A sensor needs a bias voltage of +3.000 volts. The DAC producing the voltage is 12 bits with a 3.3 volt reference. What is the DAC input code? 1241 3724 3103 3541
A sensor needs a bias voltage of +3.000 volts. The DAC producing the voltage is 12 bits with a 3.3 volt reference. What is the DAC input code? 1241 3724 3103 3541
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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![**Question:**
A sensor needs a bias voltage of +3.000 volts. The DAC producing the voltage is 12 bits with a 3.3 volt reference. What is the DAC input code?
- ○ 1241
- ○ 3724
- ○ 3103
- ○ 3541
**Explanation:**
To find the correct DAC input code, use the formula for a Digital-to-Analog Converter (DAC):
\[ \text{Output Voltage} = \left(\frac{\text{Digital Code}}{2^n - 1}\right) \times \text{Reference Voltage} \]
Where:
- \( n \) is the number of bits (12 bits in this case).
- Digital Code is what we are trying to find.
- Reference Voltage is 3.3 volts.
Rearranging for the Digital Code gives:
\[ \text{Digital Code} = \left(\frac{\text{Output Voltage} \times (2^n - 1)}{\text{Reference Voltage}}\right) \]
Plug in the values:
\[ \text{Digital Code} = \left(\frac{3.000 \times (2^{12} - 1)}{3.3}\right) \]
Calculate:
\[ 2^{12} - 1 = 4095 \]
\[ \text{Digital Code} = \left(\frac{3.000 \times 4095}{3.3}\right) \]
\[ \text{Digital Code} \approx 3724 \]
Therefore, the correct answer is:
- ○ 3724](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78790db4-d72e-4785-a32a-a8a933d5b327%2Fab887157-c975-4f89-8720-ca4dad15af3f%2Fw1s42g_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
A sensor needs a bias voltage of +3.000 volts. The DAC producing the voltage is 12 bits with a 3.3 volt reference. What is the DAC input code?
- ○ 1241
- ○ 3724
- ○ 3103
- ○ 3541
**Explanation:**
To find the correct DAC input code, use the formula for a Digital-to-Analog Converter (DAC):
\[ \text{Output Voltage} = \left(\frac{\text{Digital Code}}{2^n - 1}\right) \times \text{Reference Voltage} \]
Where:
- \( n \) is the number of bits (12 bits in this case).
- Digital Code is what we are trying to find.
- Reference Voltage is 3.3 volts.
Rearranging for the Digital Code gives:
\[ \text{Digital Code} = \left(\frac{\text{Output Voltage} \times (2^n - 1)}{\text{Reference Voltage}}\right) \]
Plug in the values:
\[ \text{Digital Code} = \left(\frac{3.000 \times (2^{12} - 1)}{3.3}\right) \]
Calculate:
\[ 2^{12} - 1 = 4095 \]
\[ \text{Digital Code} = \left(\frac{3.000 \times 4095}{3.3}\right) \]
\[ \text{Digital Code} \approx 3724 \]
Therefore, the correct answer is:
- ○ 3724
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