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.I= Into + Io = Liitusion = Iphoto =o (157 — 1) .
(25.17)

VIPSolutions · Accepted Answer

Let's go through the calculations step by step using the photodiode equation:

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

where:  
- $ I_{\text{photo}} $ is the photocurrent (light-generated current)
- $ I_0 $ is the dark saturation current
- $ V $ is the voltage across the cell
- $ e $ is the elementary charge
- $ k_B $ is Boltzmann's constant ($ 8.617 \times 10^{-5} $ eV/K)
- $ T $ is the temperature in Kelvin (assume **T = 300 K**)

### 1. Find $ I_{\text{photo}} / I_0 $ at Open Circuit ($ V = V_{oc} $)

At open circuit, $ I = 0 $:

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

Now, plug in the numbers:

#### For $ V_{oc} = 0.7 $ V:

$$
\frac{e V_{oc}}{k_B T} = \frac{(0.7)}{8.617 \times 10^{-5} \times 300} \approx \frac{0.7}{0.02585} \approx 27.08
$$
$$
\frac{I_{\text{photo}}}{I_0} = e^{27.08} - 1 \approx 5.24 \times 10^{11}
$$

#### For $ V_{oc} = 0.5 $ V:

$$
\frac{e V_{oc}}{k_B T} = \frac{0.5}{0.02585} \approx 19.34
$$
$$
\frac{I_{\text{photo}}}{I_0} = e^{19.34} - 1 \approx 2.48 \times 10^{8}
$$

#### For $ V_{oc} = 0.86 $ V:

$$
\frac{e V_{oc}}{k_B T} = \frac{0.86}{0.02585} \approx 33.28
$$
$$
\frac{I_{\text{photo}}}{I_0} = e^{33.28} - 1 \approx 2.93 \times 10^{14}
$$

---

### 2. Power as a function of voltage

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

---

### 3. Fill Factor (FF)

The **fill factor** is defined as the ratio of the maximum power to the product of $ I_{\text{photo}} $ and $ V_{oc} $:

$$
\text{FF} = \frac{P_{max}}{I_{\text{photo}} V_{oc}}
$$

To find the maximum power, you need to maximize $ P(V) $ with respect to $ V $:

Set $ \frac{dP}{dV} = 0 $:

$$
\frac{dP}{dV} = \frac{d}{dV} \left( I_{\text{photo}} - I_0 (e^{eV/k_BT} - 1) \right) V
= -I_0 \frac{e}{k_B T} e^{eV/k_B T} V + I_{\text{photo}} - I_0 (e^{eV/k_B T} - 1) = 0
$$

Define $ x = \frac{eV}{k_B T} $, $ I_{\text{photo}}/I_0 = e^{x_{oc}} - 1 $.

At maximum power point:

$$
I = I_{\text{photo}} - I_0 (e^x - 1)
$$
$$
P(x) = I_0 [e^{x_{oc}} - 1 - (e^x - 1)] \cdot \frac{k_B T}{e} x
$$

The maximum power point occurs at $ x_{mp} $ where:

$$
\frac{dP}{dx} = 0
$$
$$
0 = -I_0 e^x \frac{k_BT}{e} x + I_0 [e^{x_{oc}} - e^x] \frac{k_B T}{e}
$$
$$
0 = -e^x x + (e^{x_{oc}} - e^x)
$$
$$
e^{x_{oc}} = e^x (x + 1)
$$
$$
e^{x_{oc} - x} = x + 1
$$
$$
x_{oc} - x = \ln(x + 1)
$$
$$
x = x_{oc} - \ln(x + 1)
$$

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

---

#### Calculate for Each $ V_{oc} $:

- $ x_{oc} = \frac{e V_{oc}}{k_B T} $

##### For $ V_{oc} = 0.7 $ V:

$ x_{oc} = 27.08 $

Solve:
$$
x = 27.08 - \ln(x + 1)
$$

Start with $ x \approx 27 $:

$$
\ln(28) \approx 3.33
$$
$$
x = 27.08 - 3.33 = 23.75
$$

Try again:
$$
\ln(24.75) \approx 3.21
$$
$$
x = 27.08 - 3.21 = 23.87
$$

Iteratively, $ x_{mp} \approx 23.8 $

Now, calculate the fill factor:

$$
\text{FF} = \frac{P_{max}}{I_{\text{photo}} V_{oc}}
$$

At maximum power point,
$$
I_{mp} = I_{\text{photo}} - I_0 (e^{x_{mp}} - 1)
$$
But $ I_{\text{photo}} = I_0 (e^{x_{oc}} - 1) $, so
$$
I_{mp} = I_0 (e^{x_{oc}} - 1 - (e^{x_{mp}} - 1)) = I_0 (e^{x_{oc}} - e^{x_{mp}})
$$

So,
$$
P_{max} = I_{mp} V_{mp} = [I_0 (e^{x_{oc}} - e^{x_{mp}})] \cdot \frac{k_B T}{e} x_{mp}
$$
$$
I_{\text{photo}} V_{oc} = I_0 (e^{x_{oc}} - 1) \cdot \frac{k_B T}{e} x_{oc}
$$

$$
\text{FF} = \frac{(e^{x_{oc}} - e^{x_{mp}}) x_{mp}}{(e^{x_{oc}} - 1) x_{oc}}
$$

Plug in values for $ x_{mp} \approx 23.8 $, $ x_{oc} = 27.08 $:

$$
e^{x_{oc}} \gg 1
$$
So,
$$
\text{FF} \approx \frac{e^{x_{oc}} - e^{x_{mp}}}{e^{x_{oc}}} \cdot \frac{x_{mp}}{x_{oc}} = \left[1 - e^{-(x_{oc} - x_{mp})} \right] \frac{x_{mp}}{x_{oc}}
$$

$$
x_{oc} - x_{mp} \approx 3.3 \implies e^{-3.3} \approx 0.037
$$
$$
\text{FF} \approx (1 - 0.037) \frac{23.8}{27.08} \approx 0.963 \times 0.879 = 0.847
$$

**So, for $ V_{oc} = 0.7 $ V:**
- $ I_{\text{photo}}/I_0 \approx 5.24 \times 10^{11} $
- **Fill Factor $ \approx 0.85 $**
- Maximum power as a fraction of $ I_{\text{photo}} V_{oc} $: **0.85**

---

##### For $ V_{oc} = 0.5 $ V:

$ x_{oc} = 19.34 $

Solve $ x = 19.34 - \ln(x + 1) $:

Trial with $ x = 19 $:

$$
\ln(20) \approx 3.0
$$
$$
x = 19.34 - 3.0 = 16.34
$$

Try $ x = 16.34 $:
$$
\ln(17.34) \approx 2.85
$$
$$
x = 19.34 - 2.85 = 16.49
$$

So, $ x_{mp} \approx 16.4 $

Now,
$$
x_{oc} - x_{mp} = 2.94 \implies e^{-2.94} \approx 0.052
$$
$$
\text{FF} \approx (1 - 0.052) \frac{16.4}{19.34} = 0.948 \times 0.848 = 0.804
$$

**For $ V_{oc} = 0.5 $ V:**
- $ I_{\text{photo}}/I_0 \approx 2.48 \times 10^8 $
- **Fill Factor $ \approx 0.80 $**

---

##### For $ V_{oc} = 0.86 $ V:

$ x_{oc} = 33.28 $

Solve $ x = 33.28 - \ln(x + 1) $:

Try $ x = 33 $:

$$
\ln(34) \approx 3.53
$$
$$
x = 33.28 - 3.53 = 29.75
$$

Try $ x = 29.75 $:
\[
\ln(30.75) \app

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