Electronic – Is boost converter enough to provide pulse currents without drawing too much current from a coin cell

boostcoin-celldecoupling-capacitorhigh-currentvoltage-regulator

I am making a RF circuit which could operate using a 3V coin cell battery like CR2032 or CR2450. In sleep mode it consumes around 10-20uA but during Tx/Rx which lasts for 10ms it consumes ~300mA-ms. My circuit works in 2.8-3.6V range. To supply this surge current I have tried using a 1000uF capacitor and it has performed well except when handling the startup inrush current.

Now to increase battery usage I am planning to use a boost converter with low Iq like : http://www.ti.com/lit/gpn/tps61322. I couldn't use 2 coin cells (6V) and an LDO as I want to keep the form factor smaller and battery replacement easier.

My doubt is if I am using a boost converter, would I still require a capacitor parallel to the battery so that battery is not subjected to high current surges or a boost converter can inherently take care of such surges without drawing too much current (~10-15mA) from the battery.

More details:
Attached is the graph showing current vs time during the Tx/Rx operation. In total it lasts for approx 10ms and consumes about 300mA-ms. Once this Tx/Rx is done the circuit goes to sleep and consumes 10-20uA Current Profile for Tx/Rx

The overall current profile of the RF circuit looks like:

Complete current profile including sleep and Tx/Rx

Best Answer

First some rules of physics.

  • All loads have an equivalent impedance for each state.
  • All batteries like electrolytic capacitors have an Effective Series Resistance (ESR) and Capacitance , C [F] where the change in capacitance can be measured in Ah or mAh
  • the product of ESR *C is a function of cell chemistry and bulk size and may be used to compare.
  • any DC supply voltage in a somewhat linear mode will drop voltage according to the resistive divider or Load Regulation formula.
    • Vdrop = I * ESR = Vo-Vi = ESR/(ESR+Load).
    • where desired is ESR < 1% R load for performance reasons.
    • ESR rises rapidly below 10% state of charge (SoC) and rises slowly with aging effects ( heat, under/over charge , cycles, ambient temp)
  • for any given chemistry (primary or secondary) the ESR is inverse to Ah capacity ( for a single cell)
  • ESR multiplies for cells in series *(S) and divides for cells in parallel (P) and all must be matched.
  • Hybrids may be S x P arranged for suitable voltage and current and thus source impedance.
  • All electrolytic caps and batteries have memory such that after a short, they return to a resting voltage. - This is a secondary ESR2*C2 equivalent circuit. There may be more.
  • Compute the energy loss in a large cap and see if that is worth buffering the ESR.
    • E=1/2 CV^2=VI*t

Now go choose a battery that has an acceptable Vdrop and Capacity at desired load current..

Converters do not add energy, mainly change impedance ratio. BE conservative.

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