batteriescapacitorpower supplyvoltage-regulator
I have a long-term application with very low quiescent current and infrequent pulses. I've spec'ed high capacity, low pulse current batteries that will give me the lifetime I need, and I want to charge a capacitor to handle the infrequent high current (regulated) loads.
Can I put the cap directly in parallel with my batteries? Will the voltage drop from the current pulse have a negative effect on the battery? Or would I have to regulate before as well as after the cap?
The application note is clear on ceramic vs. electrolytic capacitors:
"Factor 5" (bootstrap capacitor leakage current) "... is only relevant if the bootstrap capacitor is an electrolytic capacitor, and can be ignored if other types of capacitor are used."
The application note also refers to DT98-2a (which itself refers to DT04-04) which despite focusing on IGBTs has this to say:
"To size the bootstrap capacitor, the first step is to establish the minimum voltage drop (\$ \Delta V_{BS}\$) that we have to guarantee when the high side IGBT is on.
If \$V_{GE(min)}\$ is the minimum gate emitter voltage to maintain, the voltage drop must be:
\$ \Delta V_{BS} ≤ V_{CC} −V_F −V_{GE(min)} −V_{CE(on)} \$
under the condition:
\$V_{GE(min)} > (V_{BS(UV)}- \$)
where \$V_{CC}\$ is the IC voltage supply, \$V_F\$ is bootstrap diode forward voltage, \$V_{CE(on)}\$ is emitter-collector voltage of low side IGBT and (\$V_{BS(UV)}-\$) is the high-side supply undervoltage negative going threshold."
With some interpretation, we can see that \$V_{GE(min)} => V_{GS(min)}\$ and \$V_{CE(min)} => V_{DS(min)}\$.
The calculation is straightforward. The capacitor size is simply a question of how much voltage drop you can tolerate over the duration of the pulse. The average current from the battery is a function of the duty cycle.
ΔV = I × Δt / C
Solving for C gives:
C = I × Δt / ΔV
Let's assume you can allow ΔV = 0.1V. For your first example, this works out to:
C = 25 mA × 25 ms / 0.1 V = 6.25 mF
The average current draw is 25 mA * 25 ms / 2.5 s = 0.25 mA.
For the second example, the numbers work out to:
C = 50 mA × 100 ms / 0.1 V = 50 mF
Average current = 50 mA * 100 ms / 1.0 s = 5 mA.
Best Answer
Even "directly in parallel with the batteries" isn't really directly in parallel with the batteries, thanks to wiring resistances.
The capacitor should have the closest and most direct connection to the load, then this pair should be connected to the battery via wiring which gives you some control of the current drawn from the battery.
Find the maximum recommended current (Imax) from the battery, probably from its datasheet.
Calculate the voltage drop (dV = dQ/C) across the capacitor under load, where dQ = (Iload-Imax) * pulse width - or more conservatively, Iload * pulse width.
Calculate the resistance R = dV/Imax you need in the connection to the battery. This resistance limits the current from the battery to Imax when the capacitor voltage dips by dV.
And ensure the battery wiring (including any fuse) has at least that resistance.
If dV and thus R are too high to make this all work, increase capacitance C and try again.