Question 8.5: For the circuit shown in Fig. 8.14, determine the RMS output......
For the circuit shown in Fig. 8.14, determine the RMS output and input-referred noise over a bandwidth from DC to 1 kHz. Verify your answer with simulations.
The PSD of the noise currents are given by (again noting, as seen in Fig. 8.6, that we apply zero volts to the input when calculating output noise PSD, that is, we ground the input)
I_{\mathrm{10k}}^{2}(f)={\frac{4\cdot(13.8\times10^{-24})\cdot(300)}{10,000}}=1.66\times10^{-24}\,{\frac{A^{2}}{H z}}\to I_{\mathrm{10k}}(f)=1.29\times10^{-12}{\frac{A}{\sqrt{H z}}}
{ I}_{1k}^{2}(f)=16.66\times10^{-24}\,\frac{A^{2}}{Hz}\rightarrow I_{1k}(f)=4.1\times10^{-12}\frac{A}{\sqrt{H z}}
The output thermal noise PSD due to the 10k resistor is
{I}_{10k}^{2}(f)\cdot\left[\frac{1 k\cdot10k}{1 k+10k}\right]^{2}~\mathrm{~units~}V^{2}/Hz
In order to avoid analyzing circuits in a different way, (for example, V^{2}=I^{2}R^{2}\,\ {i}_{d}^{2}= g_{m}^{2}\nu_{g s}^{2},\ \ V^{2}=I^{2}\big(R_{1}^{\phantom{\leftrightarrow}}+R_{2}\big)^{2}, etc.), we can use the voltage or current spectral densities (square-root of PSD) of the noise when doing circuit noise calculations
V_{10k}(f)=I_{10k}(\mathord{\mathord{f}})\cdot\frac{1k\cdot\mathrm{{10}}k}{1 k+10k} units V/{\sqrt{Hz}}
which evaluates to 1.2\ n V/\sqrt{Hz}. The output noise voltage-spectral density V_{1k}(f) from the 1k resistor is 3.7\ nV/\sqrt{Hz}\;. The total output noise PSD, {V^{2}}_{o n o i s e}(f), is the sum of {{V_{10k}^{2}(f)}} and {{V_{1k}^{2}(f)}}. Again, we sum the PSDs (or power) but not the voltage-spectral densities (or the RMS voltages).
The mean-squared output noise voltage over a bandwidth of 1 kHz is
V_{o n o i s e,R M S}^{2}=\int\limits_{f_{L}}^{f_{H}}V_{o n o i s e}^{2}(f)\cdot d f=\int\limits_{0}^{1kHz}\left(V_{10k}^{2}(f)+V_{1k}^{2}(f)\right)\cdot d f=15.1\times10^{-15}~V^{2}
The RMS output noise voltage is
V_{o n o i s e,R M S}=123\ nV
The input-referred noise voltage is
V_{i n o i s e,R M S}=123nV\cdot\frac{10k+1 k}{1 k}=1.35\ \mu V
SPICE simulation gives an output mean squared noise of 1.5053 \times 10^{-14}\ V^{2}\ (=123\ nV\ \ RMS) and an input-referred noise referenced to an input voltage of 1 V of 1.8215 \times 10^{-12}\ (=1.35\ \mu V\ \ RMS). The SPICE netlist is shown below.
| *** Example 8.5 CMOS: Circuit Design, Layout, and Simulation *** .control destroy all run print all .endc .noise v(2,0) Vin dec 100 1 1k R1 1 2 10k R2 2 0 1k Vin 1 0 dc 0 ac 1 .print noise all .end The SPICE output is seen below TEMP=27 deg C Noise analysis … 100% inoise_total = 1.821510e-12 onoise_total = 1.505380e-14 |
A bandwidth of 1 to 1,000 Hz was used in this simulation rather than DC to 1 kHz. Also note that the reference supply, Vin in the netlist, was a voltage, so the units of the SPICE output are V². The input AC source has a magnitude of 1 V, so the input noise SPICE gives is divided by 1 V squared. If we had used a 1 mV AC supply for Vin, then the inoise_total above would be 1.8215\times 10^{-6}\ V^{2}.
Here, in SPICE, we used a single input-referred voltage source to model the input-referred noise, see Eq. (8.22) and the associated discussion. In SPICE we always refer the input noise to a source (voltage or current) and not to a node (like we do in our analysis). This means that the input-referred noise in SPICE is always a single source in series (voltage input) or parallel (current input) with the input of the circuit.
V_{i n o i s e ,R M S}^{2}=V_{i n o i s e , R M S1}^{2}+V_{i n o i s e,R M S2}^{2}/A_{1}^{2}+V_{i n o i s e,R M S3}^{2}/(A_{1}A_{2})^{2} (8.22)
