Questions about MadVapes 5v circuit

[Digs]

So madvapes has this schematic on how to implement a LDO 5v regulator into a mod:

I'm learning the basics of electronic circuits and there's a couple things that I don't understand about this circuit that I'm hoping someone can explain.

It looks like a 470 ohm resistor is being used to limit the current to the switch. What I don't understand is the way it's connected. One side of the resistor is connected to the same wire as the switch, and the other side is connected to a ground wire.

Now if I hadn't seen this and was going to build this based on my own very basic knowledge of circuits I would have made it like this:

Or I may have also placed the resistor after the switch (between it and the regulator). I don't think that would make a difference. What i wouldn't have done is connected one side of the resistor to a ground. Why does the madvapes schematic do that?

My second question is about the capacitor. I understand capacitors will store electrical charge up to the amount its rated for. So if I understand the MadVapes circuit right, when the user hits the switch the capacitor will charge up and then the atty will fire up. As soon as the switch is released, the capacitor will discharge its stored charge to the atty, keeping the atty on for a little bit after the button was released. Why would you want to do that? Wouldn't you want the atty to turn off right away after releasing the button?

Please someone explain this to me! I hate it when I don't understand something!

 

[petercro]

Hi
I not 100% sure about the answers but, I will try.
The ctrl pin is itself like a capacitor, (it holds a charge).
So when you push the switch the ctrl pin powers up.
But when you let go of the switch the ctrl pin stays powered up, because it is holding
a charge. So the 470 Ohms resistor is to simply discharge the residual charge within
the ctrl pin, thus turning off the circuit after you release the switch.
If it wasn't there the atty would continue to burn after you released the switch!

You may find that the capacitor is there to simply smooth the output voltage.

(OR)
Am I totally wrong on both counts!!!!!!!!!!!.

 

[candre23]

(OR)
Am I totally wrong on both counts!!!!!!!!!!!.

Nope, you're right on both. The resistor prevents the control pin from being stuck on, or even firing randomly from acquired static voltage. You can use any value you have lying around. The cap is not strictly necessary (the atty doesn't care if the output is noisy), but it does make live easier for the regulator, so you should use one for longevity's sake.

 

[Rocketman]

Yee-ha,
This be the place to learn.
Pull down resistor on the regulator control pin, yup.
Other types of adjustable regulators have a resistor to the return for voltage programming, not the same.

The capacitor on the output is primarily to prevent dynamic loads from being reflected backwards into the regulator circuitry. Using this to run an IPOD, with loads depending on music content would be a dynamic load. Short rise/fall time output current changes can cause internal (to the regulator) loads impedances that would be outside the capabilities of the regulator. Regulator would try to compensate to load changes but would be too late and compensation might be at wrong time.
Capacitor energy storage in the order of microseconds would probably go unnoticed

 

[Digs]

You guys are great. I really appreciate all your help with this. The capacitor makes sense now and I understand the reason for way the resistor is wired, though it's going to take some more understanding of pull-down resistors, control pins, and whatever else before I truly understand how it works. But at least I know the "why" now!

I just ordered everything I need to get started from MadVapes, including a cheapo $5 soldering iron. I decided not to spend the a lot of $ on a nice one, at least not yet. Sometimes I just see something nice and think I have to have it! Then the next day I realize I don't really need it.

I guess my only question now is whether the resistor is having any effect on the current running through the switch. I've read that the whole point of using a control pin is to be able to use a small low amp switch. So is the resistor playing any role in that? If I removed it would the current going through the switch increase?

I wish I could just test it myself but my circuit design and analysis software doesn't have regulators with control pins

 

[petercro]

The ctrl pin is that of a logic circuit (Transistor).
Your talking a few miliamps no more.

 

[hoogie76]

In short the resistor ties the ctl pin to gnd to keep the device off. If it the pins float it could be unpredicatable.

If power were applied to turn the control pin (via the switch) and there were no resistor then there would be a direcit short to gnd.. 470 ohm is probably a bit underrated but I've got 18000 of them

 

[petercro]

470 ohm is probably a bit underrated but I've got 18000 of them

That is probably the best answer to a question I have ever heard!!!!!!!