CSE 2317 · W2 C1 · The BJT switch: load lines and switching times

01 Cold open 0–3 min

The switch that won't let go

You tell a transistor to switch OFF. For the next hundred nanoseconds it keeps conducting as if nothing happened. Then it collapses.

Pop culture

Wile E. Coyote runs off the cliff and hangs in mid-air, legs still spinning, until he looks down. A saturated transistor does the same thing. Today we measure that hang time, give it a name (storage time) and learn to shorten it.

Think deeper

Guess before we start: for a transistor switch driven hard into saturation, which is slower, turning ON or turning OFF?

Argue it in pairs for two minutes. There may be more than one defensible answer.

Closer to home: an IPS keeping your fan running through load-shedding switches its power devices thousands of times a second. Every slow edge is wasted as heat.

02 Concept 3–8 min

One line, two addresses: the DC load line

KVL around the collector loop of a common-emitter switch gives the only (I_C, V_CE) pairs the circuit allows:

V_{CC} = I_C R_C + V_{CE}\;\;\Longrightarrow\;\; I_C = \frac{V_{CC}-V_{CE}}{R_C}

A straight line on the output characteristics. The transistor (through I_B) picks where on it you sit: the Q-point.

End	V_CE	I_C	Switch
Cutoff	V_CC	≈ 0	OFF (open)
Saturation	≈ 0.2 V	(V_CC − 0.2)/R_C	ON (closed)

Pro tip

Amplifiers live in the middle of the line; switches live only at its two ends. They should race through the middle as fast as possible, because that is where the transistor is both carrying current and dropping voltage, so it heats.

Gotcha

The ideal intercept is V_CC/R_C at V_CE = 0. The real ON point stops just short of it, at V_CE(sat) ≈ 0.2 V. Use I_C(sat) = (V_CC − 0.2)/R_C in calculations.

03 Live lab 8–14 min

Slide the Q-point along the line

Load-line explorer Change I_B and the Q-point slides along the load line. A switch should only ever sit at the two ends.

Base current I_B = 60 µA V_CC = 10.0 V R_C = 1.0 kΩ

Active · amplifier territory

V_CE3.64 V

I_C6.36 mA

β·I_B6.00 mA

I_B(min)98 µA

Overdrive0.61×

P = V_CE·I_C23.1 mW

04 Concept 14–18 min

How hard to push: saturation and overdrive

To reach the ON point, the base must supply at least what the load needs, divided by the worst β you might get:

I_B \;\ge\; I_{B(min)} = \frac{I_{C(sat)}}{\beta_{min}}

How far past that minimum you go is the overdrive factor:

\text{ODF} = \frac{I_B}{I_{B(min)}} = \frac{\beta_{min}I_B}{I_{C(sat)}},\qquad \beta_{forced} = \frac{I_{C(sat)}}{I_B} < \beta_{min}

Last class: I_C(sat) = 4.8 mA, β_min = 50 ⇒ I_B(min) = 96 µA; with I_B = 0.43 mA, ODF = 0.43/0.096 ≈ 4.5.

Gotcha

In saturation, I_C ≠ βI_B. β·I_B is what the transistor could carry; R_C decides what it does carry. The ratio I_C/I_B in saturation is the forced β, always below β.

Pro tip

Why β_min? One reel of BC547s can range from about 110 to 800 in β, and β drops at low temperature and at high current. Designs use 2–5× overdrive for margin. Why not 50×? Slide 14.

05 Your turn 18–23 min

Your turn · 1: read the load line

04:00

1 Exam CQ

V_CC = 9 V, R_C = 1.5 kΩ, β = β_min = 100, V_CE(sat) = 0.2 V.

Find the load-line end points and the ON point.
For I_B = 30 µA find the Q-point. Switch or amplifier?
Find I_B(min) and the ODF at I_B = 100 µA.
Compare the transistor's power at (b) and in saturation.

Work it out first.
Press Solve when the class has an answer.

06 Worked example 23–29 min

Worked example: an ESP32 drives a 12 V relay

Given

A rooftop water-pump controller in a Dhaka apartment: an ESP32 pin (V_OH = 3.3 V) switches a 12 V relay coil (R_coil = 240 Ω) through an NPN with β_min = 50. V_BE(sat) = 0.8 V, V_CE(sat) = 0.2 V.

Find

R_B (standard E12 value) for at least 2× overdrive, the resulting ODF, pin current and ON-state dissipation.

Method: saturated-switch design (Boylestad, transistor switching networks; Millman & Taub, transistor switch)

Solution

Predict the answer first. Then reveal one step at a time.

07 Live lab 29–33 min

Design lab: one pin, three loads

Base-resistor designer Pick a board and a load, then find an R_B that saturates the transistor without cooking the pin.

R_B = 10 kΩ (E12) β_min = 50

Flyback diode across the coil

I_C(sat) needed68.6 mA

I_B from pin0.26 mA

Overdrive0.19×

I_C actual13.0 mA

P_Q53 mW

2× design820 Ω

Not saturated. β_minI_B = 13.0 mA < 68.6 mA. V_CE = 4.09 V: the load is starved and the transistor heats.

08 Concept 33–37 min

Four times that steal your speed

t_d delay: input edge → i_C reaches 10 %. Charging the junction capacitances up to cut-in.
t_r rise: 10 % → 90 % of I_CS. Building up base charge through the active region.
t_s storage: input falls → i_C drops to 90 %. Removing the excess charge stored in saturation.
t_f fall: 90 % → 10 %. Back through the active region.

t_{on} = t_d + t_r \qquad t_{off} = t_s + t_f

Why saturation stores charge

In saturation both junctions are forward-biased, so the collector junction injects too and the base fills with minority carriers beyond what I_CS needs. That excess Q_S must drain before the collector junction can reverse-bias; until then i_C stays stuck at I_CS.

09 Live lab 37–44 min

The switching oscilloscope

Switching oscilloscope Watch the beam: i_C lags the input on both edges. Push the overdrive up and see which bracket grows.

Turn-on drive I_B1 = 0.40 mA · overdrive 2.0× Reverse drive I_B2 = 0.20 mA

Speed-up capacitor

t_d14 ns

t_r14 ns

t_s21 ns

t_f14 ns

t_on27 ns

t_off35 ns

f_max16.0 MHz

β = 50, I_CS = 10 mA, τ = 25 ns, τ_s = 50 ns · leaves saturation after 20 ns (t_s formula: 20 ns); readout t_s is to the 90 % point

10 Concept 44–48 min

Charge control: formulas you can use

While active, i_C relaxes toward βi_B with time constant τ. From 0 toward βI_B1:

i_C(t) = \beta I_{B1}\left(1-e^{-t/\tau}\right)\;\Rightarrow\; t_r = \tau\ln\frac{\beta I_{B1}-0.1I_{CS}}{\beta I_{B1}-0.9I_{CS}}

Excess charge drains with time constant τ_s, pulled by the reverse current I_B2:

t_s = \tau_s\ln\frac{I_{B1}+I_{B2}}{I_{CS}/\beta + I_{B2}}

Same idea for the fall, now heading toward −βI_B2:

t_f = \tau\ln\frac{0.9I_{CS}+\beta I_{B2}}{0.1I_{CS}+\beta I_{B2}}

The trade-off

More I_B1 (overdrive): βI_B1 ≫ I_CS, so t_r shrinks. But I_B1 sits in the numerator of t_s: storage grows.

More I_B2 (reverse drive): shrinks t_s and t_f. Costs a negative supply or a clever circuit.

Gotcha

If βI_B1 ≤ 0.9 I_CS the log's argument goes negative: i_C never reaches 90 %. Check βI_B1 > I_CS (saturation) before using any of these.

11 Worked example 48–53 min

Worked example: rise, storage and fall

Given

β = 50, τ = 20 ns, τ_s = 40 ns, I_CS = 10 mA. Turn-on drive I_B1 = 0.5 mA; turn-off reverse drive I_B2 = 0.25 mA.

Find

t_r, t_s and t_f. Then: what happens if you double I_B1?

Method: charge-control model (Taub & Schilling; Millman & Taub, switching times)

Solution

Predict the answer first. Then reveal one step at a time.

12 Your turn 53–58 min

Your turn · 2: how much reverse drive?

04:00

2 Exam CQ

β = 40, τ = 15 ns, τ_s = 60 ns, I_CS = 20 mA, I_B1 = 1.0 mA, I_B2 = 0.5 mA.

Confirm saturation and find the ODF.
Find t_r and t_s.
What I_B2 would cut t_s to 10 ns?

Work it out first.
Press Solve when the class has an answer.

13 Concept 58–61 min

Two cheap fixes for storage time

(a) At each edge C is momentarily a short: a big current spike pours charge in on the rising edge and yanks it out on the falling edge. In steady state C is an open circuit, so R_B alone sets a modest I_B1 and little charge is stored. Both edges get faster. Rule of thumb: R_BC ≈ τ.

(b) A Schottky diode (V_F ≈ 0.3–0.4 V) from base to collector turns on before the collector junction (≈ 0.5 V) can. Extra base current is diverted into the collector, so the transistor sits at the edge of saturation instead of deep in it.

V_{CE} \approx V_{BE} - V_{SBD} \approx 0.3\text{–}0.4\ \text{V}

No deep saturation, so almost no excess charge: t_s ≈ 0.

Pro tip

This is the “S” in 74S and 74LS TTL, where every transistor that could saturate carries a built-in Schottky clamp. Week 3.

14 Think deeper 61–64 min

Critical thinking · two arguments

Think deeper

“More base current is always safer. Let's drive the base with 50× overdrive.” Argue against it.

Argue it in pairs for two minutes. There may be more than one defensible answer.

Think deeper

A Schottky diode is still a diode. Why doesn't the clamp diode itself suffer storage time?

Argue it in pairs for two minutes. There may be more than one defensible answer.

15 Your turn 64–67 min

Your turn · 3: the dim LED

3 Trap

An Arduino (5 V pin) drives a red LED (V_F = 2.0 V) with a 150 Ω series resistor from +5 V through an NPN (β_min = 100). A student picks R_B = 100 kΩ “because the base only needs microamps”.

Is the LED fully on? Find I_C, V_CE and the transistor's power. Then fix R_B.

Work it out first.
Press Solve when the class has an answer.

16 Concept 67–70 min

Where the switch gets hot

ON: tiny voltage, full current. OFF: full voltage, no current. Both are cheap:

P_{ON} \approx V_{CE(sat)}\,I_{C}\cdot D\ \ (D=\text{duty}),\qquad P_{OFF}\approx 0

Every edge passes through the middle of the load line where V and I are both large. For a resistive load with linear edges each transition costs V_CCI_Ct/6:

P_{sw} \approx \tfrac{1}{6}\,V_{CC}\,I_{C}\,(t_r+t_f)\,f

Example: 12 V LED strip, 0.5 A, PWM at 50 % duty

V_CE(sat) = 0.2 V, t_r + t_f = 1 µs (a slow power BJT).

P_ON = 0.2 × 0.5 × 0.5 = 50 mW

At 20 kHz: P_sw = ⅙ × 12 × 0.5 × 10⁻⁶ × 20 000 = 20 mW

At 200 kHz: 200 mW. Switching loss now dominates.

Pro tip

Faster edges = cooler switch. That is why t_r and t_f matter even in power electronics, not just in logic.

17 Real world 70–73 min

The valve that killed its transistor

BME corner

A dialysis machine or infusion pump opens and closes a 12 V solenoid pinch valve on the tubing (coil 24 Ω, L ≈ 50 mH, I = 0.49 A). A transistor switches it. On the bench it works perfectly. After about a week in the ward, the transistor is dead.

At turn-off the coil insists on keeping its current: v = L di/dt. Interrupting 0.5 A in 1 µs asks for 0.05 × 0.5/10⁻⁶ = 25 kV. The transistor avalanches at its breakdown voltage instead and absorbs the coil's energy:

E = \tfrac{1}{2}LI^2 = \tfrac{1}{2}(0.05)(0.5)^2 = 6.25\ \text{mJ per turn-off}

Think deeper

Why does it survive the bench test but die after a week? And what is the fix?

Argue it in pairs for two minutes. There may be more than one defensible answer.

18 Pop quiz 73–78 min

Pop quiz

Score 0

Question 1 / 5

On the DC load line, where does a BJT switch sit when it is ON?

19 Recap 78–80 min

Three things to remember

01

A switch lives at two ends of the load line: OFF (V_CC, 0) and ON (0.2 V, I_C(sat)). Saturate with I_B ≥ I_C(sat)/β_min, times a 2–5× overdrive.

02

t_on = t_d + t_r, t_off = t_s + t_f. Overdrive speeds the rise but stores charge and lengthens t_s; reverse drive drains it.

03

Fixes: speed-up capacitor, Schottky clamp (74LS, Week 3), or a device with no stored charge: the MOSFET, next class. Always put a diode across a coil.

HW Before next class

β = 60, τ = 25 ns, τ_s = 50 ns, I_CS = 12 mA, I_B1 = 0.6 mA, I_B2 = 0.3 mA, t_d = 5 ns. Find t_r, t_s, t_f, t_on, t_off and the rough maximum switching frequency 1/(t_on + t_off).

Work it out first.
Press Solve when the class has an answer.