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Art of Electronics (1st Ed) ex 1.28 - am I right

Started by Unknown July 23, 2015
Hello, Technician's Apprentice Here,
I was given this book as a source to expand my knowledge, and not being able to flip to the back and verify my answers is troubling but not impassable.

Please help me check my grasp on the material.  The subject is Diode gates.
The sample is an AC fed +15v DC power supply through the anode of a diode to a +12V to +15V electronic clock with the cathode common to the clock's supply and the battery backup's cathode.

Ground is common to the DC supply, the Battery(-), and the Clock.

The Battery's Positive is connected to the Anode of the battery's Diode.


EXercise 1.28:
Make a simple modification to the circuit so that the battery is charged by the DC supply (when power is on, of course) at a current of 10mA (such a circuit is necessary to maintain the battery's charge).

My answer:
add a third diode with the anode at the node with the two other diodes' cathodes and the clock's input voltage, with the cathode attached in series with a 180 ohm resister to the battery's positive terminal.

My methodology is:
The DC source is 15V minus 1.2V (2x0.6V diode drops) minus the 12V (assumed battery charge/capacity), thus leaving a 1.8V trickle.
With the stated goal of a 10mA current, I divided the 1.8V by the 10mA, to arrive at my 180 ohm resister.

Is this correct?

Thanks in Advance,
JW David Asher, FC1(SW)
US Navy         Retired
On Thursday, July 23, 2015 at 10:39:23 AM UTC-4, J W David Asher, FC1(SW) USN Retired wrote:
> Hello, Technician's Apprentice Here, > I was given this book as a source to expand my knowledge, and not being able to flip to the back and verify my answers is troubling but not impassable. > > Please help me check my grasp on the material. The subject is Diode gates. > The sample is an AC fed +15v DC power supply through the anode of a diode to a +12V to +15V electronic clock with the cathode common to the clock's supply and the battery backup's cathode. > > Ground is common to the DC supply, the Battery(-), and the Clock. > > The Battery's Positive is connected to the Anode of the battery's Diode. > > > EXercise 1.28: > Make a simple modification to the circuit so that the battery is charged by the DC supply (when power is on, of course) at a current of 10mA (such a circuit is necessary to maintain the battery's charge). > > My answer: > add a third diode with the anode at the node with the two other diodes' cathodes and the clock's input voltage, with the cathode attached in series with a 180 ohm resister to the battery's positive terminal. > > My methodology is: > The DC source is 15V minus 1.2V (2x0.6V diode drops) minus the 12V (assumed battery charge/capacity), thus leaving a 1.8V trickle. > With the stated goal of a 10mA current, I divided the 1.8V by the 10mA, to arrive at my 180 ohm resister. > > Is this correct? > > Thanks in Advance, > JW David Asher, FC1(SW) > US Navy Retired
That's okay but it looks like all you need is a resistor in parallel with the existing battery diode. And at 10mA level, the diode drops are more like 0.7V.
On Thursday, July 23, 2015 at 10:22:20 AM UTC-7, bloggs.fred...@gmail.com wrote:
> On Thursday, July 23, 2015 at 10:39:23 AM UTC-4, J W David Asher, FC1(SW) USN Retired wrote: > > Hello, Technician's Apprentice Here, > > I was given this book as a source to expand my knowledge, and not being able to flip to the back and verify my answers is troubling but not impassable. > > > > Please help me check my grasp on the material. The subject is Diode gates. > > The sample is an AC fed +15v DC power supply through the anode of a diode to a +12V to +15V electronic clock with the cathode common to the clock's supply and the battery backup's cathode. > > > > Ground is common to the DC supply, the Battery(-), and the Clock. > > > > The Battery's Positive is connected to the Anode of the battery's Diode. > > > > > > EXercise 1.28: > > Make a simple modification to the circuit so that the battery is charged by the DC supply (when power is on, of course) at a current of 10mA (such a circuit is necessary to maintain the battery's charge). > > > > My answer: > > add a third diode with the anode at the node with the two other diodes' cathodes and the clock's input voltage, with the cathode attached in series with a 180 ohm resister to the battery's positive terminal. > > > > My methodology is: > > The DC source is 15V minus 1.2V (2x0.6V diode drops) minus the 12V (assumed battery charge/capacity), thus leaving a 1.8V trickle. > > With the stated goal of a 10mA current, I divided the 1.8V by the 10mA, to arrive at my 180 ohm resister. > > > > Is this correct? > > > > Thanks in Advance, > > JW David Asher, FC1(SW) > > US Navy Retired > > That's okay but it looks like all you need is a resistor in parallel with the existing battery diode. And at 10mA level, the diode drops are more like 0.7V.
Thank You! One question though wouldn't just using a resistor in parallel cause the battery to source all/most of the time with a potential discharging trickle - causing unnecessary battery cycling? Or would the greater potential from the DC supply always force the current direction. I ask because why wouldn't they use the resistor only in the original design - did they make it overly complicated to make an exercise out of it?
On Thursday, July 23, 2015 at 3:44:25 PM UTC-4, J W David Asher, FC1(SW) USN Retired wrote:
> On Thursday, July 23, 2015 at 10:22:20 AM UTC-7, bloggs.fred...@gmail.com wrote: > > On Thursday, July 23, 2015 at 10:39:23 AM UTC-4, J W David Asher, FC1(SW) USN Retired wrote: > > > Hello, Technician's Apprentice Here, > > > I was given this book as a source to expand my knowledge, and not being able to flip to the back and verify my answers is troubling but not impassable. > > > > > > Please help me check my grasp on the material. The subject is Diode gates. > > > The sample is an AC fed +15v DC power supply through the anode of a diode to a +12V to +15V electronic clock with the cathode common to the clock's supply and the battery backup's cathode. > > > > > > Ground is common to the DC supply, the Battery(-), and the Clock. > > > > > > The Battery's Positive is connected to the Anode of the battery's Diode. > > > > > > > > > EXercise 1.28: > > > Make a simple modification to the circuit so that the battery is charged by the DC supply (when power is on, of course) at a current of 10mA (such a circuit is necessary to maintain the battery's charge). > > > > > > My answer: > > > add a third diode with the anode at the node with the two other diodes' cathodes and the clock's input voltage, with the cathode attached in series with a 180 ohm resister to the battery's positive terminal. > > > > > > My methodology is: > > > The DC source is 15V minus 1.2V (2x0.6V diode drops) minus the 12V (assumed battery charge/capacity), thus leaving a 1.8V trickle. > > > With the stated goal of a 10mA current, I divided the 1.8V by the 10mA, to arrive at my 180 ohm resister. > > > > > > Is this correct? > > > > > > Thanks in Advance, > > > JW David Asher, FC1(SW) > > > US Navy Retired > > > > That's okay but it looks like all you need is a resistor in parallel with the existing battery diode. And at 10mA level, the diode drops are more like 0.7V. > > Thank You! > One question though wouldn't just using a resistor in parallel cause the battery to source all/most of the time with a potential discharging trickle - causing unnecessary battery cycling? Or would the greater potential from the DC supply always force the current direction. I ask because why wouldn't they use the resistor only in the original design - did they make it overly complicated to make an exercise out of it?
The idea is when the line operated 15V supply is on, the voltage at the clock node always exceeds the battery voltage, so the battery will sink and not source current. It's just an exercise, and an old one at that given the battery and power supply voltages quoted. Modern stuff is all 3V and less.
On Thursday, July 23, 2015 at 1:30:32 PM UTC-7, bloggs.fred...@gmail.com wrote:
> On Thursday, July 23, 2015 at 3:44:25 PM UTC-4, J W David Asher, FC1(SW) USN Retired wrote: > > On Thursday, July 23, 2015 at 10:22:20 AM UTC-7, bloggs.fred...@gmail.com wrote: > > > On Thursday, July 23, 2015 at 10:39:23 AM UTC-4, J W David Asher, FC1(SW) USN Retired wrote: > > > > Hello, Technician's Apprentice Here, > > > > I was given this book as a source to expand my knowledge, and not being able to flip to the back and verify my answers is troubling but not impassable. > > > > > > > > Please help me check my grasp on the material. The subject is Diode gates. > > > > The sample is an AC fed +15v DC power supply through the anode of a diode to a +12V to +15V electronic clock with the cathode common to the clock's supply and the battery backup's cathode. > > > > > > > > Ground is common to the DC supply, the Battery(-), and the Clock. > > > > > > > > The Battery's Positive is connected to the Anode of the battery's Diode. > > > > > > > > > > > > EXercise 1.28: > > > > Make a simple modification to the circuit so that the battery is charged by the DC supply (when power is on, of course) at a current of 10mA (such a circuit is necessary to maintain the battery's charge). > > > > > > > > My answer: > > > > add a third diode with the anode at the node with the two other diodes' cathodes and the clock's input voltage, with the cathode attached in series with a 180 ohm resister to the battery's positive terminal. > > > > > > > > My methodology is: > > > > The DC source is 15V minus 1.2V (2x0.6V diode drops) minus the 12V (assumed battery charge/capacity), thus leaving a 1.8V trickle. > > > > With the stated goal of a 10mA current, I divided the 1.8V by the 10mA, to arrive at my 180 ohm resister. > > > > > > > > Is this correct? > > > > > > > > Thanks in Advance, > > > > JW David Asher, FC1(SW) > > > > US Navy Retired > > > > > > That's okay but it looks like all you need is a resistor in parallel with the existing battery diode. And at 10mA level, the diode drops are more like 0.7V. > > > > Thank You! > > One question though wouldn't just using a resistor in parallel cause the battery to source all/most of the time with a potential discharging trickle - causing unnecessary battery cycling? Or would the greater potential from the DC supply always force the current direction. I ask because why wouldn't they use the resistor only in the original design - did they make it overly complicated to make an exercise out of it? > > The idea is when the line operated 15V supply is on, the voltage at the clock node always exceeds the battery voltage, so the battery will sink and not source current. It's just an exercise, and an old one at that given the battery and power supply voltages quoted. Modern stuff is all 3V and less.
Thank You for the clarification