Diode

Diode

Definition

A two-terminal device that carries current when its voltage is positive and no current when its voltage is negative.

Diode 1

I_x = I_s (e^{V_x \over V_T} -1)

不同半导体材料构成的二极管的反向饱和电流I_s不同 ：

Diode 2

因此， I_s是二极管的固有属性，与电压电流无关。

Note

If the forward bias voltage V_x > 4V_T, I_x \approx I_s e^{V_x \over V_T}

(通常情况下，我们采用此式来代表二极管两端电压和电流的关系)

Quiz

Diode 3

Since,
I_x \approx I_s e^{V_x \over V_T}
Therefore,
V_x = V_T \ln {I_x \over I_s}

Examples

• If I_s = 10^{-16} A and we measure a forward-bias current of 1 mA, what's voltage is applied to the diode?

V_F = V_T \ln {I \over I_s} = 778mV

P.S. e^{778mV \over 26mV} >> 1

• How much should the voltage increase to raise the current by a factor of 10?
  V_{F1} = V_T \ln{I_1 \over I_s}
  V_{F2} = V_T \ln{I_2 \over I_s}
  V_{F2} - V_{F1} = V_T \ln {I_2 \over I_1}
  For 10x increase in current, we need to increase the voltage by V_T \ln 10 \approx 60 mV

P.S. V_T = {k_T \over q} \rightarrow 26mV @ T = 300k

A beautiful result : If we have an exponential device which is the case with a diode, in order to increase its current by a factor of 10, all we need to do is increase its voltage by 60 mV.

Typically, forward-bias voltages for diodes are in the range of 700 - 800 mV