EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: smaroukis on January 20, 2023, 01:32:57 am
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Diodes always get me, are they a fixed voltage drop? But forward voltage drop depends on current, but the current depends on the voltage drop (Shockley's equation).
Without _too_ much math how do you usually analyze diode circuits if you need to get specific?
Main question in this diagram (From Practical Electronics for Inventors, 4ed):
Why is it 1.82mA over R_Load and not 1.8ma? (1.8V/1KOhm)
Here is my circuit analysis:
① 3 Si diodes (Vf = 0.6V ea) forces V_out = 1.8V
② This forces a fixed current over R_S of (5-1.8)/1K
(should be 3.2mA not 3.18mA?)
③ in ①, the current over R_Load is also forced to 1.8V/1KOhm
④ The diode current will be the resulting 3.2mA - 1.8mA
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Why is it 1.82mA over R_Load and not 1.8ma? (1.8V/1KOhm)
I think someone messed up and assumed a total voltage drop of 1.82V for the 3 diodes instead of 1.80V ( 3x 0.6V )
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The forward voltage depends on temperature(the Vf of the 1n4148 can be used as a temperature sensor for example), but assuming 0.6 is a good start. Have a look at figure 1 in this datasheet, how current and temp affects the forward voltage: https://www.vishay.com/docs/81857/1n4148.pdf (https://www.vishay.com/docs/81857/1n4148.pdf)
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But forward voltage drop depends on current, but the current depends on the voltage drop (Shockley's equation).
In most cases, like this one, there is significant series resistance which cases the voltage available at the diode to decrease as the current through the resistor increases. There is a balance point where the current through resistor and diode is such that the voltage "remaining" for the diode is exactly equal to the voltage which causes the diode to conduct that current.
At typical diode currents the current changes greatly even for small changes in voltage. This happens at about 0.6 Volts for a silicon diode. In many cases just using that value to analyse a circuit will give accurate enough "engineering" answers.
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Why is it 1.82mA over R_Load and not 1.8ma? (1.8V/1KOhm)
The resistor value is 1 significant digit. Thus it is completely bullocks to put three significant digits in the current.
The proper calculation is 1.8V over 1K gives 2mA.
Okay, the resistor is probably within 5% tolerance from 1.0000k, thus 1.8mA is also valid. :-)