Diode Circuit Analysis • Goal: find quiescent operating point (Q-point) of the diode ( , ) • Analytical tools: – Kirchhoff’s voltage law (KVL) – Kirchhoff’s current law (KCL) – Element relations • Solution methods (Often can’t solve analytically due to non-linearity) – Graphical methods – Numerical iteration
Trang 1Diode Circuit Analysis
¢ Goal: find quiescent operating point (Q-point) of
the diode (Ip, Vp}
¢ Analytical tools:
— Kirchhoff’s voltage law (KVL)
— Kirchhoff’s current law (KCL)
— Element relations
¢ Solution methods (Often can’t solve analytically due to
non-linearity)
— Graphical methods
— Numerical iteration
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 2Diode Circuit Analysis
¢ Example: Given vp, R, and parameters for the diode
(Ic, Vr), find Q-point Up, Vp)
| +
Vo = IpR + Vp
¢ Nonlinear part (diode)
Up
Íp = ls(@”T — 1)
¢ Two equations, two
unknowns Ip, Vp
linear ! nonlinear
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 3Load Line Analysis (Graphical)
¢ Write the two equations in the form of Lp v.s Vp
p =— +; Íp = lc(eŸ7r — 1)
¢ Plot them onasame graph, find the intersection
Up
Íp = ls(@”T — 1)
linear nonlinear
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 4Numerical Analysis
¢ Combining:
p =— +; in = lc(e T— 1)
° We have: ——> + -5 = Is(e’r — 1) => vp = Vrln (1 + S72)
© Let R = 1k0,I> = 107*°A, v; = 26 mV, V; = 10 V we have
vp = 0.026 In (1 + ae)
¢ Make an initial guess (vp = 0.5 V) and solve by iteration:
in = — 2 + ¬ =9522mA Q-point = (9.522 mA, 0.4777 V)
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 5Numerical Analysis
¢ Now use Expect diode to be reverse biased
¢ From graphical analysis, Thus
¢ Q-point = (107° A, -10 V)
¢ Be careful of round-off error:
' should be 9.999999999 V `
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 6Simplified Diode Models
General diode Ideal diode Constant voltage drop
“ON”
Up = 0
a,
lp = 0
> ON: short circuit ©——O OQN: voltage source o~=É~o
OFF: open circutO 9% O OFF:opencircuit O %$ O
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 7Diode Circuit Analysis —
simplified model
¢ Analysis Method:
— Guess a state for the diode, ON or OFF
— Replace the diode by its equivalent model for this
state
— Analyze the circuit using this equivalent model
— Verify that the guess was correct
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 8Diode Circuit Analysis
Ideal model
đỗ Vp Vs V„=0
Since /, = 10 mA 2 0, D= ON correct
ak ,
“ON” condition: v) =0,ip >O |
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 9Diode Circuit Analysis
LVD model
4 |" 4 |"
*10V +10 V +0.6 V
Von = 0.6V since I, = 9.4 mA 2 0, D = ON correct
“ON” condition: vp =Von,ip = 0 |
Guess OFF: 8 1kO
Ve + Vọ= V O, since V,=10V 2 V,,, D = OFF wrong!
+10V\ - = 10V 7 “OFF” condition: ip = 0, vp S Von “OFF” condition: ip = 0, Vp < Von
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 10Multiple Diode Circuits
>
Vì
là /\ D, en O 20k
D,, D, = deal diodes
Textbook Schematic
D,
Engineering Schematic +
10kO V -15V
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 11Multiple Diode Circuits
V 15V
Guess: D,, D, both ON
Check:
, |
R5
both D, and D, are ON Correct! 10kO
Vo, -15V
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 12Multiple Diode Circuits
V 15V
Now: Swap resistors
Guess: D,, D, both ON
Check:
=> Contrary to D, ON!
20kO
V.„ -15V
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham
Trang 13Multiple Diode Circuits
V 15V
Guess: D, OFF, D, ON
C )
Check:
=> D, off Correct
20kO
Vo, -15V
EE 331 Spring 2012 Microelectronic Circuit Design © UW EE Chen/Dunham