Question
1. Clock signal for timer Tim3 is 80 MHz. Channel 1 of the timer is configured in input capture counter reset mode (rising edge polarity). The value of TIM3_PSC is 79. The value of Tim3_CCR1 register determined in input capture interrupt handling function is 2000. Determine the period of the incoming signal. Show your work.
2. Reference voltage for a 12-bit ADC configured in single-channel, single-ended, right-aligned mode is 3.0v. Determine ADC conversion result if the input voltage is 0.85 V. Show your work.
3. Clock signal for timer TIM3 is 40 MHz. Channel 1 of the timer is configured in PWM generation mode. The value of Tim3_PSC is 19. The value of TIM3_CCR1 register value is 2000. The value of TIM3_ARR is 19999. Determine the period and the duty cycle of the PWM signal. Show your work.
4. ADC has been configured with 12-bit resolution, 12.5 ADC clock cycles sampling time, and 40 MHz ADC clock. Determine conversion timer. Show your work.
2. Reference voltage for a 12-bit ADC configured in single-channel, single-ended, right-aligned mode is 3.0v. Determine ADC conversion result if the input voltage is 0.85 V. Show your work.
3. Clock signal for timer TIM3 is 40 MHz. Channel 1 of the timer is configured in PWM generation mode. The value of Tim3_PSC is 19. The value of TIM3_CCR1 register value is 2000. The value of TIM3_ARR is 19999. Determine the period and the duty cycle of the PWM signal. Show your work.
4. ADC has been configured with 12-bit resolution, 12.5 ADC clock cycles sampling time, and 40 MHz ADC clock. Determine conversion timer. Show your work.
Answers
GPT-4o mini
Let's tackle these problems step by step.
### Problem 1:
Given:
- Clock signal for Timer TIM3 = 80 MHz
- TIM3_PSC = 79
- TIM3_CCR1 value = 2000
1. **Calculating Timer Clock Frequency:**
The timer clock frequency can be determined using:
\[
f_{timer} = \frac{f_{clock}}{TIM3\_PSC + 1}
\]
Substituting values:
\[
f_{timer} = \frac{80 \, \text{MHz}}{79 + 1} = \frac{80 \, \text{MHz}}{80} = 1 \, \text{MHz}
\]
2. **Period of Timer Clock:**
The period \( T_{timer} \) of the timer clock is:
\[
T_{timer} = \frac{1}{f_{timer}} = \frac{1}{1 \, \text{MHz}} = 1 \, \mu s
\]
3. **Calculating Input Capture Length:**
The value in TIM3_CCR1 represents the number of timer counts until the next capture occurs. Therefore, the total period of the incoming signal can be calculated as:
\[
T_{signal} = T_{timer} \times TIM3\_CCR1
\]
Substituting values:
\[
T_{signal} = 1 \, \mu s \times 2000 = 2000 \, \mu s = 2 \, ms
\]
Thus, the period of the incoming signal is **2 ms**.
### Problem 2:
Given:
- ADC resolution = 12 bits
- Reference voltage = 3.0 V
- Input voltage = 0.85 V
1. **Calculating ADC Count:**
The ADC count can be calculated using the formula:
\[
ADC_{count} = \left(\frac{Input \, Voltage}{Reference \, Voltage}\right) \times (2^{Resolution} - 1)
\]
Where \( Resolution = 12 \).
Plugging the values:
\[
ADC_{count} = \left(\frac{0.85}{3.0}\right) \times (2^{12} - 1)
\]
Calculate \( 2^{12} - 1 = 4096 - 1 = 4095 \):
\[
ADC_{count} = \left(\frac{0.85}{3.0}\right) \times 4095 \approx 0.2833 \times 4095 \approx 1166.95
\]
Since ADC counts must be an integer, rounding gives:
\[
ADC_{count} \approx 1167
\]
Therefore, the ADC conversion result is **1167**.
### Problem 3:
Given:
- Clock signal for Timer TIM3 = 40 MHz
- TIM3_PSC = 19
- TIM3_CCR1 = 2000
- TIM3_ARR = 19999
1. **Calculating Timer Clock Frequency:**
\[
f_{timer} = \frac{f_{clock}}{TIM3\_PSC + 1} = \frac{40 \, \text{MHz}}{19 + 1} = \frac{40 \, \text{MHz}}{20} = 2 \, \text{MHz}
\]
2. **Period of Timer Clock:**
\[
T_{timer} = \frac{1}{f_{timer}} = \frac{1}{2 \, \text{MHz}} = 0.5 \, \mu s
\]
3. **Calculating Period of PWM Signal:**
The period of the PWM signal is determined by the TIM3_ARR:
\[
T_{PWM} = T_{timer} \times (TIM3\_ARR + 1) = 0.5 \, \mu s \times (19999 + 1) = 0.5 \, \mu s \times 20000 = 10 \, ms
\]
4. **Calculating Duty Cycle:**
The duty cycle can be calculated using:
\[
Duty \, Cycle = \left(\frac{TIM3\_CCR1}{TIM3\_ARR + 1}\right) \times 100\%
\]
Substituting values:
\[
Duty \, Cycle = \left(\frac{2000}{20000}\right) \times 100\% = 0.1 \times 100\% = 10\%
\]
Therefore, the period of the PWM signal is **10 ms** and the duty cycle is **10%**.
### Problem 4:
Given:
- ADC resolution = 12 bits
- ADC clock cycles sampling time = 12.5 cycles
- ADC clock = 40 MHz
1. **Calculate time per ADC clock cycle:**
\[
T_{ADC\, clock} = \frac{1}{f_{ADC}} = \frac{1}{40 \, \text{MHz}} = 25 \, ns
\]
2. **Calculate Conversion Time:**
The conversion time is determined by the number of clock cycles used for sampling:
\[
Conversion \, Time = ADC\, clock \, cycles \times T_{ADC\, clock} = 12.5 \times 25 \, ns = 312.5 \, ns
\]
Thus, the conversion time is **312.5 ns**.
### Problem 1:
Given:
- Clock signal for Timer TIM3 = 80 MHz
- TIM3_PSC = 79
- TIM3_CCR1 value = 2000
1. **Calculating Timer Clock Frequency:**
The timer clock frequency can be determined using:
\[
f_{timer} = \frac{f_{clock}}{TIM3\_PSC + 1}
\]
Substituting values:
\[
f_{timer} = \frac{80 \, \text{MHz}}{79 + 1} = \frac{80 \, \text{MHz}}{80} = 1 \, \text{MHz}
\]
2. **Period of Timer Clock:**
The period \( T_{timer} \) of the timer clock is:
\[
T_{timer} = \frac{1}{f_{timer}} = \frac{1}{1 \, \text{MHz}} = 1 \, \mu s
\]
3. **Calculating Input Capture Length:**
The value in TIM3_CCR1 represents the number of timer counts until the next capture occurs. Therefore, the total period of the incoming signal can be calculated as:
\[
T_{signal} = T_{timer} \times TIM3\_CCR1
\]
Substituting values:
\[
T_{signal} = 1 \, \mu s \times 2000 = 2000 \, \mu s = 2 \, ms
\]
Thus, the period of the incoming signal is **2 ms**.
### Problem 2:
Given:
- ADC resolution = 12 bits
- Reference voltage = 3.0 V
- Input voltage = 0.85 V
1. **Calculating ADC Count:**
The ADC count can be calculated using the formula:
\[
ADC_{count} = \left(\frac{Input \, Voltage}{Reference \, Voltage}\right) \times (2^{Resolution} - 1)
\]
Where \( Resolution = 12 \).
Plugging the values:
\[
ADC_{count} = \left(\frac{0.85}{3.0}\right) \times (2^{12} - 1)
\]
Calculate \( 2^{12} - 1 = 4096 - 1 = 4095 \):
\[
ADC_{count} = \left(\frac{0.85}{3.0}\right) \times 4095 \approx 0.2833 \times 4095 \approx 1166.95
\]
Since ADC counts must be an integer, rounding gives:
\[
ADC_{count} \approx 1167
\]
Therefore, the ADC conversion result is **1167**.
### Problem 3:
Given:
- Clock signal for Timer TIM3 = 40 MHz
- TIM3_PSC = 19
- TIM3_CCR1 = 2000
- TIM3_ARR = 19999
1. **Calculating Timer Clock Frequency:**
\[
f_{timer} = \frac{f_{clock}}{TIM3\_PSC + 1} = \frac{40 \, \text{MHz}}{19 + 1} = \frac{40 \, \text{MHz}}{20} = 2 \, \text{MHz}
\]
2. **Period of Timer Clock:**
\[
T_{timer} = \frac{1}{f_{timer}} = \frac{1}{2 \, \text{MHz}} = 0.5 \, \mu s
\]
3. **Calculating Period of PWM Signal:**
The period of the PWM signal is determined by the TIM3_ARR:
\[
T_{PWM} = T_{timer} \times (TIM3\_ARR + 1) = 0.5 \, \mu s \times (19999 + 1) = 0.5 \, \mu s \times 20000 = 10 \, ms
\]
4. **Calculating Duty Cycle:**
The duty cycle can be calculated using:
\[
Duty \, Cycle = \left(\frac{TIM3\_CCR1}{TIM3\_ARR + 1}\right) \times 100\%
\]
Substituting values:
\[
Duty \, Cycle = \left(\frac{2000}{20000}\right) \times 100\% = 0.1 \times 100\% = 10\%
\]
Therefore, the period of the PWM signal is **10 ms** and the duty cycle is **10%**.
### Problem 4:
Given:
- ADC resolution = 12 bits
- ADC clock cycles sampling time = 12.5 cycles
- ADC clock = 40 MHz
1. **Calculate time per ADC clock cycle:**
\[
T_{ADC\, clock} = \frac{1}{f_{ADC}} = \frac{1}{40 \, \text{MHz}} = 25 \, ns
\]
2. **Calculate Conversion Time:**
The conversion time is determined by the number of clock cycles used for sampling:
\[
Conversion \, Time = ADC\, clock \, cycles \times T_{ADC\, clock} = 12.5 \times 25 \, ns = 312.5 \, ns
\]
Thus, the conversion time is **312.5 ns**.