S-Parameters – They Are Easy!

Home , Gain (electronics), Reflection coefficient

ECE145A/218A Notes Set #4 1

2–Port Parameters

Two-ways of describing device:

A. Equivalent - Circuit-Model

• Physically based • Includes bias dependence • Includes frequency dependence • Includes size dependence - scalability • Ideal for IC design • Weakness: Model necessarily simplified; some errors. Thus, weak for highly resonant designs

B. 2–Port Model

• Matrix of tabular data vs. frequency • Need one matrix for each bias point and device size • Clumsy – huge data sets required • Traditional microwave method • Exact

2 Port descriptions

These are black box (mathematical) descriptions.

I I 1 2

+ port port V + V 1 – 1 2 – 2

Inside might be a transistor, a FET, a transmission line, or just about anything.

The terminal characteristics are V1 V2 I1 & I2 – there are 2 degrees of freedom.

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 2

Admittance Parameters

⎡ I1 ⎤ ⎡Y 11 Y12⎤ ⎡V 1⎤ = ⎢ I ⎥ ⎢Y Y ⎥ ⎢V ⎥ ⎣ 2 ⎦ ⎣ 21 22⎦ ⎣ 2 ⎦

Example: Simple FET Model

C gd g V m gs

+ Cgs V Rds – gs

By inspection:

⎡ jωCgs + jωCgd −jωCgd ⎤ Y = ⎢ g − jωC G + jωC ⎥ ⎣ m gd ds gd⎦

Easy!

II YY==11 11VV 12 12VV21=00=

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 3

Impedance Parameters

⎡V 1 ⎤ ⎡ Z11 Z12⎤ ⎡ I1 ⎤ = ⎢V ⎥ ⎢ Z Z ⎥ ⎢ I ⎥ ⎣ 2 ⎦ ⎣ 21 22⎦ ⎣ 2 ⎦

Example

R 1 R 2

R 3

By inspection

⎡ R1 + R3 R3 ⎤ Z = ⎢ R R + R ⎥ ⎣ 3 2 3 ⎦

VV ZZ==12 11II 21 11II22=00=

But, y, z, and h parameters are not suitable for high frequency measurement.

Problem: How can you get a true open or short at the circuit terminals? Any real short is inductive. Any real open is capacitive.

To make matters worse, if you are trying to measure a high freq. active device, a short or open can make it oscillate!

Solution: Use termination in Z0 instead!

Broadband. Not very sensitive to parasitic L,C Kills reflections.

Redefine parameters to use fwd. and rev. voltage waves.

Measurement can use directional couplers.

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 4

S–Parameters

Z 0 Z 0

a 1 a 2 b 1 b2 Zo Zo z = 0 z = 0

input reflection coeff a2 =0 rev. transm. gain a1=0

⎡ b1⎤ ⎡ S11 S12 ⎤ ⎡ a1 ⎤ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎣ b 2 ⎦ ⎣S 21 S22 ⎦ ⎣ a2 ⎦

fwd transm. gaina =0 output Γ 2 a1=0

Note that Z0 must be defined. We don’t really need transmission lines.

Our objective now is to de-mystify S-parameters – they are easy!

Recall

V(x) = V +(x) + V −(x) phasor quantities. V +(x) V −(x) I(x) = − amplitude, not rms values. Z0 Z0

We can normalize the amplitude of waves to Z0:

V +(x) a(x) = forward wave Z0

V −(x) b(x) = reverse wave Z 0 1 Why? So that a(x)a* (x) = power in forward wave. 2 if a = 1.414 then power in wave is 1 watt. (or a rms = 1)

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 5

likewise, b(x)b*(x)/2 is the power in the reverse wave

So, in terms of total voltage V(x) and current I(x),

Vx( ) vx()= = ax()+ bx() Z0 = = − ix () Z0 Ix() ax() bx()

or,

1 1 a(x) = []v(x) + i(x) = []V (x) + Z0I(x) 2 2 Z0

1 1 b(x) = []v(x) − i(x) = []V (x) − Z0I(x) 2 2 Z0

Reflection

So, how is Γ defined in terms of the S parameters? At port 1,

b1 Γ1 = a1 But,

bSaSa1111122= +

We need to eliminate a2. How?

a2 If ZL = Zo, Γ=L 0 = so, therefore a2 = 0 if port 2 is terminated in Zo. b2

b Γ=1 =S 111a 1 a2 =0 Same with at port 2 with S22: b S = 2 =Γ 22a 2 2 a1=0

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 6

Transmission

bSaSa2211222= +

So, the forward transmission S21 can be found by setting a2 = 0 (terminate output)

b S = 2 21 a 1 a2 =0

Reverse transmission, similarly, is found by setting a1 = 0 (terminate input in Zo)

bSaSa1111212= +

b S = 1 12 a 2 a1=0

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 7

Some comments on power measurement:

Power can vary over a large range, therefore it is often specified on a logarithmic scale. There must be a point of reference on the scale; the power measurements are usually with reference to 1 mW.

The unit is called dBm meaning dB relative to 1 mW of power. Thus,

0 dBm = 1 mW 10 dBm = 10 mW -10 dBm = 0.1 mW etc.

To convert mW to dBm:

dBm = 10 log10 (P)

To convert dBm to mW:

P = 10dBm/10

What is the difference between dB and dBm? dB is a power ratio – used to describe a gain or loss for example.

G = 10 log10 (Pout/Pin) dB Return Loss = - 20 log10 |Γ| dB

But, dB says nothing about the absolute power level. Don’t confuse their usage!

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 8

Now, define available power:

PAVS = max power output from a source with impedance Zs that can be absorbed into a load.

* let Z S =Z0, ZL =ZS = Z0 (in this case)

because maximum power transfer occurs when we have a conjugate match

Vgen Vgen/2 Z0

generator load

2 1 Vgen PPload== AVS 8 Z0 Or, in terms of a and b:

a 1

b 1 + Z 0 + b = 0 V V Z gen ~ 0 –

+ VVgen+−⎛⎞Z0 a11==== and b0; V Vgen ⎜⎟ and V = 0 Z0 ⎝⎠ZZ00+ 2

So, 2 1 * Vgen PPload== AVS aa11 = 28Z0

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 9

We see that the available power is independent of load impedance. Even if the load is not matched, available power remains constant. Actual power in the load is reduced however.

Generator output power is calibrated and displayed as available power.

Actual Load Power 11122 P=−=ab Re⎡ IV* ⎤ Load222 1 1⎣ 1 1 ⎦ or

2 PPLoad=− AVS (1 S11 )

Reflected Power b1 = a1 S11

11222 2 Pb== aSPS = RAVS22111111 2 Power reflected from input b 2 = = 1 S11 2 Power incident on input a1

2 2 Power reflected from network output b S = = 2 22 Power incident on output 2 a2 Similarly, 1 2 a2 = Power incident on output 2 = Reflected power from load

1 2 b1 = Power reflected from input port 2

1 2 b2 = Power incident on load from the network 2

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 10

a → 1 ← a 2 → b 1 ← b 2

Pload Also, by definition, transducer gain = = GT even if Pavs 1. load isn’t matched to network and 2. input of network not matched to generator

22 Here, PbLoad=−Γ2 (1 L )

S21 is defined in terms of transducer gain for the special case of where Z L = Z0 :

2 2 b2 S21 = 2 a1 a2 =0

1 2 b = power incident on load (and is absorbed since ΓL=0) 2 2 1 2 a = source available power 2 1

2 = transducer gain with source and load S 21 Z 0

Similarly,

2 = reverse transducer power gain S 12

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 11

Reference Planes

B C Microwave transistor in package

On board:

B C

⎡ S11 S12⎤ []S = ⎢ S S ⎥ ⎣ 21 22⎦

connection to instruments hereDefine defining x = 0 at zboth = 0 portshere.

Defining the reference planes differently changes the S-parameters.

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 12

phase shifts! 50 Ω microstrip transmission lines

⎡ S e j 2θ1 S e j(θ1+θ2)⎤ a S˜ = 11 12 b 1 2 ⎢ j()θ1+θ2 j 2θ2 ⎥ ⎣ S21e S22e ⎦ b2 a 1 1 2 1 x =− x = − θ1 = 2π 1 1 = 2 2 λ x 1 0 x 2 = 0 2 θ2 = 2π λ connections to instruments here

2π θβ==−x 1 11 λ

2π θβ==−x 2 22 λ

⎛ 2θ1 j(θ1+θ2) ⎞ ' ⎜ S11 e S12e ⎟ S = ⎜ j(θ1+θ2) j2θ2 ⎟ ⎝ S21e S22e ⎠

The reflection parameters are shifted in phase by twice the electrical length because the incident wave travels twice over this length upon reflection. The transmission parameters have the sum of the electrical lengths, since the transmitted wave must pass through both lengths.

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 13

Comment on electrical length:

The microwave literature will say a line is 43° long at 5GH z . What does this mean? f ref

Electrical length = E = ⋅ 360° λ ref

Recall f ⋅ λ = v so fref λref = υ

→ E = ⋅360°= ⋅ f ⋅ 360° v / f v ref ref

E = T ⋅ f ⋅ 360 ref

a line which is 1 ns long has an electrical length E = 360° at fref =1 GHz

and

an electrical length E = 36° at F = 100 MHz ref

Why not just say Τ= 1ns?

…you should be conversant with both terminologies.

Converting to physical length

f λref= v p v λ = p ref f

E(deg)λref thus: physical length = = Electrical length (in wavelengths) λ 360 ref

or:

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 14

How to Calculate S-Parameters Quickly

First Comment

b1 S11 = a1 a2 =0

b1 = S11a1 + S12a2

(We must kill a2 in order to measure or calculate S11) Γ L

S Z L

→ b 2 ← a 2

if Z L = Z0 , then Γ L is zero and so a =Γ b = 0. 2 L 2

b1 S11 = a1 ZL =Z0

So if we say that Zin is the input impedance with Z0 = ZL ZL =Z0 then

Z − Z in ZL =Z0 0 S11 = =Γin Zin + Z0 ZL =Z0 or:

1 + S Z = 11 in ZL =Z0 1 − S11

The same comment clearly applies for S 22 . The Smith Chart is often used to plot S 11, S22 .

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 15

Example:

4 Ω

Given Z 0 = 50Ω , what is S 11?

4 Ω

50 Ω

Z = 54Ω in ZL =Z0 54 − 50 4 S11 = = 54 + 50 104

4 Similar arguments give S22 = . 104

Find S21 b S = 2 | 21a2 = 0 a1

ΓS Z S = Z0 a a 1 2 → ← Z L = Z0 ~ V ← → gen b 1 b 2

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 16

What is a1 in this case?

+ V1 + Vgen We know that: a1 = and V1 = Zo 2

So, Vgen a1 = 2 Zo

Vout Consider the load: b2 = Why? Z0

b2 → ← + a 2 Z L V out _

a 2 = ΓLb2

But, Γ L = 0 because Z L = Z0 , so a 2 = 0.

+ − Vout = V + V = Z0 a2 + Z0 b2

= Z0 b2

Now, calculate Vout/Vgen: Vout = Z0 b2 = Z0 (S21a1 + S22a2 )

But, a2 = 0 because the load impedance = Z0, so

Vout = Z0 S21 a1

Substitute for a1:

Vgen a1 = 2 Z0 so, V Z S S out = 0 21 = 21 Vgen 2 Z0 2

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 17

2Vout thus, S21 = when ZL = ZS = Z0 Vgen

Why the factor of 2?

Vgen Vgen/2 Z0

generator load

We see that the generator voltage is split between the source and load in the matched case. Here, 2 we see that Vout/Vgen = ½, but the transducer gain must be equal to 1. (PLOAD/PAVS). |S21| is the transducer gain in this situation. If we insert an amplifier into the network, the signal has been increased by an amount S21.

Vgen Vout = S21 Vgen/2 Z0

generator load

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 18

2 So, |S21| is the FORWARD INSERTION GAIN or FORWARD TRANSDUCER GAIN in a system of impedance Z0.

EXAMPLE: Find S21

50 4

Vgen 50 S21= 2 Vout/Vgen

Vout/Vgen = 50/104 = 0.48 S21 = 0.96 0

OR, we could let Vgen = 2. Then, S21 = Vout.

What about a reference plane extension?

X1 = 0 X2 = 0 X1 = - l1 X2 = - l2

50 4

Vgen 50 S21= 2 Vout/Vgen

Θi = 2π li /λ

2jΘ 2jΘ S11’ = S11 e 1 S11 = ΓIN(0) S22 = ΓOUT(0) S22’ = S22 e 2 and

22π π θβ=− =− θ =− β =− 11λλ 1 2 2 2

' j()θ12+θπ−+ 2j()/ 12λ SSe21== 21 Se 21

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 19

EXAMPLE: Find the 4 S parameters of the following circuit:

C Z0 Vgen

S11: Find Zin (with ZL = Z0), then calculate input reflection coefficient.

Z | =1/(sC + 1/ Z ) IN ZL = Z0 0

Z IN −1 Z − Z Z S = IN 0 = 0 11 Z Z IN + Z0 IN + 1 Z0 turning the crank,

− jωCZ0 / 2 S11 = 1+ jωCZ0 / 2

S22 will be the same due to symmetry. Note that we calculated ZIN with port 2 terminated in Z0. This is part of the definition of S11 so is essential.

Rev.11/07 Prof. S. Long/ECE/UCSB ECE145A/218A Notes Set #4 20

Now find S21: first use Thevenin – Norton transformation:

Vout

Vgen/Z0 Z0 C

Vgen 1 Vout = =I/Y Z0 2 + sC Z0

2Vout 1 S21 = = = S12 Vgen 1 + jωCZ0 / 2

Rev.11/07 Prof. S. Long/ECE/UCSB