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What are the essential considerations of a BUCK converter with a battery(lead-acid) as a load? Here are some basic questions which i need some clarification:

1.Is it better to use a synchronous buck(free-wheeling diode used) or a non-synchronous buck(free-wheeling diode replaced by a FET)?

2.Is there a requirement of the output capacitor,as the battery itself would act as a large capacitor in this case.

3.Assuming that the step-ed down voltage is always greater than the battery voltage,How to calculate inductor value which would ensure continuous current flow from the BUCK to the battery,in order to charge it.

4.Would it be better to user a low side shunt or a high side shunt to monitor charging current.

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  1. You terminology is reversed. A synchronous buck with the second active switch has the potential of less loss because the switch can have lower voltage drop than a diode.

  2. Copied from your comment -- L = (Vin - Vout)/(Vin * frequency * r * Iout_max). r would be the inductor ripple current ratio. So (r * Iout_max) would be the inductor ripple current. I would say set r to the lower ratio unless the inductor gets too large. Notice that when Iout_actual (if you need to throttle the output current for some reason) drops below (0.5 * r * Iout_max), for a synchronous buck, the inductor current actually reverses; for a non-sync buck, the inductor current becomes discontinuous.

    You do want an output capacitor for at least these reasons -- to prevent reverse current to the battery in case of synchronous buck and throttling the current is part of the scheme; to reduce the inductive effect of the output wires and load; to reduce EMI -- because otherwise the still relatively large inductor ripple current would go straight out. The output capacitor helps to reduce the ripple current going out to a fraction of the inductor ripple current.

  3. If the battery is always floating then do which ever is easier, which is usually dictated by the regulator IC that you use. If the battery is connected, then you may have to use high side shunt to avoid a ground loop issue.


New edit to add info on output capacitor and output ripple current:

$$ I_{out\_ripple} = \frac{Z_c}{Z_{load} + Z_c} I_{ind\_ripple}$$

So the first order estimate is just a simple current divider relationship, where \$Z_{load}\$ would be the ESR of the battery in this case, and \$Z_c\$ the impedance of the output capacitor:

$$ Z_c = \frac{1}{2 \pi \times frequency \times C_{out}} $$

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  • \$\begingroup\$ I plan on using a diode first,then change it to a switch later.I would like to keep the inductor operating in continuous mode for as long as possible. Is there any mathematical formula which is to be used for calculating the output capacitor? \$\endgroup\$ – SuperHuman Dec 5 '15 at 5:21
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  1. synchronous buck is better (the one with transistor, not with a freewheeling diode) becsuse more effective
  2. Yes capacitor is needed , low ESR because the battery isn't a capacitor even if it acts like a big one.
  3. You would need a charging procedure, with tracking voltage and current. Inductor is calculated by knowing duty cycle ratio and current.
  4. I would preffer a hall sensor instead, but you didn't specified what battery and size.
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  • \$\begingroup\$ could you please expand on the 3rd point,about inductor calculation.I had referred some application notes on TI website,i got the following formula to calculate the inductor value:L = (Vin - Vout)/(Vin * frequency * r *Iout).Where r is a constant between 0.2 to 0.5 \$\endgroup\$ – SuperHuman Dec 4 '15 at 10:58
  • \$\begingroup\$ @SuperHuman That's the hardest point. Wait for someone with better experience. \$\endgroup\$ – Marko Buršič Dec 4 '15 at 11:00

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