Transistor Switching

A BJT used as a switch must transition between the cutoff region and the saturation region. Unlike resistive switching, the transistor does not switch instantaneously. Minority carrier storage in the base region introduces time delays during both turn-on and turn-off, which directly limits the maximum operating frequency of digital circuits built with BJTs.

Core Concept Explanation

When a step input is applied to the base of a BJT to turn it on, the collector current does not rise immediately. First, the delay time (td) elapses. During this period, the input charges the emitter-base junction capacitance and the base current is building up, but VBE has not yet crossed the threshold needed to inject minority carriers. No significant collector current flows during td.

After the delay, minority carriers begin to enter the base and the transistor moves from cutoff toward active mode. The rise time (tr) is the time for IC to rise from 10 percent to 90 percent of its final saturation value. This time is governed by the minority carrier transit time through the base and the charging of junction capacitances.

Turn-off is more problematic. When the base drive is removed or reversed, the transistor does not immediately leave saturation. This is because the base region is flooded with excess minority carriers stored during saturation. The storage time (ts) is the time required to remove these excess carriers before the transistor can begin to leave saturation. Storage time is the dominant bottleneck in BJT switching speed.

Once the excess carriers are swept out and the transistor enters active mode, IC starts to fall. The fall time (tf) is the time for IC to drop from 90 percent to 10 percent of its saturation value. The total turn-off time is toff = ts + tf and is usually much larger than ton = td + tr.

Mathematical Expression

The storage time can be approximately expressed using the minority carrier lifetime in the base region. If the transistor is driven into saturation with a base overdrive factor, the storage time ts = τs × ln((IB1 + IB2) / (IB2 − IC/βF)), where τs is the effective minority carrier lifetime, IB1 is the saturation drive current, IB2 is the reverse base current applied to aid turn-off, and βF is the forward current gain.

The rise and fall times depend on the transit frequency fT of the transistor. Higher fT means smaller minority carrier transit time in the base, which leads to faster switching. Transistors used in high-speed digital logic are designed with very thin bases to maximize fT and minimize tr and tf.

A useful approximation is tr ≈ 0.35 / fT for the rise time limited by transit frequency. Saturation must be avoided in high-speed circuits because ts dominates total switching time.

Practical Understanding

Schottky clamping is the standard technique to prevent deep saturation in BJT switching circuits. A Schottky diode is connected between the base and collector. When the collector voltage tries to fall below the base voltage (entering deep saturation), the Schottky diode clamps the collector voltage. Since the diode has a low forward voltage of about 0.3 V, it prevents the transistor from entering deep saturation, drastically reducing ts and allowing faster switching.

In GATE problems, you may be asked to identify which time component is responsible for slow turn-off. The answer is always storage time ts. You may also be asked to compare switching times of BJT versus MOSFET, where MOSFET wins because it has no minority carrier storage.

Given:
τs = 50 ns (minority carrier lifetime in base)
IB1 = 0.5 mA (base drive during ON state)
IB2 = 0.3 mA (reverse base current for turn-off)
IC(sat) = 5 mA, βF = 50

Why this formula applies:
Transistor was in saturation (IB1 > IC/βF = 0.1 mA), so excess charge exists.

Formula:
ts = τs × ln((IB1 + IB2) / (IB2 − IC/βF))

Substitution:
IC/βF = 5 mA / 50 = 0.1 mA
Numerator: IB1 + IB2 = 0.5 + 0.3 = 0.8 mA
Denominator: IB2 − IC/βF = 0.3 − 0.1 = 0.2 mA

Calculation:
ts = 50 ns × ln(0.8 / 0.2) = 50 × ln(4) = 50 × 1.386

Final Answer:
ts ≈ 69.3 ns

Exam Tip: In GATE, storage time ts is always the largest switching delay in a saturated BJT. To reduce ts, either avoid saturation (Schottky clamp) or increase reverse base drive IB2 to sweep out stored minority carriers faster.

Mechanism: Carrier Storage During Switching

• Turn-on phases: delay time (td) for EBJ charging, then rise time (tr) as IC climbs from 10 to 90 percent of IC(sat).
• Turn-off phases: storage time (ts) to remove excess base charge, then fall time (tf) as IC drops from 90 to 10 percent.
• Storage time ts is the dominant bottleneck. It exists only because the transistor was in saturation. Avoiding saturation eliminates ts entirely.
• Schottky clamping prevents deep saturation by diverting excess base current through a Schottky diode before the CBJ becomes strongly forward biased.
• Higher transit frequency fT and thinner base both reduce tr and tf. Storage time is reduced by applying reverse base current (active turn-off).

Quick Revision

• Four switching time intervals: td (delay), tr (rise), ts (storage), tf (fall).
• ton = td + tr, toff = ts + tf. Turn-off is usually slower because of ts.
• Storage time ts exists because excess minority carriers are stored in the base during saturation.
• ts = τs × ln((IB1 + IB2) / (IB2 − IC/βF)). Larger reverse base drive IB2 reduces ts.
• Schottky transistor: Schottky diode clamps the CBJ to prevent deep saturation and eliminates storage time.
• Exam trap: MOSFET has no minority carrier storage, so it switches faster than BJT in saturation. BJT in active mode (not saturated) can also be fast.
• Rise and fall times scale inversely with fT. Thinner base, higher doping contrast, and smaller capacitances all increase fT.