AMME1705 · Introduction to Electromechanical Systems
Transistors, PWM & Motor Drive
In AMME1705 Introduction to Electromechanical Systems at the University of Sydney, this chapter is the power interface between a low-current control signal and a real motor. You learn the bipolar transistor as a current amplifier and switch (IC = β·IB, VBE ≈ 0.7 V), the emitter-follower and Darlington buffers, saturation, the diodes that protect and steer current (including the flyback diode), and pulse-width modulation with the H-bridge for bidirectional drive.
It maps to the Week 8 “Delivering Power” material and is a signature multi-part item on the paper-based final worth 33% of the unit.
What this chapter covers
- 01Read the BJT current-gain law I_C = beta*I_B and the V_BE = 0.7 V base-emitter drop, with units and symbol definitions
- 02Tell apart the three operating regions: cut-off (off), forward-active (amplifier), and saturation (fully-on switch)
- 03Size an emitter-follower (V_out = V_in - 0.7 V) and a Darlington pair (V_out = V_in - 1.4 V) and compare their current gains
- 04Use a diode's forward drop (~0.7 V silicon, ~2 V LED) and a series resistor to set an LED current
- 05Explain why an inductive motor needs a flyback (freewheeling) diode and which way round it goes
- 06Convert between duty cycle and average voltage: D = t_on/T and V_avg = D*V_HIGH
- 07Trace an H-bridge's three valid states and avoid same-leg shoot-through
- 08Work the full transistor-driven-motor chain end to end, including the saturation-onset speed limit
NPN transistor driving a DC motor, from control voltage to saturation speed
- +1(a) The base-emitter junction holds V_BE = 0.7 V, so the rest of V_in appears across the base resistor: V_Rb = V_in - V_BE = 2.5 - 0.7 = 1.8 V.
- +1(b) Ohm's law on R_b gives I_B = V_Rb / R_b = 1.8 / 2200 = 8.18e-4 A = 0.818 mA. In the forward-active region the collector (motor) current is I_C = beta*I_B = 100 x 0.818 mA = 81.8 mA, i.e. about 82 mA.
- +1(c) The motor current flows through the armature resistance: V_Ra = I_C x R_a = 0.0818 A x 3.0 Ohm = 0.245 V = 245 mV.
- +1(d) Torque is the torque constant times the current: tau = K_t x I_C = 25 mN.m/A x 0.0818 A = 2.05 mN.m, i.e. about 2.0 mN.m.
- +1(e) Use the collector-to-rail balance V_rail = I_C*R_a + V_EMF + V_CE. At the edge of saturation V_CE = 0.7 V, so V_EMF = 12 - 0.245 - 0.7 = 11.055 V. The speed is Omega = K_v x V_EMF = 382 rpm/V x 11.055 V = 4223.0 rpm, i.e. about 4223 rpm.
Key terms
- Current gain (beta, h_FE)
- The common-emitter current gain of a bipolar transistor: the factor by which base current is multiplied to give collector current, I_C = beta*I_B. It is dimensionless (a ratio of amps to amps), typically in the range 100-300, and applies only in the forward-active region.
- Base-emitter drop (V_BE)
- The roughly fixed voltage across the base-emitter junction of a conducting silicon transistor, about 0.7 V. You subtract it from the control voltage before finding the base current.
- Saturation
- The fully-on region where the transistor's collector-emitter voltage V_CE is small and the load (not beta) sets the collector current. Driving a motor, saturation onset is taken at V_CE about 0.7 V in this unit; beyond it the transistor no longer controls the current.
- Emitter-follower
- A single-transistor current buffer whose output tracks the input one diode drop lower, V_out = V_in - 0.7 V, while multiplying the current the source must supply by beta. Voltage gain is about 1; the benefit is current gain.
- Darlington pair
- Two transistors cascaded so the current gains multiply (I_out = beta1*beta2*I_in) at the cost of two base-emitter drops, V_out = V_in - 1.4 V. Used when a single stage cannot supply the load current.
- Flyback (freewheeling) diode
- A diode placed across an inductive load such as a motor, cathode to the positive rail, so it is reverse-biased in normal running. When the switch opens it gives the collapsing inductor current a safe loop to decay through, clamping the voltage spike that would otherwise destroy the transistor.
- Duty cycle (D)
- The fraction of each PWM period for which the signal is high, D = t_on / T (a number between 0 and 1, or a percentage). The load sees an average voltage V_avg = D*V_HIGH, so duty sets the effective drive level without wasting power as heat.
- H-bridge
- Four switches arranged around a motor so that closing one diagonal pair (e.g. S1+S4) drives current one way and the other diagonal (S2+S3) reverses it. Its three valid states are off and the two diagonals; closing two switches on the same leg causes destructive shoot-through.
Transistors, PWM & Motor Drive FAQ
Is the transistor current gain I_C = beta*I_B always valid?
No. It holds only in the forward-active region. Once the transistor is driven into saturation - as happens when a motor speeds up and its back-EMF pulls the collector voltage down toward V_CE about 0.7 V - the external circuit (the load resistance and the supply) sets the collector current, and pushing more base current no longer raises it. Cut-off is the other extreme, where I_B = 0 and no collector current flows.
How is duty cycle related to the average voltage a motor sees?
PWM switches the supply fully on and off, and the motor responds to the average. With the supply at V_HIGH and a duty cycle D = t_on/T, the average is V_avg = D*V_HIGH, so D = V_avg/V_HIGH. Remember D is a ratio: 60% means D = 0.60, so a 12 V rail at 60% duty gives an average of 7.2 V. Because a saturated switch drops almost no voltage, this wastes far less power than dropping the extra voltage across a resistor.
Can AI help me with transistors, PWM and motor drive 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 subtract V_BE before finding the base current, how to set up the collector-to-rail balance V_rail = I_C*R_a + V_EMF + V_CE, or why the flyback diode faces the way it does - and can re-derive a worked chain 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 do the problem under exam conditions. Always check any explanation against your current Canvas notes and lecturer's conventions.
Studying with AI? Sia — free AI electrical engineering tutor works through AMME1705 step by step.
Exam move
Learn the chapter as one chain, not a list of facts. From a control voltage, subtract V_BE (~0.7 V), divide by the base resistor for I_B, multiply by beta for the motor current, then use tau = K_t*I for torque and the three-term balance V_rail = I_C*R_a + V_EMF + V_CE for the saturation-limited speed. Keep the input (base) loop and the output (collector) loop separate. Rehearse the small standard sub-questions too - duty cycle D = V_avg/V_HIGH, the LED series-resistor current (V_supply - V_F)/R, the emitter-follower and Darlington drops, and the H-bridge's two legal diagonals. Because the final awards marks line by line and to a stated precision, write every relation 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.
