Thursday, 29 November 2018

High efficiency LED drive + energy feedback; n > 100%

the switching circuit shown below in Fig 1. operates at approx 130 kHz, in 2 half-cycles a) & b):-

Fig. 1

a) the transistor switches on, its collector current illuminates the output LED(s) and also stores energy in the coil field and electrolytic cap; when the coil saturates then the transistor turns off;  enough energy remains in the electrolytic cap to supply the output LED(s) until the next cycle (the cap voltage is DC with minimal ripple)

b) current supplied from the energy stored in the collapsing coil-field and the electrolytic cap is fed back to the battery (while the transistor is off) via the feedback LED - IFF the feedback LED is illuminated then the battery is receiving charge in this half-cycle (in this case, equivalent to approximately one-fifth of the output current)

(the transformer here is approx 1cm^3 volume of ferrite, solenoidal wound, almost completely enclosed by the ferrite;  the wire is 2x45x0.45cm diam. insulated copper, 5:5 turns)

for these tests, the battery is an LIR providing 4.15V;  supply current is 5.4mA

pk voltage of feedback pulse across feedback LED + battery is approx 7V;  approx 20% duty cycle

Fig 2. Voltage pulse across Feedback LED

so-called 'conventional' current flows OUT of the positive terminal of the battery in half-cycle a) and INTO that terminal in half-cycle b)

voltage across 1ohm Current Sensing Resistor in +ve supply line shows approx same peak magnitude current flows, into and out of the battery, matching the duty cycle for the respective average current flows

Fig 3. Voltage across 1ohm CSR in +ve supply line

(NB.  all values measured using a True RMS meter and confirmed in order of magnitude using a 'scope)

feedback current:  1.18mA (True RMS)
feedback LED load:  approx 2.6V * 1.18mA = 3mW approx

main LED branch current:  6.5mA (True RMS)
main LED branch load:  4.15V * 6.5mA = 26.98mW

voltage across electrolytic//main LEDs is 2.7V DC
main LED load:  2.7V * 6.5mA = 17.55 mW

switching cct load:  26.98 - 17.55 = 9.43mW

total LED load:  17.55 + 3 = 20.55mW

total power load:  20.55 + 9.43 = 29.98mW

supply current:  5.4mA (True RMS)
total power supply:  4.15 * 5.4mA = 22.41mW 

Efficiency  n  =  (total load / supply)  =  (29.98 / 22.41) = 134%

results shown are instanteous power, mW (these are proportional to the energy being converted, mWh)