From the wall socket to the motor

1. RMS means Root of the Mean of the Square: square the wave, average it, then square-root it. We square because heating power goes as V².
2. The wave is V = V_pk sin θ, so V² = V_pk² sin² θ.
3. Over a whole cycle, sin² θ averages to exactly ½. It spends as long above ½ as below it.
4. So the mean of the square is ½ V_pk², and its square root is:

1. Average height of one hump = its area ÷ its width.
2. Area under one hump of sin θ, from 0 to π, is exactly 2 (∫ sin θ dθ = [−cos θ] = 1 + 1). Width is π.
3. So the average is 2/π = 0.637 of the peak.
4. RMS is 0.707 of the same peak. Divide one by the other and the peak cancels:

1. Read each diode as a rule. D1 allows A → +. D2 allows B → +. D3 allows − → A. D4 allows − → B.
2. Spot the pattern. Two arrows point into +, two point out of −. So current can only arrive at + and leave from −. It always goes through the load from + to −.
3. Positive half: A pushes. A → + through D1, through the load, then − → B through D4. D1 and D4 on, D2 and D3 off.
4. Negative half: B pushes. B → + through D2, through the load, then − → A through D3. D2 and D3 on, D1 and D4 off.
5. Result: the load sees the same polarity every half-cycle. The output is a row of positive humps, 100 per second (2 per AC cycle).
6. Cost: current always passes through exactly two diodes, so the output peak is V_pk − 2V_F, and two diodes are always making heat.

1. Time between refills: 2 humps per cycle × 50 cycles = 100 a second, so the gap is Δt = 1 / (2 × 50) = 10 ms.
2. Charge drained in that gap: current is charge per second, so ΔQ = I × Δt.
3. Voltage drop: a capacitor obeys Q = CV, so ΔV = ΔQ / C.

In words: dip = drain rate ÷ (refills per second × tank size). Bigger C, smaller ripple.

1. The capacitor charges to the top of each hump.
2. The top of the AC is V_pk = 325.3 V, but the current had to pass through two diodes to get there, each taking V_F.

1. Power = voltage × current. A conducting diode has a fixed V_F across it, so its power is V_F × i at every instant.
2. Average over time: V_F is constant, so P = V_F × (average current). Diodes use the average current. Resistors and MOSFETs use I²R with RMS.
3. Two diodes are always on, so P = 2 × V_F × I_avg.
4. Convert to the RMS you're given: I_avg = 0.9 × I_RMS (chapter 01).

1. Fully on, the MOSFET is just a resistor of value R_DS(on).
2. Power in a resistor = I²R (or equivalently V × I with V = I × R_DS(on)).

Because of the square, double the current gives four times the conduction loss. That's why high-power systems use higher voltages, and why MOSFETs are put in parallel to share current.

1. Look at the turn-on in the graph. First the current climbs from 0 to I_D while the voltage is still at V_DD. Then the voltage falls to almost 0 while the current stays at I_D.
2. In each part, power = V × I rises from 0 to V_DD × I_D or falls back. It's a triangle on the power graph.
3. Energy = area under the power graph. A triangle's area is ½ × height × width, so across the whole switching time t:

4. The datasheet gives the turn-on time in two parts, delay + rise; turn-off is delay + fall:

5. This happens once on and once off every cycle, F_SW times a second. Energy per second is power:

The full formula on the slide also has a Q_rr × V_DD term for the diode's recovery charge. It's ignored in these problems.

1. The same heat P flows through every step, like current through series resistors.
2. Each step adds ΔT = P × R_th. Start from the air, which you know, and climb:

3. Limits: the absolute max T_J is 175 °C. Design for 100–120 °C and never above 120 °C.
4. Safety margin compares T_J with 175 °C. The target is at least 30%:

1. Rearrange the line above for the one resistance you get to choose:

2. Pick your target T_J (100 or 120 °C), put in P and the two datasheet values, and you get the largest R_th3 allowed. Choose a heatsink at or below it. A lower °C/W means a bigger heatsink.
3. If the answer is negative: even a perfect heatsink (R_th3 = 0) can't keep it cool, because R_th1 + R_th2 alone already make it too hot. Fixes: a better thermal interface (grease, 0.24, instead of an insulating pad), less loss (lower F_SW, a lower-R_DS(on) part, MOSFETs in parallel), lower ambient, or forced air or liquid cooling.

1. The key fact: in steady state, the inductor's average voltage over a cycle must be zero. If it weren't, its current would keep growing forever. So (voltage × time while on) + (voltage × time while off) = 0. This is called volt-second balance.
2. Buck, switch on for D × T: the inductor sits between V_in and V_out, so V_L = V_in − V_out.
   Buck, switch off for (1 − D) × T: the diode connects it to ground, so V_L = −V_out.

3. Boost, switch on: the inductor is straight across the battery, V_L = V_in.
   Boost, switch off: it sits between V_in and V_out, V_L = V_in − V_out (negative).

1. Inductor: V = L × (change in current ÷ time). While off, V_L = V_out for a time (1 − D)/F_SW, and the current changes by ΔI_L. Rearrange for L:

2. Capacitor: it absorbs the inductor's triangle-shaped ripple current. The charge in one half of that triangle sets the voltage ripple:

Slide check: for 24 V → 12 V at 10 A, 25 kHz, D = 0.5 and 10% ripple, the slide puts 24 V in the L formula and gets 480 µH. The formula says V_out = 12 V, which gives 240 µH. The calculator shows both. Ask which he wants.