Question 8.9: Calculate the input-referred noise, F, and SNRs for the circ......

Calculate the input-referred noise, F, and SNRs for the circuit seen in Fig. 8.21.

8.21
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

Let’s begin by adding the noise voltage spectral density to the circuit, Fig. 8.22a. The output noise PSD is

V_{o n o i s e}^{2}(f)=4k T R_{s}\biggl[{\frac{R_{i n}}{R_{i n}+R_{s}}}\biggr]^{2}+4k T R_{i n}\biggl[{\frac{R_{s}}{R_{i n}+R_{s}}}\biggr]^{2}

To determine V_{o n o i s e,R M S}, we integrate this PSD over the bandwidth of interest B or

V_{o n o i s e,R M S}^{2}=\int\limits_{fL}^{fH}{V^2_{onoise}} (f)\cdot d f=4k T B R_{s}\Bigl\lbrack\frac{R_{i n}}{R_{i n}+R_{s}}\Bigr\rbrack^{2}+4k T B R_{i n}\Bigl\lbrack\frac{R_{s}}{R_{i n}+R_{s}}\Bigr\rbrack^{2}

Noting our gain A (=V_{out}/V_{in}\ \text{not}\ V_{out}/V_{s}) is one, we can use the model shown in Fig. 8.22b. To determine the input-referred noise sources, we can use Eq. (8.32) and the results in Ex. 8.7. To determine V_{inoise,RMS}, we short the input to ground (R_s = 0 in Fig. 8.21 and the equation above), Fig. 8.22c, and equate the circuit output to V_{onoise,RMS}, This gives

V_{o n oi s e,R M S}^{2}=4k T R_{s}B\cdot\left\lgroup\frac{A R_{i n}}{R_{s}+R_{i n}}\right\rgroup ^{2}+I_{i n o i s e,R M S}^{2}\cdot\left\lgroup\frac{A R_{s}R_{i n}}{R_{s}+R_{i n}}\right\rgroup ^{2}+V_{i n o i s e,R M S}^{2}\cdot\left\lgroup\frac{A R_{i n}}{R_{s}+R_{i n}}\right\rgroup ^{2}          (8.32)

V_{o n o i s e,R M S,Rs=0}=V_{i n o i s e,R M S}=0

To determine I_{inoise,RMS}, we open the input (R_s=\infty ), Fig. 8.22d, and equate the circuit’s output to V_{onoise,RMS} (from the equation above). This gives

R_{i n}^{2}\cdot I_{i n o i s e,R M S}^{2}=V_{o n o i s e,R M S,Rs=\infty }^{2}=4k T B R_{i n}\rightarrow I_{i n o i s e,R M S}=\sqrt{\frac{4k T B}{R_{i n}}}

The input SNR is given in Eq. (8.29). The output SNR, Fig. 8.22e, is

SNR_{in}=\frac{V^2_{s,RMS}\cdot \left[\frac{R_{in}}{R_{in}+R_{s}} \right]^2 }{4kTR_sB\ \cdot \left[\frac{R_{in}}{R_{in}+R_{s}} \right]^2}=\frac{V^2_{s,RMS}}{4kTR_sB}          (8.29)

S N R_{o u t}={\frac{V_{s,R M S}^{2}\cdot\left[{\frac{R_{i n}}{R s+R_{i n}}}\right]^{2}}{V_{o n o i s e,R M S}^{2}}}={\frac{V_{s,R M S}^{2}}{4k T B\cdot R_{s}(1+R_{s}/R_{i n})}}          (8.40)

The noise factor is then

F=1+\frac{R_{s}}{R_{i n}}          (8.41)

To minimize the NF, we can decrease R_s or increase R_{in}. Decreasing R_s causes SNR_{in} and SNR_{out} to increase, as seen in Eqs. (8.29) and (8.40). At the same time, increasing R_{in} causes SNR_{out} to move towards SNR_{in}, Eq. (8.40), resulting in F moving towards 1.

8.22a
8.22b
8.22c
8.22d
8.22e

Related Answered Questions

Question: 8.4

Verified Answer:

The term kT has units of Joules (or Watt-sec) and ...
Question: 8.11

Verified Answer:

Again, as determined in Ex. 8.5, the total output ...
Question: 8.18

Verified Answer:

The schematic used for simulations is seen in Fig....