply voltage you have, the time to get to a final value of the current is 250nS (in your example).
Can we link it to the voltage slew rate required for an amp? Let us take the proverbial 100W/8ohm amp that delivers 50V/uS output voltage slew rate.
That 100W in 8ohms requires 28V across that 8ohms, or 3.5A through those 8ohms.
So to support that 50V/uS output voltage slew rate, the amp should also be able to deliver a 3.5A/uS current slew rate.
Am I right?
Jan
Can we link it to the voltage slew rate required for an amp? Let us take the proverbial 100W/8ohm amp that delivers 50V/uS output voltage slew rate.
That 100W in 8ohms requires 28V across that 8ohms, or 3.5A through those 8ohms.
So to support that 50V/uS output voltage slew rate, the amp should also be able to deliver a 3.5A/uS current slew rate.
Am I right?
Jan
It depends on the circuit configuration. Often high feedback amps are more sensitive to capacitive loads and then you need a small (few uH) coil to isolate them from the amp.
Jan
In many cases the output coil will be outside the NFB loop. That means that it is simply a linear component in series with the speaker. It may 'limit' the slew rate, but in a non-distorting way - just a low pass filter, and as (almost) everyone knows a linear filter does not introduce distortion.
Slew rate is redundant for audio amps
I admit I haven't read this huge thread, but here's my opinion about slew rate:
If your music amplifier is capable to output a full power, undistorted signal at the full audio spectrum (20-20Khz), (incl. a 20Khz sine wave at max amplitude), then slew rate as a parameter is completely redundant to characterize that amplifier and even misleading. Period.
The 20Khz test at max amplitude is the worst-case evaluation of the amplifier's capability and tolerance
1) to ensure flat response at the audio band at full power,
2) because you don't want the amp to blow up if a microphone or wrong wiring produces such a feedback signal accidentally.
Even if you care about frequencies that you can't hear, THD along with frequency response is enough and far more informative. Volts per microsecond doesn't mean anything alone, as it is dependent on the specific amp's max output voltage (and power) at a specific load.
Slew rate is useful to characterize other applications, NOT audio (music) amplifiers, because other parameters like THD, frequency response and load (ohms or capacitance) are far more informative.
I admit I haven't read this huge thread, but here's my opinion about slew rate:
If your music amplifier is capable to output a full power, undistorted signal at the full audio spectrum (20-20Khz), (incl. a 20Khz sine wave at max amplitude), then slew rate as a parameter is completely redundant to characterize that amplifier and even misleading. Period.
The 20Khz test at max amplitude is the worst-case evaluation of the amplifier's capability and tolerance
1) to ensure flat response at the audio band at full power,
2) because you don't want the amp to blow up if a microphone or wrong wiring produces such a feedback signal accidentally.
Even if you care about frequencies that you can't hear, THD along with frequency response is enough and far more informative. Volts per microsecond doesn't mean anything alone, as it is dependent on the specific amp's max output voltage (and power) at a specific load.
Slew rate is useful to characterize other applications, NOT audio (music) amplifiers, because other parameters like THD, frequency response and load (ohms or capacitance) are far more informative.
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