got to add a 4' heat sink n a small fan to it to make it last

@@808pathfinder Yeah I noticed how fast it heats up if the load draws a lot of current even with a heatseek. Maybe a resistor on the output would help regulate the current draw.

@@johnruscigno5738 Can I realise the same connections but using IRF 540 transistor or Lm317 and B 100 k potentiometer and 10 k resistor ohms???

@@tetraederzufrequenztesla2342 that's a good idea. The lm317 is a voltage regulator that limits the current draw as well I believe. That should stop the mosfet or transistor from overheating.

Ok

Meu amigo vc utilizou à fonte dê 12 v foi nã entrada

@@lilasouza9375 na verdade eu ainda não montei, mais a saída 12v da fonte é ligada sim na entrada do circuito, assim pode sair 1~12v do circuito!

@@gabriel-dz7zg obrigado meu irmão

So if the load goes up the voltage will drop?

I'm trying to grasp the principle.

The circuit is a simple "source follower" setup where the voltage at the source follows the voltage set by the potentiometer. This leads to three problems, one minor and two major. The minor problem is that the output voltage changes with the load current, but not by much. The output impedance is the reciprocal of the forward transconductance (minimum of 19 S from the data sheet), so the output voltage will drop by about 0.05 volts per amp. The first major problem is that the output voltage changes *directly* with any change in input voltage, so there is no regulation as such, and no ripple or noise rejection. The other major problem is that without a heatsink, the IRFZ44N has a temperature rise of up to 62 C per watt. A circuit similar to one demonstrated used to drive a 12V LED from 29V supply would drop (29-12) = 17 volts across the MOSFET. If 1 amp of current were provided, it would produce (17 x 1) = 17 watts in the transistor, potentially raising its junction temperature by as much as (62 x 17) = 1,054 degrees Celsius. These MOSFETs were designed for switching operations where they are either fully on or fully off. Using them in linear applications requires a sizeable heatsink.

Olá a fonte pode ser de até 50V?

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Can suggest a transistor for 15watt... ☺️

Почему не будет ? Пару ампер пропустит

Yaxshi ishlamaydi 4 marta kuydi

@@user-gw3qc9fg9f а потенциометр выживет?

For me the best solution is to buy a switching power supply. But if u want create a circuit at home, the easiest way is a linear psu. Transformer, full bridge rectifier, capacitor and a Lm7812. But this ic can output max 1amp. U have to add an external power transistor to amplify the current capability on this circuito.

After proper cooling system it can be use. Am I right?

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👍👍👍👌

A need that answer

Se inscrevi no seu canal curto eletrônica de +

Sim, adquira seguro de vida.

Wahhhh kya knowledge hai

😂😂😂

Saludos desde honduras

En que idioma escribe?

Maximum amp 35A and It depends on your power supply current

Eeee ta capio injoma

Θα παίξει σε 12v αμεμιστηρακι PC ? θέλω να μειώσω λίγες στροφές10-20% και να εχω τον έλεγχο ?

Heatsink

Bhai ek to voltage drop half krdia upar se curent b 4 amp choti c ic ki leli garm to hogi e

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Não é componente, é só um jamper( pedaço de fio )

Good question

That's connect a random pin of potentiometer to the drain of mosfet

@@s0ssis926 a random pin you meant the one in the middle of potensiometer?

@@anismuha yes