Dual Slope ADC Peak Voltage Calculation

A Dual Slope Analog-to-Digital Converter (ADC) is a type of analog-to-digital converter that uses an integrator to convert an analog input voltage into a digital output. It operates in two main phases: a charging (integration) phase and a discharging (de-integration) phase.

Understanding Dual Slope ADC Operation

In the Dual Slope ADC, the input voltage is first applied to an integrator for a fixed period, causing the capacitor to charge and its voltage to ramp up linearly. This marks the charging phase, and the voltage reaches a certain peak value. After this, the input is switched to a known reference voltage of opposite polarity, and the capacitor discharges linearly back to zero. This is the discharging phase. The time taken for this discharge is measured, and this time is proportional to the original input voltage.

Key Parameters for ADC Calculation

Let's list the given parameters for this Dual Slope ADC problem:

We need to find the peak voltage reached by the triangular wave during charging.

Charging and Discharging Phases Analysis

During the charging phase, the capacitor charges through the resistor R due to the input voltage (which is implicitly integrated). The voltage across the capacitor ramps up. The peak voltage is reached at the end of this charging phase.

During the discharging phase, the capacitor discharges through the resistor R due to the reference voltage \(V_{ref}\). The voltage across the capacitor ramps down from its peak voltage to zero. The rate of change of voltage across a capacitor in an RC circuit is related to the current flowing through it and the capacitance.

The voltage change across the capacitor is given by the formula: \[ \Delta V = \frac{1}{C} \int I \, dt \] For a constant current \(I = V/R\), this simplifies to: \[ \Delta V = \frac{V_{source}}{RC} \Delta t \]

Peak Voltage Derivation Formula

The peak voltage (\(V_{peak}\)) reached during the charging phase is the starting voltage for the discharging phase. During discharging, the capacitor's voltage decreases from \(V_{peak}\) to 0. Therefore, the change in voltage is \(V_{peak}\).

Using the relationship for the discharging phase with the reference voltage: \[ V_{peak} = \frac{V_{ref}}{RC} \times T_{discharge} \]

Step-by-Step Peak Voltage Calculation

Let's convert the given values into their standard SI units:

• Capacitance (\(C\)) = 0.34 nF = \(0.34 \times 10^{-9}\) F
• Resistance (\(R\)) = 1 K\(\Omega\) = \(1 \times 10^3\) \(\Omega\)
• Discharging Time (\(T_{discharge}\)) = 9 ns = \(9 \times 10^{-9}\) s
• Reference Voltage (\(V_{ref}\)) = 2.5 V

Now, substitute these values into the formula for \(V_{peak}\):

\[ V_{peak} = \frac{V_{ref}}{R \times C} \times T_{discharge} \]

\[ V_{peak} = \frac{2.5 \, \text{V}}{(1 \times 10^3 \, \Omega) \times (0.34 \times 10^{-9} \, \text{F})} \times (9 \times 10^{-9} \, \text{s}) \]

First, calculate the product \(R \times C\):

\[ R \times C = (1 \times 10^3) \times (0.34 \times 10^{-9}) = 0.34 \times 10^{-6} \, \text{s} \]

Now, substitute this back into the equation for \(V_{peak}\):

\[ V_{peak} = \frac{2.5}{0.34 \times 10^{-6}} \times (9 \times 10^{-9}) \]

\[ V_{peak} = \frac{2.5 \times 9 \times 10^{-9}}{0.34 \times 10^{-6}} \]

\[ V_{peak} = \frac{22.5 \times 10^{-9}}{0.34 \times 10^{-6}} \]

\[ V_{peak} = \frac{22.5}{0.34} \times 10^{-9} \times 10^6 \]

\[ V_{peak} = 66.17647 \times 10^{-3} \]

\[ V_{peak} = 0.06617647 \, \text{V} \]

Final Result for Peak Voltage

Rounding the calculated peak voltage to three decimal places, or as suggested by the options, we get:

\[ V_{peak} \approx 0.066 \, \text{V} \]

This value represents the maximum voltage achieved by the integrator's output during the charging phase, which then serves as the starting point for the linear discharge using the reference voltage.