New for February 2009!
This page will show you why you need to worry about the equivalent
series resistance (ESR) of charge storage capacitors in pulsed applications.
In some cases, the number of capacitors you need will be driven
by heating, not by voltage droop.
In the case of electrolytic capacitors
powering a pulsed transmitter, the ESR of a single 4.7 uF capacitor
might be on the order of one ohm. The equivalent series resistance
of a capacitor can be a strong function of voltage in ceramic capacitors,
but a near constant in electrolytics. Often, in pulsed transmitters,
you will have to design a modulator circuit that supplies high peak
currents to the power amps, with fast rise and fall times. The power
supply will be located far enough from the amplifier/modulator so
that it can't source the current pulses. This is when charge storage
capacitors take over the heavy lifting.
In the chart below we examine
a charge storage system (yes, we need to post a schematic too!)
where the current from the power supply is effectively decoupled
from the charge storage capacitor bank. The caps are located right
next to the modulator circuit, which switches the transmitter on
for 10 usecs, then off for 40 usecs (20% duty cycle). The power
supply supplies direct current of 0.2 amps (peak current x duty
factor). The caps supply the other 0.8 amps during the pulse, and
charge up at -0.2 amps when the transmitter is off (so that charge
The chart below was corrected
on March 15, thanks to Steve.
Voltage drop effect
The voltage the the charge storage
capacitor bank supplies is reduced by its equivalent series resistance.
For the net ESR of the bank you must compute the ESR of all capacitors
in parallel. The usual assumption is that they are all the same
(but they are not in reality.) But capacitors are usually specified
at a maximum ESR so this calculation will be slightly pessimistic
and will give your design some margin.
In the above example, if the
charge storage presents 1 ohm of ESR, it will contribute one volt
drop in the system; thus if you wanted to drain-pulse a GaAs amplifier
at 8 volts you would need a 9 volt supply. The next part of this
page will explain how the power dissipation due to ESR is calculated.
Power dissipation effect
Dissipated power across a resistor
is equal to the RMS value of the pulsed current waveform squared
times the resistance. For N capacitors in parallel, divide the single-cap
value of ESR by N.
Next we need to compute the RMS
current of the waveform.
The art of calculus without
In order to RMS current, you
need to time-average the square of the current, then take the square-root
of the result. Most often your microwave instructor will resort
to integral signs at this point. But for square current pulses,
you can almost do this in your head and don't need to resort to
calculus! We will follow Bruce Lee's example:
The art of fighting
without fighting from Enter the Dragon
During the current pulse, it
the power supply is entirely decoupled from the charge storage,
the capacitor source the peak current minus the average current
When the transmitter is off,
the caps recharge at current I2:
Now we merely time-average the
square of the current waveform to get RMS current. The equations
below shown the calculation step-by-step.
With the RMS current now calculated,
we can now calculate the dissipation due to ESR:
Last, let's look at the RMS current
as a percentage of peak current. For the transmitter fully on or
off (DF=100% or DF=0%), the capacitors supply zero current. The
maximum current occurs when DF=50% (a square wave). At 25% or 75%
DF, the ratio is 3/16.
If anyone wants the spreadsheet
that performed this calculation, just ask.
Now let's calculate power dissipation
in the example. The peak current at 1 amp, DF at 20%, and ESR at
1 ohm, the capacitor sources 0.32 amps RMS. Its power dissipation
is 102 mW, there should not be a problem in this case. If you asked
that single capacitor to source 10 amps peak, keep the fire extinguisher
handy, it would dissipate 10 watts!