I was browsing the web, and found this:
http://electronics.howstuffworks.com/digital-electronics4.htm
"The 7805 takes in a voltage between 7 and 30 volts and regulates it
down to exactly 5 volts."
How does it do this?
I'm guessing that when it sees a higher input voltage, it increases the
current available on the output... is my guess correct?
---
Yes.
Here's how it works:
12V 5V
/ /
Vin>-----[Rs]---+---Vreg
|
[Rl]
|
GND>------------+
Where Rl is the load resistance and Rs is the equivalent series
resistance of the regulator.
Let's say that Rl is 5 ohms and that the regulator is designed to
have an output voltage of 5 volts, like a 7805.
Let's also say that Vin is 12 volts.
Now, for Rl to have 5 across it Rs must drop 7V of the 12V supply
voltage and let 1 amp through, since that what the load wants:
E 5V
Il = --- = ---- = 1A
R 5R
That means that Rs must look like:
Vin - Vreg 12V - 5V
Rs = ------------ = ----------- = 7 ohms
Il 1A
and, since in a series circuit the current is everywhere the same,
the regulator will be dissipating:
P = (Vin - Vreg) * Il = 7V * 1A = 7 watts.
Now let's say that the supply voltage is 24V.
In order for the regulator to keep its output at 5V it must soak up
the difference between the supply and the output, which is 19V.
Such being the case, we now have:
24V 5V
/ /
Vin>-----[Rs]---+---Vreg
|
[Rl]
|
GND>------------+
and Rs must now look like:
Vin - Vreg 24V - 5V
Rs = ------------ = ----------- = 19 ohms
Il 1A
and the power it must dissipate (the energy it must dump as heat)
will be:
P = 17V * 1A = 17 watts
Similarly, as load current changes Rs must be varied in order to
keep Vreg constant.
Here's [very basically] what makes the regulator work:
Q1
Vin>-----+---C E-----------------+-----+---->Vout
| B | |
[R1] | [R2] [R4]
| | | |
+-----+ +--[R3]--+ |
| | | |
C | /+|--+ |
Q2 B--[R4]--+--< | |
E A1 \-|--------+
| |
| [REF]
| |
GND>-----------+-------------------------+---->GND
Q1 is the regulator's "pass transistor", and all of the load current
must pass through it. Initially it starts to conduct when current
flows into its base from R1. Q2 is initially turned off until
current starts to flow out of the emitter of Q1, at which time
current will start to flow into R2, R3, and the base of Q2, as well
as into R4 and the reference, and the load.
R2 and R3 are arranged so that When Vout = 5V the voltage on the +
input of A1 will be equal to the voltage on the - input, and for
that to happen A1 will turn on Q2 just enough to pull the base of Q1
just close enough to ground to keep its emitter at 5V.
So, you can see that if the supply voltage increases, the base of Q1
will be pulled more closely to ground and if the supply voltage
decreases, the base will be allowed to go more positive in order to
keep the emitter at 5V.
Similarly, if the load current changes, Qi's emitter voltage will
try to change, but A1 and Q2 will hold it in check and keep it at
5V.
