modified 9 years ago

If you want a current source constant to 1 percent over a load voltage range of 0 to 10V how large a voltage source must use in series with a single resistor

01:13:22

Art of Electronics 3rd ed. Exercise 2.9, 2.10

(The current source in this exercise is a voltage source plus resistor)

1. When Vout=0
I1=Vin/R

2, When Vout=10V
I2=(Vin-Vout)/R

I1/I2=100
100Vin/R=(Vin-Vout)/R
100Vin=Vin-Vout
99Vin=Vout
99*10=Vout = 990V

Exercise 2.10
Let I=10mA, how much power is dissipated in the series resistor? How much gets into the load?
I1=10mA, R=99K
Pr=I*U=0.01*990=9.9W on the resistor
Pload=I*10V=0.01*10V=0.1W on the load

published 9 years ago

2ctiby

9 years ago

Should we assume here that the mentioned 1% means ± 0.5% .....because if it means ± 1% then I think the voltage source answer would be 490V rather than 990V

nemethakos

9 years ago

Solve the exercise yourself and show your calculations!

2ctiby

9 years ago

I have done the calculations, that is why I am asking you a straightforward question about my query. Are you just copying interesting snippets from a book or are you able to deal with sensible issues about them....in which case please answer my query about your 1% statement.

nemethakos

9 years ago

I don't see any of your calculations here.

2ctiby

9 years ago

I am not asking you to judge my calculations, and I am not disputing the 1% statement...I simply would like your opinion on whether that stated 1% is meant to represent 2 halves (each part either side of the current amp value...thus making the one percent), or whether that shown 1% is meant to represent a whole one percent either side of the source current amp value....the shown statement does not clearly tell us which of those it is representing.... a ±% statement would be clearer.

nemethakos

9 years ago

I think it is easy. The book both uses the "simple percent" and the "plus-minus percent" terms, e.g.: "resistors typically have tolerances of ±5% or ±1%;" or "Although the “official” line voltage is 120V ±5%, you’ll sometimes see 110V, 115V, or 117V." or "Temperature coefficient ±0.4%/◦C". So when it is a simple percent I'm sure I should interpret "constant to 1%" as 1-Ilow/Ihigh (since the minimum current is no less than 99% of the maximum)

2ctiby

9 years ago

Your mention of a "simple percent" is to me a total nonsense....It is time that you should admit that you are uncertain of something rather than pretending and making ridiculous sidetracking...then we can help you learn...as the rest of us are doing....I will show the calculation here only when you admit that.

nemethakos

9 years ago

simple percent = %

2ctiby

9 years ago

Here is a very brief introduction to error estimation https://www.google.co.uk/url?sa=t&source=web&rct=j&url=http://www.animations.physics.unsw.edu.au/sf/toolkits/Errors_and_Error_Estimation.pdf&ved=0ahUKEwiw5enUpKjSAhWrB8AKHdEEAs8QFggaMAA&usg=AFQjCNGZTFvSEntQaJDkyDSZgaO_4MZTgw&sig2=z2oTIn2Iqh8J3PaqZdW9vQ

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