You are off by at least a factor of 100, try again. Look at energy stored after 2 us and before with 2A charge current and sufficient time to restore 48V will be what duty cycle?There is something strange in your simulation. The resistor R1 can just sink some of the current from C1 subtracting it from that going to the MOS. Could you post you LT project ?
In order to calculate the capacitor you shoud write down the equations of the circuit, in any case you could estimate the value using the condition that the drawn current is almos constant over the period you need. In this case we know that the energy stored in the capacitor is:
Es = 0.5*C*V^2
energy is power*time. Then under the simplified hypothesis that I=constant, the required energy over a time period "t" is:
Er = V*I*t
equating now Es = Er and solving with respect to C:
C = 2*I*t/V
in your case I=30A, V=45V, t=2us that is
C= 2*30*2u/45 = 2.7 uF
However in real life you must consider also the output impedance of the generator as well as the ESR of the capacitor.
Not sure to have correctly understood your comment, however I'm referring to a one shot discharge of a fully charged capacitor. I didn't read about repetitive pulses.You are off by at least a factor of 100, try again. Look at energy stored after 2 us and before with 2A charge current and sufficient time to restore 48V will be what duty cycle?
Ok I lost a maximum dV=0.5v was required. Where is written ?The question was what C permits a 0.5V sag in 2 us from a 30A load (45V/1.5 ohm)?
C = Ic dt/dV = 30A *2e-6s/0.5V = 120 uF ideal
With a 2 Adc supply that requires dt=C*dV/Ic= 120e-6 * 0.5V/2A = 30 us if the current is constant to restore the 0.5V.
A good low ESR Cap is around Tau = 1 us , so 8 mOhm * 30A = 240 mV means you need two Caps. to meet the 500 mV spec. or 240 uF.
so you were off by ~ 100
Post #5Ok I lost a maximum dV=0.5v was required. Where is written ?
Post #5Ok I lost a maximum dV=0.5v was required. Where is written ?
Not quite.I don't understand your math (possibly you are considering a different circuit from mine). In any case If you have:
Ic costant current drawn
V voltage across the ideal capacitor
Vc voltage across the series C+ESR
and we suppose negligible the current from the power supply (i.e.: considering just the discharge of the capacitor with constant current Ic) equations are:
Ic = C*dV/dt
Vc = V - Esr*Ic --> V = Vc + ESR*Ic
that is:
ESR*Ic + dV <= 0.5 or ESR*Ic + Ic*dt/C <=0.5
solving with respect to C:
C >= I*dt/(0.5 - ESR*Ic)
of course must be ESR*Ic < 0.5 that is: ESR < 0.5/Ic
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