BJT input resistance

The BJT input resistance $r_{\pi }$ is the small-signal resistance seen looking into the base of a BJT in active mode:

$r_{\pi }=\frac{v_{be}}{i_b}=\frac{\beta }{g_m}=\frac{\beta V_T}{I_C}$

Unlike the MOSFET, whose gate is insulated by oxide and draws no DC current (infinite input resistance), the BJT base passes a real base current. Apply a small AC voltage $v_{be}$ across base-emitter and an AC base current $i_b$ flows; their ratio is a finite resistance.

Derivation

The collector responds to $v_{be}$ through the transconductance: $i_c=g_m{v}_{be}$. The base only ever carries $1/\beta$ of the collector current (that is the definition of the Common-emitter current gain $\beta =i_c/i_b$), so

$i_b=\frac{i_c}{\beta }=\frac{g_m{v}_{be}}{\beta }$

Therefore the resistance looking into the base is

$r_{\pi }=\frac{v_{be}}{i_b}=\frac{v_{be}}{g_m{v}_{be}/\beta }=\frac{\beta }{g_m}$

Substituting $g_m=I_C/V_T$ gives the equivalent form $r_{\pi }=\beta V_T/I_C$.

BJT input resistance at the base: $r_{\pi }=\beta /g_m=\beta V_T/I_C$. Finite (unlike the MOSFET), a few kΩ at standard bias currents.

Numbers

At $I_C=1\text{mA}$ with $\beta =100$ and $V_T=25\text{mV}$:

$r_{\pi }=\frac{(100)(25\text{mV})}{1\text{mA}}=2.5\text{k}\Omega$

A few kilo-ohms matters: many real signal sources have output impedances comparable to or larger than this, so a Common-emitter amplifier loads its source noticeably. This is one of the few genuine disadvantages of the BJT versus the MOSFET in amplifier design. A common-source MOSFET stage presents an essentially infinite input resistance, a common-emitter BJT stage only $r_{\pi }$.

$r_{\pi }$ scales with $\beta$ and inversely with $I_C$, so bias the device at a lower current to raise its input resistance (at the cost of lower $g_m$). It’s the base-emitter element of the hybrid-π form of the BJT small-signal model. The much smaller resistance looking into the emitter instead is $r_e\approx r_{\pi }/(\beta +1)$, the Emitter resistance; the difference is which terminal carries the full current.