Charles daEngineer

Determining Noise for Current Amplifier

Noisy Amplifier Starting Point — Noisy Current Amplifier Circuit

Our first step is transforming all noise sources into equivalent noise free sources.

Here we shift the noisy voltage and current sources from the input of the nullor into the feedback loop and the input of the source.

Here we shift the voltages from the feedback loop into the node where Z1 is located.

Shifting Feedback Loop Voltages to output

We can move the voltages sources at the output of the nullor to the input, but since the nullor is all zeros we can remove them completely. The next step is transforming the voltage sources at Z1 to current sources.

Transforming Voltage sources and nulling outputs

Now we shift the current to the output of the feedback loop using current transforms.

Now we can shift the currents from the output of the feedback loop to the input of the feedback loop by multiplying by the $(\begin{matrix} A & B \\ C & D \end{matrix})$ where $A = B = C = 0$ . Lastly we also transfrom all remaining voltages at the input to current sources.

Transforming Current Sources from output to input

Finally we can find the equivalent input noise current as:

$i_{e q} = V_{n n} (1 / Z_{s} + D / Z_{1}) + V_{i n s} / Z_{s} + I_{n n} - V_{n 1} D / Z_{1} + V_{n 2} D / Z 1$

With $D = \frac{Z_{1}}{Z_{1} + Z_{2}}$ we can easily find the spectral density as:

$S_{i e q} = S_{V n n} {| 1 / Z_{s} + 1 / (Z_{1} + Z_{2}) |}^{2} + S_{V i n s} / {| Z_{s} |}^{2} + S_{I n n} + S_{V n 1} / {| Z_{1} + Z_{2} |}^{2} + S_{V n 2} / {| Z_{1} + Z_{2} |}^{2}$

Lastly the Input Refered Voltage noise is the voltage noise from the nullor.

Self Check

Lets check our answer by saying all the impedances are Resistors and simplifying.

$S_{i e q} = S_{V n n} \frac{{(R_{1} + R_{2} + R_{s})}^{2}}{R_{s}^{2} {(R_{1} + R_{2})}^{2}} + S_{V i n s} / R_{s}^{2} + S_{I n n} + S_{V n 1} / {(R_{1} + R_{2})}^{2} + S_{V n 2} / {(R_{1} + R_{2})}^{2}$

Now since the spectral noise density of the resistors are $4 k T R$ then we have:

$S_{i e q} = S_{V n n} \frac{{(R_{1} + R_{2} + R_{s})}^{2}}{R_{s}^{2} {(R_{1} + R_{2})}^{2}} + 4 k T / R_{s} + S_{I n n} + \frac{4 k T R_{1}}{{(R_{1} + R_{2})}^{2}} + \frac{4 k T R_{2}}{{(R_{1} + R_{2})}^{2}}$

SLiCAP Verification

The results from SLiCAP were

$S_{i n} = \frac{4 R_{1} T k}{{(R_{1} + R_{2})}^{2}} + \frac{4 R_{2} T k}{{(R_{1} + R_{2})}^{2}} + S_{i} + \frac{4 T k}{R_{s}} + \frac{S_{v} {(R_{1} + R_{2} + R_{s})}^{2}}{R_{s}^{2} {(R_{1} + R_{2})}^{2}} [\frac{A^{2}}{Hz}]$