Simple Circuit question

Doing self study. Diode circuit. 10V battery followed by series load resistor. Thence a resistor and diode in parallel and back to the battery. Simple to take the 0.7v drop of the diode and have 9.3V remaining on across the first resistor. This allows calculation of total current through circuit and the rest is easy. BUT Then they want you to use the "ideal diode" method to solve same problem. With 0 voltage across "ideal diode" I will have 0 voltage across parallel resistor. I.e. a short circuit. It's like those components don't esist. What am I missing?

Reply to
Ivan Vegvary
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The drop across the diode may not be 0.7.

A short has zero voltage drop. No problem, as long as the current is in the diode forward direction.

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John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

No, you can't be sure without calculating the voltage across the parallel resistor ignoring the diode. If it's less than your semi-ideal 0.7V then the diode can be replaced with an open circuit. If it's more than 0.7V AND the diode is forward biased then you replace it with a 0.7V voltage source and the currents re-calculated.

For example a 12V battery, 100K series and 1K parallel to the diode. The voltage across the 1K will be about 119mV no matter whether the diode is there or not, or if the diode is reverse biased.

Similar to the above, except replace 0.7V with 0V. If the voltage would be greater than 0V AND the diode is forward biased, replace it with a 0V source (equivalent to a short) and re-calculate. Otherwise ignore.

--sp

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Spehro Pefhany 
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Reply to
Spehro Pefhany

So many of your sentences are far from unambiguous. Still... why don't you tell us what you're missing, as you don't say what the problem is.

Diodes only drop 0.65v or so at the knee, expect 1-2v at full rated current. And due to the parallel R there may be less V across it. Or if series R is very high.

The rest of what you write is too incommunicative to comment usefully.

NT

Reply to
tabbypurr

** Nothing.

It's a trap question to catch out the unwary.

Anecdote:

A teacher who gave electronics exams to young apprentices liked to pose this question if he saw any doing last minute study before going into the room.

"How many ohms are there in a Coulomb ?".

.... Phil

Reply to
Phil Allison

Thank you Spehro.

More specific. Battery V=10. Series resistor =10k. Parallel resistor =1k. Diode Silicon =o.7v

Current through diode ergo = 0.23ma. Book then asks you to recalculate using 'ideal diode'. They then give an answer of 0.3ma through the diode. How did they come to that? Thank you.

Reply to
Ivan Vegvary

Doesn't make a lot of sense to me. The current through an ideal diode would be exactly 1mA if forward biased (2nd case)- no current would flow through the 1K resistor. To get 0.3mA the diode voltage would have to be 0.636V not 0V.

In the first case it would be 0.93mA - 0.7mA = 0.23mA, as given.

Book solutions are not always right.

--sp

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Best regards,  
Spehro Pefhany 
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Reply to
Spehro Pefhany

Spehro, thank you. I'll stop tugging on my hair a d move further along in my studies. You've been very kind with your time! Ivan Vegvary

Reply to
Ivan Vegvary

Drops vary. Silicon diodes can be around .6 to .7 volts. Germanium, .2 to .3 volts.

Reply to
Kevin Glover

It's logarithmic on current, plus an ohmic term.

--
John Larkin         Highland Technology, Inc 

lunatic fringe electronics
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Reply to
John Larkin

stor.

ss the first resistor. This allows calculation of total current through cir cuit and the rest is easy.

resistor. I.e. a short circuit. It's like those components don't esist.

The ideal diode is usually accompanied by a series reverse bias ideal volta ge source representing the forward voltage drop at the approximate circuit currents. So you would just redraw the circuit with a 0.7V battery in serie s with the diode and opposing forward current flow. Then the diode current is the current through the diode battery. You can use superposition as foll ows : short out the 10V source so the 0.7V supplies 0.7/1 +0.7/10=0.77mA through the resistors. Then short out the 0.7V so the 10V supplies 10/10=

1mA through the battery short. Superimposing the currents you get 1mA-0.77m A=0.23mA net current through the battery and hence the diode.
Reply to
bloggs.fredbloggs.fred

It could even be 0.001V with very little current, but this is textbook exercise-land where diodes behave a bit more ideally than you might expect.

--sp

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Best regards,  
Spehro Pefhany 
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Reply to
Spehro Pefhany

So with no diode you'd get 10/11v = 0.9v. At 0.9mA.

where do you get that from?

2/9 x 0.9mA, ok.

An ideal diode is 0 ohms here, so just 10v across 10k = 1mA, all of which flows through D.

NT

Reply to
tabbypurr

Nothing to around 2v would be closer to real life. They do teach some junk about diodes dropping 0.6v on eng courses.

NT

Reply to
tabbypurr

Coul question.

NT

Reply to
tabbypurr

Vt/Q ohms

Reply to
John S

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