Why is this diode here...

[Gibson486]

http://pics.bbzzdd.com/users/abchin486/circuit.jpg

At first, I thought, oh, they just want to protect the rail...but the diode is backwards....

So, I am having trouble understanding why the diode is there...

What is it supposed to do? BTW, Vo is connected to a pin on a chip. I am guessing the circuit is used to control sleep mode for the device.

 

[PottedMeat]

huh...

maybe to discharge the capacitor quickly to ~0.7v through the +5 rail when the board is turned off?

im assuming the vo input is high z and the resistor is kind of high

 

[William Gaatjes]

Here is an explanation :

When understanding these circuits, you must look at the circuit in the following steps and ask yourself what will happen :

• What does the circuit do with out +5V.
• What does the circuit do at the moment when +5V is applied.
• What does the circuit do when the +5V has been present for, for example an hour (just a random number).
• What does the circuit do when the +5V is removed.
• What does the circuit do with out +5V.

This circuit is typical for giving a low level pulse when the power is first applied. Why, the capacitor can store charge. Q = I *T.
Why will there flow a current ? Q = U * C. The voltage over the capacitor was 0.
Thus the capacitor has no charge and is empty.
The rule of thumb is that an empty capacitor can for a short moment be seen as a short circuit with an increasing resistance until there is no more voltage difference between the capacitor and the "supply" it is connected to.
Thus for a short moment we can say that the voltage over the resistor is the full +5V.
For a short moment, you have a voltage divider combination R & C.

Now, when we remove the voltage, the capacitor will discharge again through the diode. acting as a supply for surrounding electronic loads.
Since the capacitor can only store a little bit of charge, it will be empty quickly. Now, imagine that this is connected to a reset pin of some microchip.
The whole idea of reset at power up is to give the circuit a steady known state.

Ask yourself the following questions :
What will happen when you use the circuit above and power the circuit and remove power again ?
What would happen with the diode ?
What would happen without the diode ?
Think about it in units of time...


http://en.wikipedia.org/wiki/Coulomb

Charles-Augustin de Coulomb (14 June 1736 – 23 August 1806) was a French physicist. He is best known for developing Coulomb's law, the definition of the electrostatic force of attraction and repulsion. The SI unit of charge, the coulomb, was named after him.

 

[Mark R]

It's to protect the chip during power down.

When power is removed from the rail, if the output had been high, then the capacitor will be charged.

When the rail is removed, the charge in the capacitor may reverse bias circuits within the chip causing malfunction or damage. The presence of the diode ensures that the capacitor discharges once power is removed.

 

[William Gaatjes]

It's to protect the chip during power down.

When power is removed from the rail, if the output had been high, then the capacitor will be charged.

When the rail is removed, the charge in the capacitor may reverse bias circuits within the chip causing malfunction or damage. The presence of the diode ensures that the capacitor discharges once power is removed.

That can indeed be a function but most chip pins have internal ESD diodes and resistor combinations, especially logic chips. Often when a very large capacitor is used, a series resistor is advised to limit the current through the pin in combination with an external (preferably) schottky diode(lower voltage drop). Then, never a current can flow, high enough to destroy or damage the ESD D+D+R network for the pin. Usually this is done because every time an ESD D+D+R network experiences high currents or voltages(and then high currents will also flow), because of the damage, higher leakage currents will be present. Making a chip consume more current when powered down or reduce signal quality at high frequencies because of output or input impedance changes.

 

[William Gaatjes]

The answer :

Spoiler

The secret is of course, slow charge time (R*C and D non conductive because of reverse voltage) and fast discharge time. D is conducting and has much lower resistance (less then 1 Ohm) while conducting in comparison with the resistor.
When power is applied, the capacitor is slowly charged. When power is removed, the capacitor is quickly discharged. When power is again applied, the capacitor will have sufficient low amount of charge to be able to provide a low voltage level at vo (U = Q / C).

If you wish, you can calculate the RC time constant with R = 10000 Ohm. C = 100nf = 100*10^-9 F. D is a schottky diode = 0,3 V forward voltage drop and you may assume the diode is 1 Ohm when conducting. You will find a 1 to 10000 difference in time

 

[Gibson486]

no offense, william, but do you just look up random stuff to try to answer the question? I asked what the purpose of the circuit was...not what a cap does what, what discharge time is or what the function of all the parts are.

 

[PottedMeat]

just wondering - what are the values for R & C? what kind of diode is that? is Vo an analog or digital input?

 

[Mark R]

Actually, I've been thinking on the basis that Vo is an output of the chip.

If it's an input related to sleep mode, then there is an alternative explanation, particularly if there is another power rail somewhere:
If Vo is a signal that powers the device (or provides some form of control current), then the circuit will ensure that the capacitor charges relatively slowly when the power comes on. This results in a delay when the system is powered up (e.g. to ensure that the chip is activated after a small voltage stabilisation delay).

The diode ensures that the cap rapidly discharges when power is removed, causing the chip to shutdown or go into standby rapidly, when the 5V rail is lost.

It would be really helpful to have an idea of what the circuit was, or at least what the chip is (so that we know what Vo actually does).

 

[William Gaatjes]

no offense, william, but do you just look up random stuff to try to answer the question? I asked what the purpose of the circuit was...not what a cap does what, what discharge time is or what the function of all the parts are.

Woops. I was under the false impression that you wanted to know how it works.
I always give a little more information then needed, but this little more information helps to understand the circuit. I do not need to look up random stuff. The knowledge is present in my mind as with almost everything that i post. I only use the internet to lookup historic details such a dates and times or side information (such as hyperlinks) that is not directly relevant.
My mind is the cache and main memory so to say, the internet is my hard drive so to say..

Anyway, the way i explain it, is how i see it myself. I know i sometimes appear fuzzy, but that is because i always had difficulty expressing all simultaneous thoughts through one bottlenecked medium.
I am bandwidth limited... I need a better body.

 

[darthsidious]

My initial guess it that it exist to make sure Vout can never go above 5V+0.7V. So it's clamping Vo, not the rail. This makes sure that if there any 5V-ish devices connected, they can;t be damaged by overvoltage.

Or as someone else mentioned, it could be one part of an ESD protection circuit (to route current to the supply, which likely has the ESD clamp).

As for what it controls, it's hard to say. It could be an output pin, the RC timer might be used to calibrate some timer inside the chip; without knowing thr chip, it's hard to say.

http://pics.bbzzdd.com/users/abchin486/circuit.jpg

At first, I thought, oh, they just want to protect the rail...but the diode is backwards....

So, I am having trouble understanding why the diode is there...

What is it supposed to do? BTW, Vo is connected to a pin on a chip. I am guessing the circuit is used to control sleep mode for the device.

 

[nagol567]

Woops. I was under the false impression that you wanted to know how it works.
I always give a little more information then needed, but this little more information helps to understand the circuit. I do not need to look up random stuff. The knowledge is present in my mind as with almost everything that i post. I only use the internet to lookup historic details such a dates and times or side information (such as hyperlinks) that is not directly relevant.
My mind is the cache and main memory so to say, the internet is my hard drive so to say..

Anyway, the way i explain it, is how i see it myself. I know i sometimes appear fuzzy, but that is because i always had difficulty expressing all simultaneous thoughts through one bottlenecked medium.
I am bandwidth limited... I need a better body.

...movie idea

 

[William Gaatjes]

...movie idea

Not really. Experience.
But ideas can be as crossroads where creative minds find each other and connect...
History is full of examples where people from different places without ever knowing each other have the same conclusion.
Even the plot of the movie the matrix can be found in sages through history...