# Simple Circuit question

Started by November 3, 2016
```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?
```
```On Thu, 3 Nov 2016 17:58:44 -0700 (PDT), Ivan Vegvary
<ivanvegvary@gmail.com> wrote:

>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.

The drop across the diode may not be 0.7.

>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?

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

```
```On Thu, 3 Nov 2016 17:58:44 -0700 (PDT), the renowned Ivan Vegvary
<ivanvegvary@gmail.com> wrote:

>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.

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.

>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?

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
Amazon link for AoE 3rd Edition:            http://tinyurl.com/ntrpwu8
```
```On Friday, 4 November 2016 00:58:51 UTC, Ivan Vegvary  wrote:

> 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?

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
```
```Ivan Vegvary wrote:
>
> 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?
>

** 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
```
```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.
```
```On Thu, 3 Nov 2016 19:14:48 -0700 (PDT), the renowned Ivan Vegvary
<ivanvegvary@gmail.com> wrote:

>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.

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
Amazon link for AoE 3rd Edition:            http://tinyurl.com/ntrpwu8
```
```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
```
```On Thu, 03 Nov 2016 18:12:46 -0700, John Larkin
<jjlarkin@highlandtechnology.com> wrote:

>On Thu, 3 Nov 2016 17:58:44 -0700 (PDT), Ivan Vegvary
><ivanvegvary@gmail.com> wrote:
>
>>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.
>
>The drop across the diode may not be 0.7.
Drops vary. Silicon diodes can be around .6 to .7 volts. Germanium, .2
to .3 volts.
>
>>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?
>
>A short has zero voltage drop. No problem, as long as the current is
>in the diode forward direction.
```
```On Fri, 04 Nov 2016 10:10:40 -0700, Kevin Glover
<kevinfglover@yahoo.com> wrote:

>On Thu, 03 Nov 2016 18:12:46 -0700, John Larkin
><jjlarkin@highlandtechnology.com> wrote:
>
>>On Thu, 3 Nov 2016 17:58:44 -0700 (PDT), Ivan Vegvary
>><ivanvegvary@gmail.com> wrote:
>>
>>>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.
>>
>>The drop across the diode may not be 0.7.
>Drops vary. Silicon diodes can be around .6 to .7 volts.

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

--

John Larkin         Highland Technology, Inc

lunatic fringe electronics

```