1. Home
  2. Electronics Design
  3. Art of Electronics (1st Ed) ex 1.28 - am I right

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

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

Reply to
jwdavid.asher
Loading thread data ...

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.

Reply to
bloggs.fredbloggs.fred

On Thursday, July 23, 2015 at 10:22:20 AM UTC-7, snipped-for-privacy@gmail.com w rote:

able to flip to the back and verify my answers is troubling but not impass able.

tes.

de to a +12V to +15V electronic clock with the cathode common to the clock' s supply and the battery backup's cathode.

.

d 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).

cathodes and the clock's input voltage, with the cathode attached in serie s with a 180 ohm resister to the battery's positive terminal.

umed battery charge/capacity), thus leaving a 1.8V trickle.

to arrive at my 180 ohm resister.

the existing battery diode. And at 10mA level, the diode drops are more li ke 0.7V.

Thank You! One question though wouldn't just using a resistor in parallel cause the ba ttery 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 ove rly complicated to make an exercise out of it?

Reply to
J W David Asher, FC1(SW) USN Retired

ng able to flip to the back and verify my answers is troubling but not impa ssable.

gates.

iode to a +12V to +15V electronic clock with the cathode common to the cloc k's supply and the battery backup's cathode.

de.

ged by the DC supply (when power is on, of course) at a current of 10mA (su ch a circuit is necessary to maintain the battery's charge).

s' cathodes and the clock's input voltage, with the cathode attached in ser ies with a 180 ohm resister to the battery's positive terminal.

ssumed battery charge/capacity), thus leaving a 1.8V trickle.

A, to arrive at my 180 ohm resister.

th the existing battery diode. And at 10mA level, the diode drops are more like 0.7V.

battery to source all/most of the time with a potential discharging trickle - causing unnecessary battery cycling? Or would the greater potential fro m the DC supply always force the current direction. I ask because why woul dn't they use the resistor only in the original design - did they make it o verly complicated to make an exercise out of it?

The idea is when the line operated 15V supply is on, the voltage at the clo ck node always exceeds the battery voltage, so the battery will sink and no t source current. It's just an exercise, and an old one at that given the b attery and power supply voltages quoted. Modern stuff is all 3V and less.

Reply to
bloggs.fredbloggs.fred

On Thursday, July 23, 2015 at 1:30:32 PM UTC-7, snipped-for-privacy@gmail.com wr ote:

eing able to flip to the back and verify my answers is troubling but not im passable.

e gates.

diode to a +12V to +15V electronic clock with the cathode common to the cl ock's supply and the battery backup's cathode.

iode.

arged 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).

des' cathodes and the clock's input voltage, with the cathode attached in s eries with a 180 ohm resister to the battery's positive terminal.

(assumed battery charge/capacity), thus leaving a 1.8V trickle.

0mA, to arrive at my 180 ohm resister.

with the existing battery diode. And at 10mA level, the diode drops are mor e like 0.7V.

e battery to source all/most of the time with a potential discharging trick le - causing unnecessary battery cycling? Or would the greater potential f rom the DC supply always force the current direction. I ask because why wo uldn't they use the resistor only in the original design - did they make it overly complicated to make an exercise out of it?

lock 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

Reply to
J W David Asher, FC1(SW) USN Retired

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.