Thank's Tony

Wouldnt the Recifier drop 2.8 volt's 4 * 0.7 assuming it has silicon diode's

Probably need a 18 to 20 volt transformer to be safe i guess

 

Not a problem. Writing that stuff out helps me too, since I have a final exam on it within the next few weeks.

 

No, because current is only going to be flowing through 2 diodes at a time. If you want me to sketch you out a picture, it's not a problem.

 

ahh so 2 diodes are for the negative cycle and 2 diodes are for the positve cycle

a picture might help me lol im kinda one that learns with sound and images rather than words

 

Sure, I'll sketch one up for ya. Give me a few minutes. But yes, you have it right.

 

Here ya go

Attached Images

bridge-rectifier.jpg (19.5 KB, 630 views)

 

THanks

I wish you good luck on your exam

Next year in school im taking Electronic's and auto Mechanics

Im Planning on being a mechanic and doing electronic's on the side or vise versa

 

keep in mind that adding a capacitor will increase the voltage (Voltage * sqrt(2)). So if you have 12V coming out if your rectifier a large cap will make it 17V.

This means if you get a 20V transformer you'll get about 28V to feed your 7812, making it get much hotter (too hot?) then it would if you would feed it only 15V for example

 

Ya the 7812 got a 35 volt max input so at 20 volt's i probably got to slap a heat sink on it

 

Indeed. I actually don't know how transformers are specified. If they are just specified as a number of turns, then 10:1 should work beautiful for this example (since it'll have 170Vpp in, and 17Vpp out).

 

[quote="windozeuser"]
I Want To Use A step Down Transformer To Change The 120 VAC To 12 VAC Then Covert it into Pure 12 VDC


I did not get the idea.....

First, you were designing a 12 Vdc output then at the last portion it was already 15 Vdc.

If you are going to stick with th 12 Vdc output then the 12 Vac secondary output on you transformer is fine. By calculation Vdc = Vac * 1.414 = 16.96 Vdc. Minus the voltage drop on the two diodes (0.7*2=1.4Vdc) =16.96-1.4= 15.56Vdc input on your 7812 regulator.

__________________
Shalom!

 

Well It Dont Really matter how much voltage is going into the 7812 as long as it is above 12 and less than 37, the output well be at a constant 12 volts at 1 amp

 

I agree with just about everything but nobody wrote anything about
current draw or device type. What are the devices you need to power?
Notes:
If you intend to power your portable vacum cleaner from power
supply using 7812 you could be in for a nasty surprise. Read the
datasheet for the regulator.
Also about capacitance. Math is fine but it is not exactly the best
practice just to throw in insane size cap just to get ripple as low
as couple of milivolts. This is where regulators come in place.
As a rule of thumb (99% of DC devices powered by 50/60Hz)
use ca 2000uF/A (micro farad per amp).
Large capacitance also means big inrush current so be careful
in diodes selection (specially if you intent to power up unit while
load is attached).

 

And one more thing, yes it DOES matter what is the voltage in front
of the regulator. Do NOT put more than couple of volts than needed
if you want to use full current or if you are trying to use small size heatsink.
Example:
Using 7812 (12V output) at 1Amp (full current) and input voltage of
ca 15V.
Power dissipation for 7812 is (15V-12V)*1A=3Watt (not bad at all).
Now if you want to power same load and your input voltage is ca 25V
you get (25V-12V)*1A=13Watt (ouch... that is more than one of my soldering irons use!)
Did someone mention 37V on input????
According to datasheet for LM340 (this is one of the 7812 flavors)
maximum power disipation at INFINITE heatsink is under 20Watts.
(Check page 9 : http://www.national.com/ds/LM/LM340.pdf ).

So (37V-12V)*1A=25Watt (!)

 

[quote="panic mode"]And one more thing, yes it DOES matter what is the voltage in front
of the regulator. Do NOT put more than couple of volts than needed
if you want to use full current or if you are trying to use small size heatsink.
Example:
Using 7812 (12V output) at 1Amp (full current) and input voltage of
ca 15V.
Power dissipation for 7812 is (15V-12V)*1A=3Watt (not bad at all).
Now if you want to power same load and your input voltage is ca 25V
you get (25V-12V)*1A=13Watt (ouch... that is more than one of my soldering irons use!)
Did someone mention 37V on input????
According to datasheet for LM340 (this is one of the 7812 flavors)
maximum power disipation at INFINITE heatsink is under 20Watts.
(Check page 9 : http://www.national.com/ds/LM/LM340.pdf ).

So (37V-12V)*1A=25Watt (!)

I just would like to say thank you! I'm learning.....
Thanks to all and to you panic mode

__________________
Shalom!