Diode equivalent circuits

Diode equivalent circuits 1. Access resistances in vertical devices Top metal contact

There is an “access” resistance connected in series with an “ideal” diode “D” (the diode can be of any type, i.e. rectifier, LED, Varicap, Tunnel diode etc.).

Area = A

Rcp

p-type

Rsp

dp

D

Total access (series) resistance for the diode on the left:

Rs = Rcp+Rsp + Rsn + Rcn n-type

Rcn

Bottom metal contact

Rsn

dn

Rcp= ρcp/A; Rcn= ρcn/A Note the difference between lateral and vertical contacts. For lateral contacts, a unit-width contact resistance (Ω×mm or Ω×cm) is typically used. For vertical contacts, a unit-area specific resistance (Ω×mm2 or Ω×cm2) is typically used.

1. Access resistances in vertical devices (cont.) Top metal contact

The series resistance associated with the diode base regions: Rsp and Rsn:

Area = A

Rcp

p-type

Rsp

dp

1 Rsp = ρ p ; where ρ p = A q pμp is the resistivity of the p − material

dn

dn 1 Rsn = ρ n ; where ρ n = A q n μn is the resistivity of the n − material

D n-type

Rcn

Bottom metal contact

Rsn

dp

1. Access resistances in vertical devices (cont.) Diode low-frequency equivalent circuit Top metal contact

Area = A

Rcp

p-type

Rsp D

n-type

Rcn

Rs V

Rs = Rcp+Rsp + Rsn + Rcn

D

Rsn For a regular p-n junction diode, the total voltage drop across the device:

Bottom metal contact

V = VD + VRS VD = (kT/q) × ln(I/Is+1); VRs = I × Rs

1. Access resistances in vertical devices (cont.) Diode low-frequency equivalent circuit Top metal contact

5

Area = A

Ignoring Rs

Rcp

I×Rs

4

p-type

Rsp 3 I, A

D n-type

Rcn

Rs V

2

D Rsn

1 0 0

0.2

0.4

0.6

-1

Bottom metal contact

V, V

0.8

1

2. Access resistances in lateral devices Example: planar GaAs Schottky diode

Schottky

Ohmic W

n-GaAs

Rc = Rc1 /W;

i-GaAs

LSch

Lgap

Total access (series) resistance for the diode on the left: Rs = Rc+Rgap+ RSch

Rgap = Rsq × Lgap / W; LC

RSch ≈Rsq × (LSch / 3) / W; - accounts for the current spreading under the Schottky contact.

3. High-frequency equivalent circuits Example: vertical Schottky diode At high frequencies, capacitances and inductances associated with the diode itself and the package have significant impact on the device performance

ε ε A CB = S 0 Wdepl

CB

D n-type

Rs(incl. Rc)

Lp

capacitance associated with the diode depletion region Cp

Cp - the capacitance associated with the device contact pads or package

Lp - the inductance associated with the device wiring and/or package

3. High-frequency equivalent circuits Frequency dependence of the device impedance Example: Consider Schottky diode with the DC differential resistance of 10 Ω. Comment: the differential resistance, R = ∂V = ⎛ kT ⎞ I ≈ 0.026V at 300 K ; d I ∂I ⎜⎝ q ⎟⎠ Suppose: CB = 20 pF; Cp = 5 pF; Rs = 1 Ω; Lp = 0.5 nH

CB

D

Rs

Cp

The diode impedance can be found as follows: 1. Yd1 = 1/Rd + jωCB; Zd1 = 1/ Yd1; where ω=2πf; 2. Zd2 = Zd1 + Rs; Yd2 = 1/Zd2;

Lp

3. Yd3 = Yd2 + jωCp; Zd3 = 1/ Yd3; 4. Zdiode = Zd3 + jωLp;

Diode impedance frequency dependence %Diode impedance frequency dependence

Yd1 = 1/Rd + i*om*CB;

%Suppose: Rd=10 Ohm;

Zd1 = 1./Yd1;

%CB = 20 pF; Cp = 5 pF;

Zd2 = Zd1 + Rs;

%Rs = 1 Ohm; Lp = 0.5 nH

Yd2 = 1./Zd2; Yd3 = Yd2 + i*om*Cp;

Rd=10;

Zd3 = 1./Yd3;

CB = 20e-12;

Zdiode = Zd3 + i*om*Lp;

Cp = 5e-12;

Req=real(Zdiode);

Rs = 1;

Xeq=imag(Zdiode);

Lp = 0.5e-9;

figure(1);

%>>

plot(f*1e-6,Req);

f=1e6:1e6:3e9;

figure(2);

om=2*pi*f;

plot(f*1e-6,Xeq);

Diode impedance frequency dependence

CB

D

Rs

Lp

Cp

Diode impedance frequency dependence

CB

D

Rs

Lp

Cp