I was browsing the web, and found this:
"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?
Or does it just dump the excess energy as heat?
No, it uses a 'pass transistor' between input and output as a variable resistor. The control for that transistor comes from a built-in amplifier that senses the output voltage. When the output goes higher than 5V, the transistor's effective resistance is increased, and vice versa, to keep the output more or less exactly at 5V. By just looking at the output voltage, regulation is provided for input voltage changes and load current changes.
-- john
I suggest you get a data sheet and look at the internal circuit ( it should be on some )
No. In fact your premise has no sound basis.
Yes.
Graham
The input just needs to be a little higher than the output, around 2 or
3 volts higher. The regulator will get hotter as the input increases If you use a 30 volt input and the output load draws say 1 amp, the regulator must drop the extra 25 volts with 1 amp passing through, which equates to 25 watts of heat. So, in that mode without a heat sink to cool it off, the regulator would burn up.It's much better to use a lower voltage input to avoid the excessive heat.-Bill
So it's not a variable DC-DC converter... ok.
be
*Converter* ? No.You might understand better by examining a simple zener + emitter follower series pass regulator first. A 78xx is effectively a super version of that.
Graham
-- No.
-- Aaarghhh!!! (24V - 5V) * 1A = 19W, _not_ 17...
Ok, a big Thank You to all who replied.
Michael
snipped-for-privacy@gmail.com wrote:
in side there is a component know as a zener diode. this diode is used as a voltage reference and is selected to generate the desired voltage in the end. the output is compared with this voltage reference. there is a transistor that is used as the control path from the input of the raw voltages( unstable and much higher) to the smooth and stable output voltages. the base of this transistor controls the amount of voltage that appears on the output. attached to this base is the comparator components as we stated before. when output voltage is lower than reference voltage, more drive control on the base of the transistor is given until the reference and output match. if output is exceeding reference voltage than the current amount of drive on the base is reduced until output comes back into the window and then drive will balance out to maintain the current load level. excess voltage that appears on the input does not get drained off. it just does not get used. its like opening the water value on a tank of water, you can regulate the amount of flow you need, the rest of the water just sits there. when output is more than needed the valve is simply shut off until output level comes back down to spec, this means the load on the circuit generally uses the access of what is on the output. there are some regulators that actually do have a shunt in the output to help lower the voltage on the output by draining it to ground until the output is reached! this how ever is normally limited in current..
hope i kept it simply as possible.
-- Real Programmers Do things like this. http://webpages.charter.net/jamie_5
Yes, it's not. These are, however:
Very nice! Thanks for the link.
Michael
Well, tecnhically, the 780x series is supposed to go into thermal shutdown, but I get the point. :-)
Cheers! Rich
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.