Use the photodiode equation (25.17) to compute Iphoto /I0 for a silicon solar cell with Voc = 0.7 V. Write the power as a function of voltage and compute the fill factor, the maximum power I V attaina

Question

Use the photodiode equation (25.17) to compute Iphoto /I0 for a silicon solar cell with Voc = 0.7 V. Write the power as a function of voltage and compute the fill factor, the maximum power I V attainable as a fraction of IphotoVoc. Repeat the calculation for cells with Voc = 0.5 and 0.86 V and compute the maximum efficiency of a single-junction silicon photocell in these cases.
give answer in 3step with explanation at the end of each step and final answer at the endI= Into + Io = Liitusion = Iphoto =o (157 — 1) .
(25.17)

VIPSolutions · Accepted Answer

Let's solve the problem step by step, using the photodiode equation provided:

$$
I = I_\text{photo} - I_0 \left( e^{eV/k_BT} - 1 \right)
$$

where:
- $ I $ = output current
- $ I_\text{photo} $ = photocurrent (generated by sunlight)
- $ I_0 $ = dark saturation current
- $ V $ = output voltage
- $ k_B $ = Boltzmann constant ($ 8.617 \times 10^{-5} $ eV/K)
- $ T $ = temperature (assume 300 K for room temperature)
- $ e $ = elementary charge ($ 1.602 \times 10^{-19} $ C)

We are given: open-circuit voltage $ V_{oc} $ (try for 0.7 V, 0.5 V, and 0.86 V).

---

## **Step 1: Find $ I_\text{photo}/I_0 $ for each $ V_{oc} $**
At open circuit, $ I = 0 $:

$$
0 = I_\text{photo} - I_0 \left( e^{eV_{oc}/k_BT} - 1 \right)
$$
$$
I_\text{photo} = I_0 \left( e^{eV_{oc}/k_BT} - 1 \right)
$$
$$
\frac{I_\text{photo}}{I_0} = e^{eV_{oc}/k_BT} - 1
$$

Let's calculate for each $ V_{oc} $:

Let $ k_BT/e \approx 0.02585 $ V at 300 K.

### a) $ V_{oc} = 0.7 $ V
$$
\frac{I_\text{photo}}{I_0} = e^{0.7/0.02585} - 1 \approx e^{27.08} - 1 \approx 5.32 \times 10^{11}
$$

### b) $ V_{oc} = 0.5 $ V
$$
\frac{I_\text{photo}}{I_0} = e^{0.5/0.02585} - 1 \approx e^{19.33} - 1 \approx 2.48 \times 10^{8}
$$

### c) $ V_{oc} = 0.86 $ V
$$
\frac{I_\text{photo}}{I_0} = e^{0.86/0.02585} - 1 \approx e^{33.28} - 1 \approx 3.03 \times 10^{14}
$$

**Explanation:** This ratio is huge because the photocurrent is much larger than the dark current in a well-designed solar cell.

---

## **Step 2: Write Power as a Function of Voltage and Find Maximum Power Point (Fill Factor)**

The output power is:
$$
P = I V = \left[ I_\text{photo} - I_0 \left( e^{eV/k_BT} - 1 \right) \right] V
$$

To maximize $ P(V) $, set $ dP/dV = 0 $:
$$
\frac{dP}{dV} = 0 = I + V \frac{dI}{dV}
$$

Let's define $ x = \frac{eV}{k_BT} $, so $ V = x \frac{k_BT}{e} $.

$$
I = I_\text{photo} - I_0 ( e^x - 1 )
$$
$$
P = I V = [ I_\text{photo} - I_0 (e^x - 1) ] \cdot x \frac{k_BT}{e}
$$

Set derivative to zero for maximum power:

$$
\frac{dP}{dx} = \frac{k_BT}{e} \left[ I_\text{photo} - I_0 (e^x - 1) - I_0 x e^x \right] = 0
$$
$$
I_\text{photo} - I_0 (e^x - 1 + x e^x) = 0
$$
$$
\frac{I_\text{photo}}{I_0} = e^x (1 + x) - 1
$$

Recall from **Step 1** that at $ V_{oc} $, $ \frac{I_\text{photo}}{I_0} = e^{x_{oc}} - 1 $, where $ x_{oc} = \frac{e V_{oc}}{k_B T} $.

So, **to find the maximum power point $ x_{mp} $**, solve:
$$
e^{x_{oc}} - 1 = e^{x_{mp}} (1 + x_{mp}) - 1
$$
$$
e^{x_{oc}} = e^{x_{mp}} (1 + x_{mp})
$$
$$
e^{x_{oc} - x_{mp}} = 1 + x_{mp}
$$
$$
x_{oc} - x_{mp} = \ln(1 + x_{mp})
$$

This equation must be solved numerically for $ x_{mp} $.

### For $ V_{oc} = 0.7 $ V ($ x_{oc} = 27.08 $)
$$
27.08 - x_{mp} = \ln(1 + x_{mp})
$$
Guess $ x_{mp} \approx x_{oc} - \ln x_{oc} $ for large $ x_{oc} $:

- $ \ln(27.08) \approx 3.3 $
- $ x_{mp} \approx 27.08 - 3.3 = 23.78 $
- Check: $ \ln(1 + 23.78) \approx 3.2 $, so $ 27.08 - 23.78 = 3.3 $, matches.

#### Fill Factor (FF):
$$
V_{mp} = x_{mp} \frac{k_BT}{e}
$$
$$
I_{mp} = I_\text{photo} - I_0 (e^{x_{mp}} - 1)
$$
$$
I_{mp} = I_0 [ e^{x_{oc}} - 1 - ( e^{x_{mp}} - 1 ) ] = I_0 [ e^{x_{oc}} - e^{x_{mp}} ]
$$

$$
FF = \frac{V_{mp} I_{mp}}{I_\text{photo} V_{oc}}
$$
But $ I_\text{photo} V_{oc} = I_0 ( e^{x_{oc}} - 1 ) V_{oc} $.

$$
FF = \frac{ x_{mp} \frac{k_B T}{e} \cdot I_0 ( e^{x_{oc}} - e^{x_{mp}} ) }{ I_0 ( e^{x_{oc}} - 1 ) x_{oc} \frac{k_B T}{e} }
= \frac{ x_{mp} ( e^{x_{oc}} - e^{x_{mp}} ) }{ x_{oc} ( e^{x_{oc}} - 1 ) }
$$

Since $ e^{x_{oc}} \gg 1 $, $ e^{x_{oc}} - 1 \approx e^{x_{oc}} $:

$$
FF \approx \frac{ x_{mp} ( 1 - e^{x_{mp}-x_{oc}} ) }{ x_{oc} }
$$
But $ e^{x_{mp}-x_{oc}} = 1/(1 + x_{mp}) $ from earlier.

So,

$$
FF \approx \frac{ x_{mp} ( 1 - \frac{1}{1 + x_{mp}} ) }{ x_{oc} }
= \frac{ x_{mp} \frac{x_{mp}}{1 + x_{mp}} }{ x_{oc} }
= \frac{ x_{mp}^2 }{ x_{oc} ( 1 + x_{mp} ) }
$$

Plug in numbers.

- $ x_{oc} = 27.08 $
- $ x_{mp} = 23.78 $
- $ 1 + x_{mp} = 24.78 $

$$
FF = \frac{ (23.78)^2 }{ 27.08 \times 24.78 } \approx \frac{565.5}{671.4} \approx 0.842
$$

---

## **Step 3: Repeat for $ V_{oc} = 0.5 $ V and $ V_{oc} = 0.86 $ V**

### **a) $ V_{oc} = 0.5 $ V**
- $ x_{oc} = 0.5 / 0.02585 = 19.33 $
- $ x_{mp} \approx 19.33 - \ln(19.33) \approx 19.33 - 2.96 = 16.37 $
- $ 1 + x_{mp} = 17.37 $

$$
FF = \frac{ (16.37)^2 }{ 19.33 \times 17.37 } \approx \frac{268.1}{336.0} \approx 0.798
$$

---

### **b) $ V_{oc} = 0.86 $ V**
- $ x_{oc} = 0.86 / 0.02585 = 33.28 $
- $ x_{mp} \approx 33.28 - \ln(33.28) \approx 33.28 - 3.51 = 29.77 $
- $ 1 + x_{mp} = 30.77 $

$$
FF = \frac{ (29.77)^2 }{ 33.28 \times 30.77 } \approx \frac{886.3}{1024.3} \approx 0.865
$$

---

### **Maximum Efficiency**
The maximum *fraction* of $ I_\text{photo} V_{oc} $ attainable is the fill factor ($ FF $), so:

- For $ V_{oc} = 0.7 $ V: $ FF \approx 0.84 $
- For $ V_{oc} = 0.5 $ V: $ FF \approx 0.80 $
- For $ V_{oc} = 0.86 $ V: $ FF \approx 0.87 $

---

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