BJT PNP Operation

The 2N3906 PNP transistor works like a mirror image of its NPN counterpart. Here the emitter is the most positive terminal, and conventional current flows from emitter to collector when the base is pulled low.

Core Concept

A PNP transistor has a P-type emitter, an N-type base, and a P-type collector. The emitter is connected to the most positive supply rail. To turn the transistor ON, the base must be pulled to a voltage at least 0.7V below the emitter voltage.

When the base is pulled low enough, the emitter-base junction becomes forward biased. Holes are injected from the emitter into the thin N-type base. Most holes drift across the base and are swept into the collector by the reverse-biased collector junction. A small fraction recombines in the base, forming the base current.

The 2N3906 is the PNP complement to the 2N3904 NPN. It handles up to 200 mA collector current and has VCEO of 40V. In circuit schematics, you always draw the PNP emitter at the top (connected to VCC) and the collector pointing downward toward the load.

Key Equations

For a PNP transistor in active region, the magnitudes of the currents follow the same β relationship: IC = β × IB. All three currents flow INTO the device for a PNP (by convention, positive current enters emitter, exits base and collector).

The turn-on condition is: VEB = VE - VB = 0.7V. Note this is VEB, not VBE. The emitter is more positive than the base. This is the opposite of NPN, where VBE = VB - VE = 0.7V.

Emitter current for a PNP with emitter resistor: IE = (VCC - VEB) / RE. The KVL loop starts at VCC, drops VEB across the junction, and drops the rest across RE.

Given:
  VCC = 9V
  RE = 1 kΩ
  RC = 2.2 kΩ
  VEB = 0.7V
  β = 120 (2N3906 typical)

Why this formula:
  KVL from VCC through emitter junction to ground gives IE.
  IC ≈ IE for large β.

Formula:
  IE = (VCC - VEB) / RE
  IC ≈ IE × (β / (β+1))
  VEC = VCC - IC × RC   [note: VEC for PNP, not VCE]

Substitution:
  IE = (9 - 0.7) / 1000
  IC = IE × (120/121)
  VEC = 9 - IC × 2200

Calculation:
  IE = 8.3 / 1000 = 8.3 mA
  IC = 8.3 × (120/121) = 8.3 × 0.9917 = 8.23 mA
  VEC = 9 - (8.23 × 10^-3 × 2200)
      = 9 - 18.1 = negative!

  Negative VEC means transistor is saturated.
  In saturation: VEC(sat) ≈ 0.2V
  Reduce RC to 560Ω and recalculate:
  IC = 8.23 mA, VEC = 9 - 8.23e-3 × 560 = 9 - 4.61 = 4.39V

Final Answer:
  With RC = 560Ω: IC = 8.23 mA, VEC = 4.39V (active region confirmed)

Exam Tip: The most common PNP mistake is writing VBE = 0.7V instead of VEB = 0.7V. In a PNP, the emitter is positive. Write KVL as VCC - VEB - IE×RE = 0. If you write it the NPN way (VBE), you will get a negative IB and everything falls apart. Also note VEC for PNP, not VCE.

Key Properties

• PNP is ON when the base is pulled at least 0.7V below the emitter. The emitter is the most positive terminal.
• 2N3906 has β (hFE) typically 100 to 300, VCEO = 40V, IC(max) = 200 mA, and power dissipation of 625 mW.
• Conventional current enters the emitter of a PNP. It exits through base and collector terminals.
• The arrow on the PNP emitter symbol points INTO the base, opposite to the NPN arrow. This arrow shows the direction of conventional current.
• PNP transistors are often used as high-side switches, connecting the load between the collector and ground, with VCC at the emitter side.
• Hole mobility is lower than electron mobility, so PNP transistors are inherently slightly slower than equivalent NPN devices.

Quick Revision

• PNP: P-emitter (top/positive), N-base, P-collector.
• Turn-on condition: VEB = 0.7V (emitter more positive than base).
• Current arrows: into emitter, out of base and collector.
• Same β formula: IC = β × IB. Same α formula: IC = α × IE.
• 2N3906 is the PNP complement of 2N3904. They share the same pin configuration.
• KVL for PNP base loop: VCC - VEB - IB×RB = 0 (when base is connected through RB to ground).
• Exam trap: Writing VBE = 0.7V for a PNP transistor instead of VEB = 0.7V. This changes the sign of VB and gives completely wrong results for IB and IC.