Suppose that we are designing a cardiac pacemaker circuit. The circuit is requir
ID: 1831705 • Letter: S
Question
Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1-ms duration to the heart, which can be modeled as a 500 ohm resistance. The peak amplitude of the pulses is required to be 5 volts. However, the battery delivers only 2.5 volts. Therefore, we decide to charge two equal-value capacitors in parallel from the 2.5-V battery then switch the capacitors in series with the heart during the 1-ms pulse. What is the minimum value of the capacitances required so the output pulse altitude remains between 4.9 V and 5.0 V throughout its 1-ms duration? If the pulses occur once every second, what is the average current drain on the battery? Assume constant current during the output pulse. Find the ampere-hour rating of the battery so it lasts for five years.Explanation / Answer
Given: Question Details Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1-ms duration to the heart, which can be modeled as a 500 ohm resistance. The peak amplitude of the pulses is required to be 5 volts. However, the battery delivers only 2.5 volts. Therefore, we decide to charge two equal-value capacitors in parallel from the 2.5-V battery then switch the capacitors in series with the heart during the 1-ms pulse. What is the minimum value of the capacitances required so the output pulse altitude remains between 4.9 V and 5.0 V throughout its 1-ms duration? If the pulses occur once every second, what is the average current drain on the battery? Assume constant current during the output pulse. Find the ampere-hour rating of the battery so it lasts for five years. SOLUTION: Capacitors in series add like resistors in parallel ; so , the equivalent capacitance of the two capacitors in series is : Ceq = (1/2)C . We thus have the time constant of the two series capacitors dicharging through the 500 ohm resistor is : T = RCeq = 500(1/2)C = 250C . The voltage across the 500 ohm resistor when the capacitors are discharging is : v(t) = 5e^(-t/T) = 5e^[-t/(250C)] . During the .001s duration of the pulse , we do not want the voltage to drop below 4.9volts ; so , we want : 4.9 = 5e^[-(.001)/(250C)] e^[(.001)/(250C)] = 5/(4.9) . Take natural logs of both sides , and we get : (.001)/(250C) = ln[5/(4.9)] . Solve for C , and we get : C = (.001)/{250ln[5/(4.9)]} is approximately = 197.99 micro farads = 198 micro farads to three significant digits ( this is the minimum capacitance ) --------------------------------------------------- The average output current for the .001s of the pulse assuming constant current of 5/500 = .01amps out of every 1second cycle is : iav = 1/(1s)[(.01)(.001s)] = 10 micro amps = .00001 amps . This is the average output current at a voltage of 5volts . The average current to charge the capacitors in parallel at 2.5volts must then be . ( I will assume no power loss) : Pin = Pout 2.5 ( iavin) = 5(iavout) = 5(.00001) = .00005 ; so , the average input current is : iavin = (.00005)/(2.5) = .00002amps or 20 microamps . We thus need the battery to have an amp - hr rating of : x = (.00002amps)(5years)(365days/1year)(24hrs/1day) = .876 amp-hrs -------- solution. ------------------------------------------ Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1-ms duration to the heart, which can be modeled as a 500 ohm resistance. The peak amplitude of the pulses is required to be 5 volts. However, the battery delivers only 2.5 volts. Therefore, we decide to charge two equal-value capacitors in parallel from the 2.5-V battery then switch the capacitors in series with the heart during the 1-ms pulse. What is the minimum value of the capacitances required so the output pulse altitude remains between 4.9 V and 5.0 V throughout its 1-ms duration? If the pulses occur once every second, what is the average current drain on the battery? Assume constant current during the output pulse. Find the ampere-hour rating of the battery so it lasts for five years. SOLUTION: Capacitors in series add like resistors in parallel ; so , the equivalent capacitance of the two capacitors in series is : Ceq = (1/2)C . We thus have the time constant of the two series capacitors dicharging through the 500 ohm resistor is : T = RCeq = 500(1/2)C = 250C . The voltage across the 500 ohm resistor when the capacitors are discharging is : v(t) = 5e^(-t/T) = 5e^[-t/(250C)] . During the .001s duration of the pulse , we do not want the voltage to drop below 4.9volts ; so , we want : 4.9 = 5e^[-(.001)/(250C)] e^[(.001)/(250C)] = 5/(4.9) . Take natural logs of both sides , and we get : (.001)/(250C) = ln[5/(4.9)] . Solve for C , and we get : C = (.001)/{250ln[5/(4.9)]} is approximately = 197.99 micro farads = 198 micro farads to three significant digits ( this is the minimum capacitance ) --------------------------------------------------- The average output current for the .001s of the pulse assuming constant current of 5/500 = .01amps out of every 1second cycle is : iav = 1/(1s)[(.01)(.001s)] = 10 micro amps = .00001 amps . This is the average output current at a voltage of 5volts . The average current to charge the capacitors in parallel at 2.5volts must then be . ( I will assume no power loss) : Pin = Pout 2.5 ( iavin) = 5(iavout) = 5(.00001) = .00005 ; so , the average input current is : iavin = (.00005)/(2.5) = .00002amps or 20 microamps . We thus need the battery to have an amp - hr rating of : x = (.00002amps)(5years)(365days/1year)(24hrs/1day) = .876 amp-hrs -------- solution. ------------------------------------------Related Questions
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