Let's break down the problem and solve **step-by-step**: --- ## **Given Data** - **Common-mode input impedance of amplifier (\(Z_{CM}\))**: \(1~M\Omega\) - **Source impedance (\(Z_S\))**: \(100~\Omega\) - **Source imbalance**: \(0.5\%\) (i.e., one leg is \(100~\Omega\), the other is \(100~\Omega + 0.5\% \times 100~\Omega = 100.5~\Omega\)) - **Amplifier CMRR (\(\mathrm{CMRR_{amp}}\))**: \(80~\mathrm{dB}\) --- ## **Step 1: Convert CMRR to Ratio** \[ \mathrm{CMRR_{amp}} = 80~\mathrm{dB} \] \[ \mathrm{CMRR_{amp~(ratio)}} = 10^{(80/20)} = 10^4 = 10{,}000 \] --- ## **Step 2: Calculate Source Imbalance CMRR** Imbalance causes **differential signal** to be created from common-mode input due to mismatch in the source impedances. The mismatch in impedance is: \[ \Delta Z = 0.5\% \times 100~\Omega = 0.5~\Omega \] The **CMRR due to source imbalance** is (approximate formula): \[ \mathrm{CMRR_{source}} = \frac{Z_{CM}}{\Delta Z} \] Where \(Z_{CM}\) is the **common-mode input impedance** of the amplifier. \[ \mathrm{CMRR_{source}} = \frac{1,000,000~\Omega}{0.5~\Omega} = 2,000,000 \] --- ## **Step 3: Find **Worst-case Effective CMRR** The **worst-case effective CMRR** is determined by the **interaction of amplifier CMRR and source imbalance**, and is given by: \[ \mathrm{CMRR_{effective}} = \frac{1}{\left(\frac{1}{\mathrm{CMRR_{amp}}}+\frac{1}{\mathrm{CMRR_{source}}}\right)} \] Plug in the numbers: - \(\mathrm{CMRR_{amp}} = 10,000\) - \(\mathrm{CMRR_{source}} = 2,000,000\) \[ \frac{1}{\mathrm{CMRR_{effective}}} = \frac{1}{10,000} + \frac{1}{2,000,000} \] \[ \frac{1}{10,000} = 0.0001 \] \[ \frac{1}{2,000,000} = 0.0000005 \] \[ \text{Sum: } 0.0001005 \] \[ \mathrm{CMRR_{effective}} = \frac{1}{0.0001005} \approx 9,950 \] --- ## **Step 4: Convert Back to dB** \[ \mathrm{CMRR_{effective(dB)}} = 20 \log_{10}(9950) \] Since \(\log_{10}(9950) \approx 3.997\): \[ 20 \times 3.997 = 79.94~\mathrm{dB} \] --- ## **Step 5: Write Final Answer** \[ \boxed{\text{The worst-case effective CMRR is approximately } 9,950 \text{ (or } 79.94~\mathrm{dB}\text{)}} \] --- ### **Summary Table** | Parameter | Value | |--------------------------|--------------| | Amplifier CMRR | 80 dB (10,000) | | Source imbalance CMRR | 2,000,000 | | Effective CMRR | 9,950 | | Effective CMRR (dB) | 79.94 dB | --- ## **Final Answer** \[ \boxed{ \text{The worst effective CMRR is approximately } 9950~(79.94~\mathrm{dB}) } \]

Let's carefully analyze and solve the problem step-by-step: --- ### **Given Data:** - Common-mode input impedance of the amplifier, \( Z_{CM} = 1\, M\Omega \) - Source impedance, \( Z_S = 100\, \Omega \) - Source impedance imbalance: 0.5% - Amplifier CMRR, \( \mathrm{CMRR_{amp}} = 80\, \mathrm{dB} \) --- ### **Step 1: Convert amplifier CMRR from dB to ratio** \[ \mathrm{CMRR_{amp}} = 80\, \mathrm{dB} \Rightarrow \text{ratio} = 10^{(80/20)} = 10^4 = 10,000 \] --- ### **Step 2: Determine the source imbalance ratio** The source imbalance is 0.5%. Thus, the two source impedances are: - \( Z_{S1} = 100\, \Omega \) - \( Z_{S2} = 100\, \Omega \times (1 + 0.005) = 100.5\, \Omega \) The difference: \[ \Delta Z = Z_{S2} - Z_{S1} = 0.5\, \Omega \] --- ### **Step 3: Calculate the effective CMRR due to source imbalance** The imbalance creates a differential component from a common-mode signal, which is affected by the ratio of the common-mode impedance to the impedance mismatch: \[ \mathrm{CMRR_{source}} = \frac{Z_{CM}}{\Delta Z} \] Plugging in the values: \[ \mathrm{CMRR_{source}} = \frac{1,000,000\, \Omega}{0.5\, \Omega} = 2,000,000 \] --- ### **Step 4: Combine the effects to find the worst-case effective CMRR** The total suppression of unwanted common-mode signals (the effective CMRR) considers both the amplifier's inherent CMRR and the source imbalance: \[ \frac{1}{\mathrm{CMRR_{eff}}} = \frac{1}{\mathrm{CMRR_{amp}}} + \frac{1}{\mathrm{CMRR_{source}}} \] Substituting: \[ \frac{1}{\mathrm{CMRR_{eff}}} = \frac{1}{10,000} + \frac{1}{2,000,000} \] Calculations: \[ \frac{1}{10,000} = 0.0001 \] \[ \frac{1}{2,000,000} = 0.0000005 \] Sum: \[ 0.0001 + 0.0000005 = 0.0001005 \] Thus: \[ \mathrm{CMRR_{eff}} = \frac{1}{0.0001005} \approx 9,950 \] --- ### **Step 5: Convert the effective CMRR back to dB** \[ \mathrm{CMRR_{eff}}\, (\mathrm{dB}) = 20 \log_{10}(9,950) \] Calculate: \[ \log_{10}(9,950) \approx 3.998 \] Therefore: \[ 20 \times 3.998 \approx 79.96\, \mathrm{dB} \] --- ### **### **Final Answer:** \[ \boxed{ \text{The worst effective CMRR is approximately } 9,950 \quad \text{or} \quad \boxed{80\, \mathrm{dB}} } \] This indicates that, due to the source imbalance and amplifier characteristics, the maximum achievable suppression of common-mode signals is about **80 dB** under worst-case conditions.