# Simple Circuit question

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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?

Re: Simple Circuit question
On Thu, 3 Nov 2016 17:58:44 -0700 (PDT), Ivan Vegvary

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.

--

John Larkin         Highland Technology, Inc

lunatic fringe electronics

Re: Simple Circuit question
On Thu, 03 Nov 2016 18:12:46 -0700, John Larkin

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

Re: Simple Circuit question
On Fri, 04 Nov 2016 10:10:40 -0700, Kevin Glover

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

--

John Larkin         Highland Technology, Inc

lunatic fringe electronics

Re: Simple Circuit question
On Fri, 04 Nov 2016 10:10:40 -0700, the renowned Kevin Glover

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

--
Best regards,
Spehro Pefhany
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Re: Simple Circuit question
On Friday, 4 November 2016 17:10:46 UTC, Kevin Glover  wrote:

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

NT

Re: Simple Circuit question
On Thu, 3 Nov 2016 17:58:44 -0700 (PDT), the renowned Ivan Vegvary

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

--
Best regards,
Spehro Pefhany
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Re: Simple Circuit question
Thank you Spehro.

More specific. Battery V10%. Series resistor 10%k. 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.

Re: Simple Circuit question
On Thu, 3 Nov 2016 19:14:48 -0700 (PDT), the renowned 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

--
Best regards,
Spehro Pefhany
We've slightly trimmed the long signature. Click to see the full one.
Re: Simple Circuit question
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

Re: Simple Circuit question
On Friday, 4 November 2016 02:14:52 UTC, Ivan Vegvary  wrote:

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

Re: Simple Circuit question
On Friday, 4 November 2016 00:58:51 UTC, Ivan Vegvary  wrote:

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

Re: Simple Circuit question
Ivan Vegvary wrote:

** 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

Re: Simple Circuit question
On Friday, 4 November 2016 01:54:01 UTC, Phil Allison  wrote:

Coul question.

NT

Re: Simple Circuit question
On 11/4/2016 7:35 PM, snipped-for-privacy@gmail.com wrote:

Vt/Q ohms

Re: Simple Circuit question
On Thursday, November 3, 2016 at 8:58:51 PM UTC-4, Ivan Vegvary wrote:

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.