The ESR will limit the peak, of course. But also the Tr and Tf time. You will get ringing (under-damped) if the Ls and C are not damped... but high ESR will over-damp it. Optimum is an ESR which gives max Tr (or Tf/discharge) time or critically damped. Most of the time, large electro type caps with wiring to it and to circuit are over-damped.
You can measure the SR of current by shorting the partially charged C into itself with shortest path possible (between terminals/wires) and measure current with current probe thru the discharge path.
very high currents are generated this way so use low voltage as possible on cap. In fact mfr will not replace a cap that has been abused this way. But good for T&M purposes.
THx-RNMarsh
You can measure the SR of current by shorting the partially charged C into itself with shortest path possible (between terminals/wires) and measure current with current probe thru the discharge path.
very high currents are generated this way so use low voltage as possible on cap. In fact mfr will not replace a cap that has been abused this way. But good for T&M purposes.
THx-RNMarsh
I don't think the equivalent series resistance and inductance cause current slewing. The inductance causes an exponential change in current over time and the resistance wastes energy as the thing charges or discharges. There will be current resonance if the resistance is too small and you short the terminals, say.
Last edited:
Just read the original post. It says V/uS. Read and see, the point is obviously not what is discussed, now.
Anyway, it seems nobody reads the original post, the purpose is just to come with some clever answer.
its the psu which is feeding the power for proper slew rate. What is the point of having such an amplifier which can slew high but unable to receive faster power.
An LR circuit's time constant is L/R, in units of seconds. The complete response equation
has an exponential term of the form e^(-Rt/L). The fastest slope of the response
is for small t, where e^(-Rt/L) ~ (1-Rt/L). This is because (e^xt) ~ (1+xt) for small t.
A duration of 5 time constants, (5 x L/R) seconds, will essentially complete the response
within 1% of the final value.
Last edited:
No, not at all. The original was V/uS for power supply capacitors. Then it was clarified that this is a nonsense number because psu caps are not supposed to change in voltage level, BUT they may be called to provide rapidly changing currents which naturally leads to A/us. I think these are valuable insights and in line with the intend of the thread. All about asking, searching, and learning.
I think your last statement is unfair to all here who take the trouble to reply and provide insight.
Jan
Last edited:
Thank you! That was my reasoning but I didn't have the exact equation available. It is fully complementary to an R-C circuit where the response is essentially completed after 5RC seconds.
Jan
Agree. The slew rate for the voltage on a cap is determined by the cap and the max current available for charging, assuming that the charge current becomes constant.
Then I would think the slew rate of the current into the (parasitic) L in wiring or a cap would be determined by the max current available, assuming that current remains constant.
Let me try again. Assume a cap charged to say 50V. There is a 1A load flowing. Now assume that the load drops to required 2A to flow. How long does it take for the current to go from 1A to 2A with a specific circuit L?
Rayma? ;-)
Jan
Read the original post and the first answers, it is clear, the slew rate mentioned there is what is usually meant by slew rate. The output amp slew rate in V/uS.
You better start a carrier in politics,.
Well at least there is some technical discussion going on.
The voltage slew rate thing is a bit of a red herring anyhow. I've never built an amp that sounded bad because it didn't have enough slew rate. It sounded bad for other reasons, which those who boast about their huge slew rates don't talk about.
Don't we all.
Slew rate - Wikipedia
In this context, slew rate is specifically the maximum possible dV/dt. Not current. With a capacitor, or other passive component, there are no "maximum possibles" in theory (in reality a component may fail mechanically) so a capacitor does not have a voltage slew rate limit nor a current rate of change limit. Obviously, the rate of change of I for a delta V is limited by resistance and inductance and the size of dV.
Don't we all.
Slew rate - Wikipedia
In this context, slew rate is specifically the maximum possible dV/dt. Not current. With a capacitor, or other passive component, there are no "maximum possibles" in theory (in reality a component may fail mechanically) so a capacitor does not have a voltage slew rate limit nor a current rate of change limit. Obviously, the rate of change of I for a delta V is limited by resistance and inductance and the size of dV.
I agree slew rate is rate-of-change. But it need not necessarily be limited to voltage just because that's all most diy people ever meet. There's nothing wrong with current slew rate in current-controlled loads and it is very prevalent in many engineering areas. For instance, in SMPS's they often limit the current slew rate in order to control EMI. I think Vicor pioneered it many years ago, iirc.
In fact, if you go back far enough you will discover that the term slew rate stems from WWII when the US Navy discovered they had to greatly increase the slew rate of their AA guns to have any chance of hitting a Japanese Zero fighter. In those days, slew rate was expressed in mills/sec, where a full 360 degree circle was divided up in 6400 mills.
Just because 'meter' is often used to measure distance, that doesn't mean you can't use it to measure height ;-)
Jan
Assume the capacitor doesn't discharge significantly during this step load change.
The initial load is (50V/1A) = 50 Ohms, which changes (at t=0) to (50V/2A)=25 Ohms.
Then for t>0, we have i (t)= {2 - e^(-25t/L)} x 1A
The time constant is (L/25) in units of seconds. Then for 5 time constants
(so within 1% of the final value), the time (in seconds) is L(in Henries)/5(in Ohms).
Last edited:
OK, so assume an inductance of 100nH (not sure that is realistic or not) it would take the current to get to the new value of 2A, if my sums are correct, 2*10^-8 s
So the current slew rate is in this example is 20A/uS.
What still bothers me here is that the current rise is exponential, right? But with something slew rate limited we think of it as a constant rate, limited by the circuit.
Jan
So the current slew rate is in this example is 20A/uS.
What still bothers me here is that the current rise is exponential, right? But with something slew rate limited we think of it as a constant rate, limited by the circuit.
Jan
Last edited:
Then for 5 time constants, L/5 = (100nH/5) = 20nS, to complete the response.
The fastest rate of change of current is just after t=0.
Since i (t)= {2 - e^(-25t/L)} x 1A
then di/dt (just after t=0) = (25/L) x 1A (in Amps per second),
or (25/10^-7) = 250 A/uS maximum rate of change of current.
Last edited:
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.