There is no point putting and extra diode at the output, all it will do is drop the voltage a bit.

It will not provide any protection from reversing the battery, that would still cause extremely high currents.

All you can do to reduce the chance of damage is fit another fuse in one of the output leads.

 

There is no point putting and extra diode at the output, all it will do is drop the voltage a bit.

It will not provide any protection from reversing the battery, that would still cause extremely high currents.

All you can do to reduce the chance of damage is fit another fuse in one of the output leads.

Why , won't the diode work is it they don't make diode that are rated high enough to protect connecting the leads in the wrong way? (
Crank-amps 700amps???
I think I have to disagree with you here.

SgtWookie would be in disagreement to by what he said.

Anyway yes I can use a fuse I was think that originally but realized everytime somebody connected it in the wrong way they would have to buy a new fuse for it. Also what fuse ratings did you have in mind 250V, 3amps ... ? Or are you talking about a fuse between my transformer and the rectifier/cap or before my transformer ???

Also I figured out how much longer it would charge with pulsating dc.
This can be determined by the area under the pulsating dc curve as opposed to the area under the smooth dc line. The ratio of the areas gives approximately how much more time is needed. It is impulse function or change in momentum formula ,...etc .

But for get all that My transformer is 120vac to 12vac with max current rated at 300mA.
Curious to know when the max current will flow in secondary coil.
Is it when you directly connect the secondary wires together.

coil ratio is 120/12 = 10 turns = I2/I1 so ( I1 * 10 ) = I2
I2 has ten times as many turns in it then I1
If the max current is 300mA does that mean max current thru secondary wire or primary wires?

If secondary then I1 = .3A/10 = .03A max current in primary
If it is primary then I2 = 3A max current in secondary.

From this we can figure out the transformer max coil resistance
primary 120/max I1 = max R1
secondary 120vac/max I2 = max R2

Also Ideally Pin would equal Pout minus a small percent to heat dissipation.
But lets assume it is a perfect power in = power out for are approximations.

Using current values from above
If the I1 max current = 0.03A => 120vac * 0.03A = 3.6Watts
If the I2 max current = 3A => 12*3 = 36Watts
If the I2 max current = 300mA => 12 * .3 = 3.6 Watt *******

If ****** is true then 3.6watt/120vac = 0.03amps for primary wires.

Ok, This concludes the car battery charger power supply.

What happens if I wanted to smooth the dc so smooth that it could be used to power on an old laptop. How small would the flucations need to be made in the 12volt dc + or - what? As well what would the current have to be and how small would the flucations need to be in the current?

I have this old AC IBM ADAPTER for an old THINKPAD COMPUTER
Is say's input 100 - 240V ~ 1.2A - 0.5A 50/60hz
output 16V _= _ 3.36A LPS

I guess in this case I could use 120vac to 24vac and use resistors to drop voltage down to 16/17 volts dc then smooth it out but how smooth would be smooth enough? Also how would you get steady current?

Thanks is their regulators that you can get that regulate both the current and voltage. Better yet is their variable regulator that allow you to set both the current and voltage. Either way curious on the current issues the voltage part seems easy enough. Maybe If I connected enough resistors in parrell after the smoothing I could drop the current to the computer to 3.36A or then again If I had to raise the current I don't know how to do that without raising the voltage?

Either way I would like to create a simple ac adapter for my old IBM THINKPAD computer using what we just talked about in the last 3 pages.

The major trick would be how to increase the current while keeping the same voltage.
I know how to drop current while keeping the same voltage but raising I don't have a clue.

 

Nope, the diode won't help you if you connect the battery in reverse.

It will keep the battery from discharging through a leaky capacitor or rectifier bridge, or from charging the cap in the first place - but if you connect the battery in reverse, you will get a heavy current before something gets fried to a crisp.

Better just to use a fuse on the output of the charger.

 

OK, what ratings for the fuse i.e what voltage/current ratings?

But I am still unsure why the diode won't work it only lets in current in one direction so if they connected the wrong way nothing would happen because the diode won't let current thru? Maybe you would need 2 diode one for the + one for the - side though? Still curious on what the problem is?

Curious about the ac adapter computer questions?
How much fluctation ,...etc?

Thanks