jonathanburgos

modified 7 years ago

Protector contra inversión de polaridad

155

01:51:47

Usando JFET-P
La mayoría de la corriente fluirá a través del canal del transistor. Puesto que el canal de los J-FET presenta una resistencia muy pequeña la perdida de potencia será casi despreciable .

published 8 years ago

LeButch

8 years ago

Good. This is the most efficient way as soon as I know.

hurz

8 years ago

You really know nothing about efficency and what it is.

jonathanburgos

8 years ago

Most current will flow through the transistor channel and very little through the diode. Since the J-FET channel has a very small resistance, the power loss will be almost negligible, the diode serves to improve the polarization between gate and source.

LeButch

8 years ago

Correct. That's why it is that efficient. Hurz is not.

BillyT

8 years ago

I'd put a couple of micro ohm resistors in there and check where the current is really flowing ....

hurz

8 years ago

BillyT, forget that troll, you cant help him.

LeButch

8 years ago

Most of the current goes through the MOSFET, cause it can drop to resistances from under 1 Ohm. The diode has a fixed voltagedrop of 0.7V as far as I know.

hurz

8 years ago

Here the demosrtation drop 2.4V at the diode. LeBitch follow what BillyT said and find out by yourself. Cool efficient way to burn a diode.

LeButch

8 years ago

Dont make yourself more dump then you are. You know I am right.

hurz

8 years ago

I know you have no idea and knowledge about electronic, btw yourself said that you are a noob.

LeButch

8 years ago

Tell me, why this would burn the diode.

hurz

8 years ago

Take a diode specification and lookup the graph forward voltage versus current. And readout the current at 2.4V if this is not outside the graph LOL

hurz

8 years ago

1.5A x 2.4V is BTW 3.6Watt

hurz

8 years ago

And here the circuit how it was before the last update 40minutes ago http://everycircuit.com/circuit/6347599035760640

hurz

8 years ago

Thanx @jonathanburgos at least you are listening, but stubborn LeBitch does not as usual.

LeButch

8 years ago

Then tell me, where the 1.5A come from. The diode is paralell to the MOSFET, so most of the current will flow through the mosfet. Forst calculate the resistance of both, the FET and the diode: 1/R = 1/RFET * 1/RDiode. Then calculate the real wattage. The diode does not burn anymore, even that you did not calculate how much percent of the current runs through it.

hurz

8 years ago

Learn to read the above comments, and learn how to open links with your iPhone. It doesn't make sense to continue any discussion if you don't know how to use everycircuit.

LeButch

8 years ago

Maybe you start to be construcrive, not destructive. Also I have no IPhone, but nice that you don't know other phones.

hurz

8 years ago

BTW, what is this for nonsense "1/R = 1/RFET * 1/RDiode." "1/siemens quadrat"?, LOL. Your math skills are on troll level

LeButch

8 years ago

Thats the combined resistance of the FET (RFET) and the diode (RDiode) connected in paralell. You should know that...

LeButch

8 years ago

You know that I'm right, cause you do not answear. What Hurts might do now: 1) Call me a sad clown and never answears again. 2) Tries to offend me with anything. 3) Just doesn't answear to pretend he wouldn't care. Place your bets.

hurz

8 years ago

My "answear": Check the formular again my little clown before you continue. Ridiculous.

LeButch

8 years ago

It is the right one. And it was answear 2. Nice.

hurz

8 years ago

Your ignorance is infinite. Sad clown, I will help you a last time. Do you know the difference of '+' and '*' ? Troll math skills are not helpful for electronics. Cheers Werner von Siemens.