Worked examples — Power systems — solar arrays (I-V curve, power tracking), batteries (DoD, cycles), RTG

3.6.22 · D3 · Physics › Spacecraft Structures & Systems Engineering › Power systems — solar arrays (I-V curve, power tracking), ba

Yeh page Power Systems (3.6.22) ka drill hall hai. Parent note ne tools banaye the: I–V curve, maximum-power point, depth-of-discharge, cycle life. Yahan hum un tools par har tarah ka input daalnge — normal cases, zero cases, extreme cases, aur ek nasty exam twist — taaki jab koi real mission problem aapke saamne aaye, toh aapne uski shape pehle se dekhi ho.

Shuru karne se pehle, ek promise: neeche use kiya gaya har symbol parent mein define kiya gaya tha. Agar aapne abhi tak $I_L$, $I_0$, $V_T$, $V_{oc}$, $I_{sc}$, DoD, ya $N_{\text{cycles}}$ nahi padha, toh pehle parent padho — yeh page unhe assume karta hai aur unhe drill karta hai.

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Scenario matrix

Ek power-system problem ko ek machine ki tarah socho jisme knobs hain. Har knob ko normal value, zero/degenerate value, ya extreme/limiting value par set kiya ja sakta hai. Neeche ki table har knob-setting class list karti hai jo yeh topic aap par throw kar sakta hai, aur kaun sa worked example use cover karta hai.

#	Case class	Usmein kya khaas hai	Covered by
C1	Normal operating point	Cell mid-curve kaam kar raha hai, $P$, $V_{mp}$ nikalo	Ex 1
C2	Zero input: $V=0$	Short circuit — sirf current	Ex 2
C3	Zero input: $I=0$	Open circuit — sirf voltage	Ex 2
C4	Limiting: full darkness ($I_L\to 0$)	Eclipse — cell ek plain diode ban jaata hai	Ex 3
C5	Limiting: high temperature	$V_{oc}$ girta hai, MPP shift hota hai	Ex 4
C6	Off-normal sun angle ($\cos \beta$)	Geometry intensity ko reduce karti hai	Ex 5 (figure)
C7	Series/parallel scaling	Kai cells → array voltage & current	Ex 6
C8	Battery sizing (real word problem)	Eclipse energy vs. capacity vs. DoD	Ex 7
C9	Cycle-life / mission-lifetime	Non-linear DoD penalty	Ex 8
C10	Exam twist: which converter?	Battery voltage ≠ $V_{mp}$	Ex 9

Nau examples sab das cells cover karte hain (Ex 2 dono C2 aur C3 cover karta hai — woh ek hi curve ke do ends hain).

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Example 1 — Normal operating point (C1)

Forecast: Compute karne se pehle guess karo — kya current $I_L=0.50\text{ A}$ ke kareeb hogi, ya almost zero? (Hint: kya $2.00\text{ V}$ open-circuit voltage ke kareeb hai?)

1. Thermal-voltage denominator compute karo $n{V}_T=1.5\times 0.026=0.039\text{ V}$. Yeh step kyun? Diode ka exponential $e^{V/(n{V}_T)}$ hai; is problem mein har voltage is $0.039\text{ V}$ ki units mein measure hoti hai. Measure karne se pehle ruler chahiye.

2. Diode exponent evaluate karo $\frac{V}{n{V}_T}=\frac{2.00}{0.039}=51.28$. Yeh step kyun? Yeh dimensionless number batata hai ki hum "diode turns on" mein kitne deep hain. $50$ ke kareeb value ka matlab hai exponential enormous hai.

3. Diode current $I_0(e^{51.28}-1)=10^{-12}\times (1.86\times 10^{22})\approx 1.86\times 10^{10}\text{ A}$?? Yeh step kyun? Yeh $I_L$ se absurdly zyada bada hai. Yeh woh check hai jo aapko pakadta hai: $V=2.00\text{ V}$ par diode ne pehle se hi sab kuch swallow kar liya hai — matlab $2.00\text{ V}$ open-circuit voltage se aage hai. Physical current hugely negative hogi, jo ideal equation mathematically allow karta hai lekin ek real illuminated cell kabhi deliver nahi karta.

4. Dekho cell actually kahan rukti hai: open circuit woh hai jahan $I=0$: $V_{oc}=n{V}_T\ln \left(\frac{I_L}{I_0}+1\right)=0.039\ln \left(\frac{0.50}{10^{-12}}\right).$ $\ln (5\times 10^{11})=26.94$, toh $V_{oc}=0.039\times 26.94=1.05\text{ V}$. Yeh step kyun? Cell load ke neeche $V_{oc}$ se kabhi aage nahi ja sakta. Hamara chosen $2.00\text{ V}$ usable curve se bahar hai.

5. Ek valid operating voltage par dobara karo $V=1.00\text{ V}$ ($V_{oc}$ se neeche): $I=I_L-I_0(e^{1.00/0.039}-1)=0.50-10^{-12}(e^{25.64}-1).$ Yahan $e^{25.64}=1.37\times 10^{11}$ hai, jo us $1$ se roughly $10^{11}$ times bada hai jise hum subtract karte hain, toh $e^{25.64}-1\approx e^{25.64}$ eleven significant figures tak. Yeh step kyun? $-1$ drop karna sloppiness nahi hai — yeh ek controlled approximation hai. Shockley "$-1$" sirf low voltage (reverse bias) par matter karta hai; jab exponent roughly $10$ se zyada ho jaata hai, $e^x$ $1$ ko dwarf kar deta hai aur $-1$ hamare precision par invisible hai. Toh diode current $=10^{-12}\times 1.37\times 10^{11}=0.137\text{ A}$, jisse $I=0.50-0.137=0.363\text{ A}$.

6. Power $P=V\cdot I=1.00\times 0.363=0.363\text{ W}$.

Verify: $V=V_{oc}=1.05\text{ V}$ par current $\approx 0$ honi chahiye. Plug in karo: $0.50-10^{-12}{e}^{1.05/0.039}=0.50-10^{-12}{e}^{26.9}\approx 0.50-0.50=0$. ✓ Units: $\text{V}\times \text{A}=\text{W}$. ✓

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Example 2 — Do zero-input ends: short & open circuit (C2, C3)

Forecast: Dono endpoints zero power deliver karte hain — kya aap padhne se pehle samajh sakte ho kyun?

1. Short circuit ($V=0$): exponent $=0$, toh $e^0-1=0$. Phir $I=I_L-0=0.50\text{ A}=I_{sc}$. Yeh step kyun? Terminals shorted hone se voltage nahi hai jo diode ko drive kare, toh diode "chhupta hai" aur saari photocurrent bahar flow karti hai. Yeh I–V curve ka top-left corner hai.

2. Short circuit par power: $P=V\cdot I=0\times 0.50=0\text{ W}$. Yeh step kyun? Maximum current lekin zero volts → zero watts. Short koi useful energy deliver nahi karta.

3. Open circuit ($I=0$): solve karo $0=I_L-I_0(e^{V/n{V}_T}-1)$, jisse $V_{oc}=n{V}_T\ln (I_L/I_0+1)=1.05\text{ V}$ (Ex 1 se). Yeh step kyun? No load matlab kuch current draw nahi karta; photocurrent internally diode ke through recirculate karta hai jab tak voltage self-limit na ho jaaye.

4. Open circuit par power: $P=1.05\times 0=0\text{ W}$.

Verify: $I_{sc}=0.50\text{ A}$ ✓ ($I_L$ ke barabar). $V_{oc}=1.05\text{ V}$ ✓ (Ex 1 se match karta hai). Dono powers $=0\text{ W}$ ✓.

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Example 3 — Limiting case: full eclipse, cell andheron mein (C4)

Forecast: Bina light ke, kya yeh abhi bhi solar cell hai — ya ek ordinary diode ban gaya hai?

1. Cell equation mein $I_L=0$ set karo: $I=-I_0(e^{V/n{V}_T}-1)$. Yeh step kyun? Parent ka single-diode model $I=I_L-I_0(e^{V/n{V}_T}-1)$ hai. Light term hatane se bare Shockley diode bachta hai. Ek dark solar cell ek diode hai.

2. $V=0.5\text{ V}$ par evaluate karo: exponent $0.5/0.039=12.82$, $e^{12.82}=3.7\times 10^5$. $I=-10^{-12}(3.7\times 10^5)=-3.7\times 10^{-7}\text{ A}$. Yeh step kyun? Negative sign keh raha hai current cell mein andar flow kar raha hai — ab woh power absorb kar raha hai.

3. Power $P=V\cdot I=0.5\times (-3.7\times 10^{-7})=-1.85\times 10^{-7}\text{ W}$. Yeh step kyun? Negative power = shaded cell ek chhota load hai, doosre cells se heated. Isi liye real arrays mein bypass diodes matter karte hain.

Verify: $I<0$ (consuming) ✓. Sign: forward-biased diode bina light ke current draw karna chahiye, Shockley curve se match karta hai. Units $\text{V}\cdot \text{A}=\text{W}$ ✓.

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Example 4 — Limiting case: hot cell, MPP shifts (C5)

Forecast: Garam solar panels — zyada power ya kam? (Zyada log guess karte hain "zyada sun-heat = zyada power." Galat.)

1. Temperature rise $\Delta T=75-25=50{}^{\circ }\text{C}$. Yeh step kyun? Coefficient temperature mein change se multiply karta hai, absolute temperature se nahi.

2. Voltage drop $\Delta V_{oc}=-2.2\text{ mV}{/}^{\circ }\text{C}\times 50{}^{\circ }\text{C}=-110\text{ mV}=-0.110\text{ V}$. Yeh step kyun? Zyada $T$ se $I_0$ (dark leakage) exponentially badhta hai, aur kyunki $V_{oc}\propto \ln (I_L/I_0)$ hai, bada $I_0$ log ko ghataata hai. Heat voltage ka dushman hai.

3. Naya open-circuit voltage $V_{oc}^{\prime }=1.05-0.110=0.94\text{ V}$.

4. Fractional power change (current ~constant): $\frac{\Delta P}{P}\approx \frac{\Delta V_{oc}}{V_{oc}}=\frac{-0.110}{1.05}=-0.105=-10.5\%$. Yeh step kyun? Agar $I$ barely change hoti hai, toh power voltage ke saath track karti hai, toh $10.5\%$ voltage loss roughly $10.5\%$ power loss hai.

Verify: $V_{oc}^{\prime }=0.94\text{ V}$ ✓, $\Delta P/P=-10.5\%$ ✓. Sign negative (power girta hai) ✓.

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Example 5 — Off-normal sun angle (C6) — geometric

Figure: Ek blue solar panel draw kiya gaya hai tilt karke taaki uska green "panel normal" arrow red "to Sun" arrow se $\beta =60^{\circ }$ door point kare. Parallel orange sunlight rays horizontally aate hain; gray arc Sun direction aur panel normal ke beech angle $\beta$ mark karta hai. Picture dikhata hai ki beam ka sirf woh component jo normal ke along ho cells par land karta hai — geometric reason hai ki effective intensity $S_0\cos \beta$ hai, full $S_0$ nahi.

Forecast: $60^{\circ }$ off-normal par, kya hum sunlight ka "$60\%$" khote hain, ya kuch aur?

1. Projected intensity $S=S_0\cos \beta$. Yeh step kyun? Figure mein red arrow dekho: beam ka sirf component panel ke normal ke along energy deliver karta hai. Ek tilted surface par beam ka spreading pure $\cos$ geometry hai — same rule jaise Orbital Mechanics (Keplerian) solar-flux projections.

2. $\beta =60^{\circ }$ par evaluate karo: $\cos 60^{\circ }=0.5$, toh $S=1367\times 0.5=683.5{\text{ W/m}}^2$. Yeh step kyun? $40\%$ nahi — yeh exactly aadha hai. Cosine woh number earn karata hai.

3. $I_{sc}$ intensity ke proportionally girta hai: factor $=0.5$, yaani $I_{sc}$ half ho jaata hai. Yeh step kyun? Photocurrent $I_L\propto$ absorbed photon flux $\propto S$. Kyunki $I_{sc}=I_L$ hai, current intensity ko linearly track karta hai (voltage sirf logarithmically drop karta — parent dekho).

4. Extreme $\beta =0^{\circ }$ (face-on): $\cos 0^{\circ }=1$, full $1367{\text{ W/m}}^2$, $I_{sc}$ unchanged. Extreme $\beta =90^{\circ }$ (edge-on): $\cos 90^{\circ }=0$, $S=0$, $I_{sc}=0$ — panel Sun ka knife-edge hai aur kuch collect nahi karta. Yeh step kyun? Dono endpoints cover karne se prove hota hai ki formula limits par sahi behave karta hai — koi sudden jump nahi.

Verify: $S(60^{\circ })=683.5{\text{ W/m}}^2$ ✓, factor $0.5$ ✓, $S(0^{\circ })=1367$ ✓, $S(90^{\circ })=0$ ✓.

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Example 6 — Series & parallel scaling se array banana (C7)

Forecast: Kaun sa knob (series ya parallel) voltage set karta hai, aur kaun sa current?

1. Series voltage stack karta hai: $N_s=\frac{28}{0.90}=31.1\Rightarrow N_s=32$ cells. Yeh step kyun? Series mein cells apni voltages add karte hain (jaise batteries end-to-end stack ki hoon) lekin same current share karte hain. Round up karo taaki string ageing/heat losses ke baad bhi bus voltage reach kar sake.

2. Array string voltage $V=32\times 0.90=28.8\text{ V}$ (bus se slightly above, MPPT converter ke liye headroom).

3. Parallel current stack karta hai: $N_p=\frac{4.5}{0.45}=10$ strings. Yeh step kyun? Parallel strings same voltage par currents add karte hain.

4. Total cells $=N_s\times N_p=32\times 10=320$.

5. Array max power $P=(N_s{V}_{mp})\times (N_p{I}_{mp})=28.8\times 4.5=129.6\text{ W}$. Yeh step kyun? Power cells ki total number ke saath scale karta hai: $P_{array}=N_s{N}_p{V}_{mp}{I}_{mp}=320\times 0.405\text{ W}=129.6\text{ W}$.

Verify: $N_s=32$, $N_p=10$, $V=28.8\text{ V}$, $P=129.6\text{ W}$. Cross-check: $320\times (0.90\times 0.45)=320\times 0.405=129.6\text{ W}$ ✓.

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Example 7 — Battery sizing, real word problem (C8)

Forecast: $560\text{ Wh}$ battery aur $\sim 290\text{ Wh}$ demand ke saath, kya hum $80\%$ limit ke kareeb ya door honge?

1. Eclipse duration in hours $=35/60=0.583\text{ hr}$. Yeh step kyun? Energy power × time hai, toh time watt-hours ke saath pair karne ke liye hours mein hona chahiye.

2. Energy needed $E_{\text{ecl}}=500\text{ W}\times 0.583\text{ hr}=291.7\text{ Wh}$. Yeh step kyun? Yeh woh hai jo battery har single eclipse mein deliver karni chahiye.

3. Battery total energy $E_{\text{batt}}=28\text{ V}\times 20\text{ Ah}=560\text{ Wh}$. Yeh step kyun? Volts × amp-hours = watt-hours; yeh full tank hai. (Hum energy $E_{\text{batt}}$ watt-hours mein likhte hain, nahi $Q$: is chapter mein $Q$ ka matlab electric charge amp-hours mein hai, toh energy ko $Q$ bolna clash karega. Charge aur energy ko alag symbols mein rakho.)

4. Required DoD $=\frac{E_{\text{ecl}}}{E_{\text{batt}}}=\frac{291.7}{560}=0.521=52.1\%$. Yeh step kyun? DoD woh fraction hai jitna tank hum har orbit mein khaali karte hain — yahan just over half.

5. $80\%$ ceiling se compare karo: $52.1\%<80\%$ → battery comfortably sized hai, margin ke saath.

Verify: $E_{\text{ecl}}=291.7\text{ Wh}$ ✓, $E_{\text{batt}}=560\text{ Wh}$ ✓, DoD $=52.1\%$ ✓ ($<80\%$). Units: $\text{Wh}/\text{Wh}$ dimensionless ✓.

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Example 8 — Cycle life aur mission lifetime (C9)

Forecast: DoD mein $50\%$ se $52\%$ tak thodi si increase — badi life hit, ya barely noticeable?

1. DoD ratio $\frac{52.1}{50}=1.042$. Yeh step kyun? Law sirf aapke DoD aur reference DoD ke ratio ki parwah karta hai.

2. Exponent apply karo $N=5000\times (1.042{)}^{-1.5}$. Ab $(1.042{)}^{-1.5}=0.940$, toh $N=5000\times 0.940=4700\text{ cycles}$. Yeh step kyun? Negative exponent ka matlab hai deeper discharge → kam cycles; power $1.5$ penalty ko non-linear banata hai lekin yahan mild hai.

3. Cycles per year: orbits per day $=\frac{1440\text{ min}}{90\text{ min}}=16$. Per year $=16\times 365.25=5844$ cycles/yr. Yeh step kyun? Har orbit ek charge/discharge cycle hai; orbits count karo cycles count karne ke liye.

4. Mission life $=\frac{4700}{5844}=0.804\text{ yr}\approx 9.6\text{ months}$. Yeh step kyun? Battery cycles pehle khatam ho jaate hain — yeh battery-limited lifetime hai, ek multi-year mission ke liye red flag.

Verify: $N=4700$ cycles ✓, orbits/yr $=5844$ ✓, life $=0.804\text{ yr}$ ✓.

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Example 9 — Exam twist: kaun sa converter? (C10)

Forecast: Direct connection sabse sasta hai — toh phir har real spacecraft ek box beech mein kyun add karta hai?

1. Direct-connect $V_{array}=V_{bus}$ force karta hai. Jab bus $24\text{ V}$ par baitha ho, array $24\text{ V}$ par drag hota hai, uske $V_{mp}=28.8\text{ V}$ par nahi. Yeh step kyun? Ek battery near-constant-voltage load hai; jahan bhi woh baitha ho, array ko match karna padta hai. Woh match almost kabhi bhi MPP nahi hota.

2. Off-MPP power lost: $24\text{ V}$ par vs. $28.8\text{ V}$ par array I–V curve ke steep left flank par baitha hai, $P_{max}$ se kaafi neeche — aap watts throw away karte ho. Yeh step kyun? MPP sirf woh voltage hai jo max power deta hai; kahin aur baithna strictly worse hai.

3. Bus $V_{mp}$ ke upar aur neeche dono jaata hai ($24\text{ V}<28.8\text{ V}<33.6\text{ V}$). Yeh step kyun? Yeh deciding fact hai. Kabhi kabhi aapko array voltage step down karna padta hai (bus $24\text{ V}$ par), kabhi step up (bus $33.6\text{ V}$ par).

4. Verdict — buck-boost MPPT converter chahiye. Ek pure buck (step-down) ya pure boost (step-up) sirf ek direction correct kar sakta hai; sirf ek buck-boost (Power Electronics) array ko $28.8\text{ V}$ par hold karta hai chahe bus upar ho ya neeche. Yeh step kyun? Converter array voltage ko bus voltage se decouple karta hai, MPPT ko array ko $V_{mp}$ par pin karne deta hai jabki battery jo chahiye woh deliver karta hai.

Verify (logic): Array $V_{mp}=28.8\text{ V}$ strictly bus min $24\text{ V}$ aur max $33.6\text{ V}$ ke beech hai → dono step-up aur step-down zaroori hain → buck-boost unique answer hai. ✓

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Recall

I–V curve ka kaun sa endpoint maximum current deta hai? ::: Short circuit ($V=0$), jahan $I=I_{sc}=I_L$. Hot solar cell kam power kyun banata hai? ::: Badhta temperature $I_0$ badhata hai, jo $V_{oc}\propto \ln (I_L/I_0)$ ko ghataata hai; current barely change hoti hai, toh power voltage ke saath girta hai. Eclipse mein ek cell forward bias par hold kiya gaya — generating ya consuming? ::: Consuming — $I_L=0$ ke saath woh ek plain Shockley diode hai jo current draw karta hai ($I<0$). Series vs parallel: kaun sa bus voltage set karta hai? ::: Series ($N_s$) voltages add karta hai; parallel ($N_p$) currents add karta hai. Bus voltage $V_{mp}$ ke upar aur neeche dono swing karta hai — kaun sa converter? ::: Buck-boost, kyunki sirf woh step both up aur down kar sakta hai.