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I am trying to build a 5V constant 3A DC circuit. As stated in this datasheet: http://www.ti.com/lit/ds/symlink/lm338.pdf on page no.15.

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According to calculation R2= 1.28 ohms, but I don't have it I have 1.6 near to it.

I have created the circuit using R2=1.6 ohms resistor and placed a 1K resistor as load and checked the voltage across it, it is giving me 15.68V which is a problem.

How can I accomplish this?

As stated in this datasheet: ti.com/lit/ds/symlink/lm338.pdf on page no.12. I have created the circuit and it is giving me 5.1 V constant just want to ask for confirmation that current drawn through this ckt will depend on load right? and we can take out max 5A right?

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  • \$\begingroup\$ You can use a MOSFET's body diode. But may I ask why? Why not put a 1N4007 or equivalent? If your intention is power-related then it's not needed to use a high power diode. \$\endgroup\$ May 6 '19 at 8:01
  • \$\begingroup\$ I added the circuit from page 15. Note that this is simply a voltage regulator. It will deliver a constant 5 V if you choose R2 correctly. It will not deliver a constant 3 A. The current depends on the load. This circuit can deliver up to 5 A (limit of the regulator) provided it is kept cool enough. \$\endgroup\$ May 6 '19 at 8:39
  • \$\begingroup\$ I want to use IRFZ44N as a Diode and replace it in the above-mentioned circuit WHY? The IRFZ44N has a drain-body diode, do you want to use that or do you want to make a "more ideal" diode by switching on/off the MOSFET to make it behave as a diode? What would be the advantage of that? You need to explain better what you're trying to achieve and why the circuit with diodes is not usable for you. \$\endgroup\$ May 6 '19 at 8:43
  • \$\begingroup\$ Sir, I have calculated as per formula on page15. I got R2=1.28 ohms. I am thinking of using 1.2 ohms or 1.5 ohms or say max 2Ohm resistor. I was not having the diodes so I wanted to switch to MOSFET but now the retailer said it will be delivered within 2 days. So I am gonna use 1N4007 Diodes. \$\endgroup\$
    – El_Dorado
    May 6 '19 at 9:50
  • \$\begingroup\$ I have taken I(adj)= 3A \$\endgroup\$
    – El_Dorado
    May 6 '19 at 10:04
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In the comments you mentioned that you used I_adj = 3 A. Check the electrical characteristics on page 4 or 5. That current is at most 0.1 mA. I suggest reading the datasheet in more detail to get a better understanding of how the circuit operates and to be able to calculate the proper value of R_2.

EDIT: To give some more information to the answer... the integrated circuit will force 1.25 V accross R_1. The current flowing through it equals 1.25 V / R_1. The current coming from pin ADJ is supplied by the datasheet and is at most 0.1 mA. From Kirchhoff's current law, you can tell that the current flowing through R_2 is the sum of these two currents. The output voltage equals the sum of the voltage drops accross both resistors. We know that the drop accross the first resistor is 1.25 V, and the drop accross the second one is current*resistance. Knowing the current, you can calculate the output voltage from the resistance. Or you can express the resistance from the equation and calculate that from the output voltage.

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  • \$\begingroup\$ I am trying my best to understand datasheet but this is one thing that irritates me. I am good at programming devices not on making the circuit. I am still learning. Thanks for helping me, Sir. I will calculate it and post if there is any problem. \$\endgroup\$
    – El_Dorado
    May 6 '19 at 14:20
  • \$\begingroup\$ Wow, thank you so much for the explanation. I learned a lot. I am taking I_adj=0.1 and calculating the R2. \$\endgroup\$
    – El_Dorado
    May 7 '19 at 9:49
  • \$\begingroup\$ As stated in this datasheet: ti.com/lit/ds/symlink/lm338.pdf on page no.12. I have created the circuit and it is giving me 5.1 V constant just want to ask for confirmation that current drawn through this ckt will depend on load right? and we can take out max 5A right? \$\endgroup\$
    – El_Dorado
    May 7 '19 at 10:03
  • \$\begingroup\$ @El_Dorado, You can fine tune the value of R_2 so that the output is 5.0 V if you like. The datasheet states that I_adj is typically 0.045 mA and at most 0.1 mA, but the exact value varies from package to package. The current drawn through the two resistors will remain constant, but the current drawn through the integrated circuit and the load depends on the load. The integrated circuit is a voltage regulator; its purpose is to supply constant voltage under various operating conditions. Knowing the voltage is constant, the output current is simply calculated from Ohm's law. \$\endgroup\$ May 7 '19 at 10:27
  • \$\begingroup\$ @El_Dorado, of course the regulator is not ideal, so the output voltage does vary to some extent depending on the operating conditions. This is described in the electrical characteristics in the datasheet. \$\endgroup\$ May 7 '19 at 10:31

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