Is My Pull Up Resistor Calculation Correct?

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SUMMARY

The discussion centers on the calculation of resistance in a pull-up resistor configuration using a current source. The user initially calculates the current as I = 5V/8k, resulting in approximately 0.0006A, and expects a voltage drop close to 5V. However, they measure 9.83V, leading to a recalculated resistance of approximately 16.4k. The conclusion drawn is that the current source may not be precise, as real-world current sources require a minimum voltage drop to function correctly.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Familiarity with pull-up resistor configurations
  • Knowledge of current sources and their operational characteristics
  • Experience with voltage measurement techniques
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  • Research the specifications and behavior of current sources in electronic circuits
  • Learn about the impact of voltage drop across current sources on circuit performance
  • Study the principles of pull-up resistor design and implementation
  • Explore advanced voltage measurement techniques for accurate readings
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Electronics engineers, hobbyists working with microcontrollers, and anyone involved in circuit design and analysis will benefit from this discussion.

ee1215
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http://i.imgur.com/TbYIHnb.png?1

So, if I have a pull up and apply a current source. How would I need to read/calculate resistance?

V=IR
I = 5V/8k = .0006A would result in a voltage drop of close to 5V

Using a current source/Voltage reader I supply .0006A which should equate to ~5V drop

But when I read my voltage reader I get 9.83V. Is this correct?

Then my resistance would be 9.83/.0006 ~ 16.4k ?

These calculations do not seem correct.
 
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ee1215 said:
http://i.imgur.com/TbYIHnb.png?1

So, if I have a pull up and apply a current source. How would I need to read/calculate resistance?

V=IR
I = 5V/8k = .0006A would result in a voltage drop of close to 5V

Using a current source/Voltage reader I supply .0006A which should equate to ~5V drop

But when I read my voltage reader I get 9.83V. Is this correct?

Then my resistance would be 9.83/.0006 ~ 16.4k ?

These calculations do not seem correct.

Sounds like your current source is not very precise. Keep in mind that real-world current sources need to have some voltage drop across them in order to operate correctly. Typically you need a couple of volts across a current source in order for it to say "in compliance".
 

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