# Question 9.19: Impact of a Blocking Diode to Control Nighttime Battery Leak......

Impact of a Blocking Diode to Control Nighttime Battery Leakage. A PV module is made up of 36 cells, each having a reverse saturation current $I_{0}$ of $1 \ \times \ 10^{−10}$ A and a parallel resistance of 8 Ω. The PVs provide the equivalent of 5 A for 6 h each day. The module is connected without a blocking diode to a battery with voltage 12.5 V.

a. How many Ah will be discharged from the battery over a 15-h night?
b. How much energy will be lost due to this discharge?
c. If a blocking diode is added, how much energy will be dissipated through the diode during the daytime. Assume the diode while conducting has a voltage drop of 0.6 V.

Step-by-Step
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The voltage across each PV cell will be about 12.5 V/36 cells = 0.347 V. From (9.34) the current discharged from the battery while the PV is in the dark will be

$I_{B} \ = \ I_{d} \ + \ I_{R_{P}} \ = \ I_{0} \left(e^{38.9 V_{d}} \ – \ 1\right) \ + \ \frac{V_{d}}{R_{p}}$ (9.34)

$I_{B} \ = \ 10^{-10} \left(e^{38.9\times 0.347} \ – \ 1\right) \ + \ \frac{0.347}{8} \ = \ 0.000073 \ + \ 0.043 \ = \ 0.043 \ A \ = \ 43 \ mA$

a. Over a 15-h nighttime period, the loss in Ah from the battery will be

$\text{Nighttime loss} \ = \ 0.043 \ A \ \times \ 15 \ h \ = \ 0.65 \ Ah$

b. At a nominal 12.5 V, the energy loss at night will be

$\text{Nighttime loss} \ = \ 0.65 \ Ah \ \times \ 12.5 \ V \ = \ 8.1 \ Wh$

c. During the day the PVs will deliver

$\text{PV output} \ = \ 6 \ h \ \times \ 5 \ A \ = \ 30 \ Ah$

The nighttime loss without a blocking diode is 0.65 Ah/30 Ah = 0.02; that is, 2% of the daytime gains.
With the blocking diode dropping 0.6 V, the daytime loss caused by that diode is

$\text{Blocking diode loss} \ = \ 30 \ Ah \ \times \ 0.6 \ V \ = \ 18 \ Wh$

Question: 9.22

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