Design simple UPS circuit

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Merilyn Peters
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1 Sheet No of 6 Date: 23/4/28 Design simple UPS circuit Input Parameters Power supply and Pi power demand Power supply voltage V in 2V pi typical, minimum and maximum operating current UBEC details I pi.typ UBEC efficiency η u 85 % UBEC output voltage V u 5.2V Battery details battery pack number cells n c 8.3A cell nominal voltage V c.nom.2v cell nominal capacity C c 2mAhr cell charge efficiency η c 85 % target cell state of charge 7 % I pi.min.2a I pi.max 2.5A graph values for cell characteristic under discharge and charge conditions note, this is slow charge (typical.c rate or less), and fast discharge (typical C) col is state-of charge col is fast discharge voltage col 2 is slow charge voltage V c.t function V c () gives voltage at specified state-of-charge for discharge and charge respectively V c ( ) V. C discharge.c charge /4/28 2:3

2 Sheet No 2 of 6 Date: 23/4/28 Charging circuit details charging circuit comprises diode D and resistor R in series with battery charging circuit resistor R R 22Ω charging circuit diode characteristic V D.t col is milliamps col is volts V dropped in diode current (ma) (note fudged behaviour at low current <ma) Power delivery (discharging) circuit details discharging circuit comprises D2 and MOSFET in series with battery MOSFET on state equivalent R R mosfet.2ω power diode characteristic V D2.t col is milliamps col is volts V dropped in diode current (ma) (note fudged behaviour at low current <ma, but this is not used in calc unless very low Pi currents are specified) /4/28 2:3

3 Sheet No 3 of 6 Date: 23/4/28 Calculation of Charging Behaviour for a given state-of-charge, can iteratively solve for charging current since voltage drop across diode + voltage drop across resistor + battery voltage = input power supply voltage 7. % Given V D () I R R I n c V c ( ) = V in I Find() I I mA since this is iterative, cannot directly formulate function, but can do it by generating a lookup table with a relatively modest number of points and interpolating: ma charge I c ( ) 42.5mA 8 6 7% charge current is proportional to the rate of charging (i.e., rate of change of with unit time), the reciprocal of which is the time to charge fully, if charge current was constant: Δ s.c () s I c ()η s c Δ s.c ( ) C c however, since charge current and therefore rate of charge is variable with, need to integrate the reciprocal to determine the time to reach a given charge state 55.4hr t c ( ) ds Δ s.c () s t c ( ) 32.7hr hence time to charge fully (from empty): t c ( % ) 58.3hr hrs to reach from empty 6 t c ( ) 32.7hr /4/28 2:3

4 Sheet No 4 of 6 Date: 23/4/28 Calculation of Discharge Behaviour for a given state-of-charge and current drawn by Pi, can use iterative calculation to find current drawn from battery, since power out = η x power in to UBEC Given n c V c ( ) V D2 () I R mosfet I I η V u = I pi.typ u I Find() I I.873 A again, use interpolation between a number of solved points to define a function for discharge current ma discharge from battery 2 I d ( ) 873.4mA I d.max ( ) 723.7mA I d.min ( ) 33.mA current is proportional to the rate of discharge (i.e., rate of change of with unit time), the reciprocal of which is the time to discharge fully, if charge current was constant: Δ s.d () s I d () s C c Δ s.d ( ) 2.3hr however, since current is variable with, need to integrate the reciprocal to determine the time to reach full discharge from a given state of charge t d ( ) ds Δ s.d () s t d ( ) 42.5min hence time to discharge fully (from full): t d ( % ) 34.2min hrs to reach from full t d ( ) 42.5min t d.max ( ) 2.6min t d.min ( ) 278.8min /4/28 2:3

5 Sheet No 5 of 6 Date: 23/4/28 now can turn charging an discharging graphs to give state of charge against time charging since empty and time discharging since full: hours of charging / discharge state of charge after 24 hours charging (from empty) S c ( 24hr).536 S c ( 24hr) C c 7mAhr detail (discharging duration, mins) /4/28 2:3

6 Sheet No 6 of 6 Date: 23/4/28 Voltage into UBEC the voltage into the UBEC is a function of the current being drawn and (while running on batteries) also the state-of-charge of the batteries. This is included in calculations above, but is summarised here assuming the batteries are at a specified nominal state of charge 7. % cell voltage V c ( ).2 V as before, for a given state-of-charge and current drawn by Pi, can use iterative calculation to find current drawn from battery, since power out = η x power in to UBEC. However, will now maintain constant and vary pi current within specified limits current drawn from battery I d. I pi.typ 873.4mA I d. I pi.min 3.6mA 723.7mA I d. I pi.max.5.5 I pi.typ 2 having determined the current drawn, the voltage presented to the UBEC is equal to the battery voltage less the voltage drops in the relevant diode and the MOSFET range of voltage to UBEC: V d. I pi.typ 9. V V d. I pi.min 9.4 V 8.9 V V d. I pi.max I pi.typ /4/28 2:3

resistance in the circuit. When voltage and current values are known, apply Ohm s law to determine circuit resistance. R = E/I ( )

resistance in the circuit. When voltage and current values are known, apply Ohm s law to determine circuit resistance. R = E/I ( ) DC Fundamentals Ohm s Law Exercise 1: Ohm s Law Circuit Resistance EXERCISE OBJECTIVE When you have completed this exercise, you will be able to determine resistance by using Ohm s law. You will verify