DC-DC Converters and Battery Characteristics

Posted on Sep 5, 2024 in Technology

DC-DC Converter Analysis

Boost Converter

A boost DC-DC converter switches at 100 kHz. The input is +12 V and the duty ratio of the transistor is 75%. The input inductor is 25 μH, and the output capacitor is 100 μF. If the load draws 50 W, the output voltage is found to be 48 V.

Output Voltage Calculation

What is the output voltage for a 10 W load? If L > L_crit and C > C_crit, then V_out/V_in = 1/D₂. With a 50 W load, V_out = 48 V. This gives D₂ = 0.25, so it would be continuous mode. For a 10 W load, the converter could become discontinuous. At 50 W, R = (48)²/50 = 46.1 Ω. At 10 W, R = V_out2/10. Let us check L_crit at 10 W. If L ≥ L_crit, then V_out = 48 V, and R = 230 Ω. The output average current would be 0.208 A, and the input average current would be 0.833 A. When L = L_crit (#1 on, and the current starts from 0, the amplitude is 2 × average) v_L = L(di)/(dt), 12 V = L_crit (2 × 0.833 A)/[(0.75) × 10 μs] L_crit (10 W load) = 54 μH. The case is therefore subcritical for 10 W at 25 μH. Since P_out is known, we expect P_in = 10 W, and I_L ≥ 0.833 A. The peak input current: i_peak = (V_inD₁T)/L = (12 V)(0.75)(10μs)/(25 μH) = 3.6 A. The input energy is ½ × Li² = ½ (25 μH)(3.6 A)² = 162 μJ for each cycle. At 100 kHz, this is 16.2 W input. If the load draws only 10 W (100 μJ/cycle), the capacitor energy builds up at 62 μJ/cycle. In each cycle we have: ½ (100 μF(V_out + ΔV)² – ½ (100 μF)V_out2 = 62 μJ ½ (V_out2 + 2ΔVV_out + (ΔV)² – V_out2) = (62 μJ)/(100 μF) Since (ΔV)² is small comparing with 2ΔVV_out, we can write: ΔV = 0.62/V_out. In each cycle the capacitor voltage increases by this amount. In reality, the output voltage cannot grow indefinitely. Either the output power, or losses would increase and the steady state would be achieved (or the capacitor would fail).

Flyback Converter

Ballast Load and Inductor Selection

A flyback converter uses a 20:1 turns ratio to convert 150 V to 5 V output. The desired load range is 0 W to 200 W. The target switching frequency is 120 kHz. (a) Select a ballast load, then choose the smallest inductor such that L > L_crit. The ballast load should not drop the full-load efficiency by more than 1%. (b) For your ballast load, what is L_crit if the input voltage can vary by ±20%?

Ballast Load Selection

(a) Select a ballast load that will not drop full-load efficiency by more than 1%. This would limit the ballast load to 2 W. With 5 V output, R_ballast = (5 V)²/(2 W) = 12.5 Ω. To avoid tolerance issues, let us select a little higher value of 15 Ω. Now, L_crit for this choice. Consider an equivalent 1:1 turns ratio: The transformer can be eliminated. With #1 on, v_L = 7.5 V = L (di)/(dt) Minimum I_load = (5 V)/(15 Ω) = 0.333 A. Duty ratios: D₁/D₂ = V_out/V_in, D₁ = 2/5, D₂ = 3/5. I_load = D₂ I_L, I_L = I_load / D₂, with R_load = 0, minimum I_L = (1/3 A)/(3/5) = 5/9 A. Maximum allowed ΔI_L when L = L_crit is 2I_L = 10/9 A. For 120 kHz switching, D₁T = 3.33 μs. Then 7.5 V = L_crit (10/9 A)/(3.33 μs) L_crit = 22.5 μH (output side) On the 150 V side, the inductance is (N₁/N₂)² times this value, or L_m = 22.5 μH (400) = 9 mH.

Critical Inductance with Input Voltage Variation

(b) Now, allow the input to vary ±20%. For the 1:1 equivalent, this means V_in varies between 6 V and 9 V. The duty ratio ranges can be found: D₁/D₂ = V_out/V_in. 5/14 ≤ D₁ ≤ 5/11 and 6/11 ≤ D₂ ≤ 9/14. With #1 on, the limit on Δi_L: Δi_L = 2(I_load/D₂) = 2(1/3)/D₂ = 2/(3D₂). v_L = L(di)/(dt), V_in = L_crit [2/(3D₂)]/(D₁T) and D₁/D₂ = V_out/V_in. This simplifies to: L_crit = (3/2)D₂²TV_out. The highest D₂ value, 9/14, is needed, and L_crit = 25.8 μH (output side) On the input side, this requires L_m = 10.3 mH, almost 15% higher than in part (a).

Lead-Acid Battery Circuit Model

Battery Characteristics

Tests on a lead-acid battery show the following characteristics with full charge: The open-circuit terminal voltage is 13.4 V, the terminal voltage is 13.3 V with a 10 A load, and 13.5 V during 10 A charging. The manufacturer reports that the battery loses the equivalent of about 1200 C of charge each week when it is in storage. Ignoring the L and C values, determine a circuit model for this battery.

Circuit Model Determination

The open-circuit potential is 13.4 V, so V_b = 13.4 V. During charge and discharge, 10 A flow causes a 0.1 V change this means that R_s + R_int = (0.1 V)/(10 A) = 10 mΩ. R_dis is such that 1200 C are lost each week. This is: 1200 C(7 days/wk)(24 hours/day)(3600 s/hr) With V_b = 13.4 V, the value R_dis = V/I = 6.75 kΩ.

Voltage Induced in a Wire

Resonant Power Converter and Wire Voltage

Estimate the voltage induced across a one metre length of copper wire when a resonant power converter draws current i(t) = 20cos(400,000πt) through the wire. What if a steel wire with μ = 1000μ₀ is substituted?

Voltage Calculation

L_wire = (μ_wire/(8π) + μ/(2π)ln(D/R))l For typical wire runs D/R = 10 and expected L_wire is about 0.5 μH/m, since for copper μ_wire = μ₀. The impedance: |Z| = ωL = (400,000)π(0.5 × 10^-6) = 0.628 Ω. V = ZI About 8.9 V_RMS is induced. For steel wire, μ_wire = 1000μ₀ and the internal inductance is now 500 μH/m, and dominates the total. Z = (400,000)π(0.5 × 10^-3) = 628 Ω V_RMS = 628 × 20/√2 = 8.9 kV.