AMME1705 · Introduction to Electromechanical Systems
Power Supplies & Batteries
In AMME1705 Introduction to Electromechanical Systems at the University of Sydney, this chapter is the power-conversion toolkit: how to get energy into a circuit in the right form. You learn to turn mains AC into clean DC (rectify with diodes, smooth with a capacitor, then regulate), the heat penalty of a linear regulator, the more efficient switch-mode buck (step-down) and boost (step-up) converters, and the battery relations that set stored energy (E = V·Q) and discharge current (the C-rating).
It maps to the Week 11 “Power Supplies” material and appears as short numeric and multiple-choice parts on the paper-based final worth 33% of the unit.
What this chapter covers
- 01Trace the AC to DC chain: transformer steps the voltage, rectifier flips the wave, smoothing capacitor cuts the ripple, regulator holds a flat output
- 02Tell apart a half-wave rectifier (one diode) and a full-wave/bridge rectifier (four diodes), and why full-wave is easier to smooth
- 03Use the transformer turns ratio V_s/V_p = N_s/N_p, with units and symbol definitions
- 04Explain how a smoothing capacitor reduces ripple by holding the voltage up between peaks
- 05Compute a linear regulator's heat loss P = (V_in - V_out)*I and its efficiency V_out/V_in, and see why a large voltage drop is wasteful
- 06Compare switch-mode buck (V_out <= V_in, step down) and boost (V_out > V_in, step up via inductor back-EMF) and why they are efficient
- 07Find a battery's stored energy with E = V*Q, converting capacity in amp-hours to coulombs with Q = capacity(Ah) x 3600
- 08Use the C-rating to find the maximum discharge current I_max = C-rating x capacity, and know it does NOT set the energy
Stored energy and current limit of a battery pack
- +1(a) Convert the capacity to amp-hours, then to coulombs by multiplying by 3600 seconds per hour: 5000 mAh = 5.0 Ah, so Q = 5.0 x 3600 = 18 000 C (since 1 A.s = 1 C).
- +1(b) Stored energy E = V*Q = 12.0 x 18 000 = 216 000 J = 216 kJ. The C-rating is not used here. (Unit check: V x C = V.A.s = W.s = J.)
- +1(c) The C-rating scales the capacity to give the maximum safe discharge current: I_max = C-rating x capacity = 30 x 5.0 = 150 A.
- +1(d) Approximate run-time at a steady load: t = capacity / I = 5.0 Ah / 6.0 A = 0.833 h, about 50 min. The 6.0 A load is far below the 150 A limit, so it is within the C-rating.
Key terms
- Rectifier
- A diode arrangement that converts alternating current (AC) to one-polarity (pulsed) DC. A half-wave rectifier uses a single diode and passes only the positive halves; a full-wave or bridge rectifier uses diode pairs (four diodes) to flip the negative halves up too, giving twice as many pulses that are easier to smooth.
- Smoothing (reservoir) capacitor
- A capacitor placed across a rectifier's output that charges to each peak and then feeds the load from its stored charge between peaks, reducing the ripple. Because a capacitor resists a change in voltage, a larger capacitor sags less and gives smaller ripple; it is also why a device can run for a moment after being unplugged.
- Linear regulator
- A device (such as a 7805 holding 5 V) that produces a steady output voltage by dropping whatever excess is left over, acting like a self-adjusting resistor. It is simple and low-noise but dissipates the excess as heat, P = (V_in - V_out)*I, so a large input-to-output voltage drop is inefficient and needs heat-sinking.
- Regulator heat loss
- The power a linear regulator burns as heat while passing the load current at the leftover voltage: P_diss = (V_in - V_out)*I, in watts. For example, dropping 25 V to 5 V at 1 A dissipates (25 - 5)*1 = 20 W. Its efficiency is simply the output/input voltage ratio, V_out/V_in.
- Buck converter
- A switch-mode step-down converter: a transistor chops the input and an inductor-capacitor filter delivers the average, V_out = D*V_in (D is the duty cycle, so V_out is always <= V_in). Because the switch barely dissipates, it is far more efficient than a linear regulator for a large voltage drop.
- Boost converter
- A switch-mode step-up converter that raises the output above the input (V_out > V_in) using an inductor's back-EMF: when the switch opens, the collapsing magnetic field keeps current flowing the same way, so the inductor voltage adds to the input, V_out = V_in + V_L. A transformer cannot do this on DC and a linear regulator never can.
- Battery stored energy (E = V*Q)
- The energy a battery holds, in joules: E = V*Q, where V is the nominal voltage and Q is the charge in coulombs. A capacity quoted in amp-hours converts to coulombs with Q = capacity(Ah) x 3600, since 1 A.s = 1 C. A higher voltage or a larger charge stores more energy.
- C-rating
- A number that sets a battery's maximum safe discharge current, I_max = C-rating x capacity(Ah). A 5 Ah pack rated 20C can supply up to 100 A. The C-rating tells you how hard the pack can be discharged; it says nothing about the stored energy, which comes only from E = V*Q.
Power Supplies & Batteries FAQ
Does a battery's C-rating tell me how much energy it stores?
No. The C-rating sets the maximum safe discharge current, I_max = C-rating x capacity(Ah) - for example a 5 Ah pack rated 30C can deliver up to 150 A. The stored energy is a separate figure, E = V*Q, where Q = capacity(Ah) x 3600 is the charge in coulombs. So a 12 V, 5000 mAh pack stores 12 x (5 x 3600) = 216 000 J = 216 kJ regardless of its C-rating. When a question gives you a C-rating and asks for energy, the C-rating is a distractor.
Why is a linear regulator inefficient, and when should I use a switch-mode buck instead?
A linear regulator drops the excess voltage as heat: P_diss = (V_in - V_out)*I, and because it passes essentially the same current in and out, its efficiency is just V_out/V_in. Dropping 12 V to 5 V at 0.8 A already wastes 5.6 W as heat while delivering only 4.0 W. A switch-mode buck converter instead chops the input and filters it to the average V_out = D*V_in, wasting very little, so its efficiency stays high even with a big voltage drop. Choose linear for a small, clean drop where simplicity and low noise matter; choose a buck when the drop is large or efficiency and heat are a concern.
Can AI help me with power supplies and batteries in AMME1705?
Yes, as a study aid rather than an answer service. Sia is an AI tutor that explains the method step by step - for example, why you rectify before you smooth, how to set up the regulator heat loss P = (V_in - V_out)*I and its efficiency, or how to convert amp-hours to coulombs before using E = V*Q - and can re-derive a worked problem with you until you can reproduce it yourself. It will not hand you exam answers or promise a grade; the aim is to build the understanding you need to work these parts under exam conditions. Always check any explanation against your current Canvas notes and your lecturer's conventions.
Studying with AI? Sia — free AI electrical engineering tutor works through AMME1705 step by step.
Exam move
Learn the AC-to-DC chain as a fixed sequence - transformer (step the voltage), rectifier (flip the wave with diodes), smoothing capacitor (fill the gaps between pulses), regulator (flatten to a steady rail) - and be able to say what each stage does and why the order cannot change. For the numeric parts, drill three small formulas until they are automatic: the linear regulator heat loss P = (V_in - V_out)*I with efficiency V_out/V_in; the battery energy E = V*Q with Q = capacity(Ah) x 3600; and the C-rating current limit I_max = C-rating x capacity. The most common slips are unit conversions (amp-hours to coulombs is x3600, and joules to kilojoules) and treating the C-rating as if it were energy - it only caps the current. For the concept questions, match the direction word to the maths: buck steps down (V_out <= V_in), boost steps up (V_out > V_in). Because the final awards marks line by line and to a stated precision, write every relation and unit conversion out before plugging in numbers and round only at the end. Pace yourself at a steady rate of about 1.5 minutes per mark (a 2-hour writing time over an 80-mark paper), and confirm your paper's mark total, timing and open/closed-book status on your current Canvas.
