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schematic

simulate this circuit – Schematic created using CircuitLab

I checking the output of the circuit's current by attaching multimeter(ON 10 A) in series with a load of 3 Ohms resistor. But I am getting a 0.22mA current output.

Here's the diagram (LM338K circuit) :

enter image description here

This is the circuit I am trying to build:

http://www.ti.com/lit/ds/symlink/lm338.pdf on page no.12.

I connected my multimeter's red wire to capacitors (almost red color in the picture) and black to 3 Ohms resistor.

Please tell me what I am doing wrong. According to schmetic, I should already get 5V output. But I am not getting that too.

PS: Every 1.5 + 1.5 Ohms resistor I add as a load after taking out from circuit it doesn't show any resistance on multimeter.

Hi, I have not got a solution for this please help me regarding this.

If I use LM2576 12-V version to get 5.1V with 2.2A will this be fine?

Photo : enter image description here

enter image description here

I am getting 15.68 V without Load resistor and with load resistor same voltage and this burns resistors. Please help.

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    \$\begingroup\$ What is Ckt? And more imporantly, can you add a little bit of context? What is this device for? Also, your multimeter fails to measure resistance, so why do you trust its current measurement? \$\endgroup\$
    – Dmitry Grigoryev
    Commented May 21, 2019 at 7:09
  • \$\begingroup\$ Acutally My resistors are being burned down. I don't know why but till morning it was working great. And Now my Ckt is not working as usual. It is not giving me 5V constant output. I am creating this ckt to use as a power supply for my Raspberry pi 3B+. \$\endgroup\$
    – El_Dorado
    Commented May 21, 2019 at 7:21
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    \$\begingroup\$ What is the measured output voltage of the regulator? Is the fuse in you multimeter blown? \$\endgroup\$
    – Peter Karlsen
    Commented May 21, 2019 at 7:29
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    \$\begingroup\$ How much cooling do you have on the regulator? Sounds like the regulator is shorted from input to output. \$\endgroup\$
    – Peter Karlsen
    Commented May 21, 2019 at 7:41
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    \$\begingroup\$ Ballpark figures you have configured this linear regulator to drop about 10V then supply about 2A. That's about 20W. This part may be specified for 5A but it still needs help to get rid of all that heat. \$\endgroup\$
    – user133493
    Commented May 21, 2019 at 7:44

2 Answers 2

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OK, you're working on a breadboard.

Breadboards have so much contact resistance that it will be impossible to make your circuit work properly. You expect a couple of Ampere to flow, that's not going to happen in this setup.

I would only rely on a breadboard for currents up to 100mA, for higher currents the voltage drop across wires and contacts is simply too high preventing the circuit from behaving as it is designed to.

Also you're not using any heatsink on the voltage regulator so even if a high current would flow it would heat up and limit the current to protect itself.

Your circuit's schematic looks OK, you just need to build it on some prototype PCB (or veroboard) so that you have soldered contacts instead of plug-in contacts as on a solderless breadbord.

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  • \$\begingroup\$ Ok, I will try that. It has an LM7805 ckt also which supplies 900mA current. This ckt was working fine till now what happened I don't know :( \$\endgroup\$
    – El_Dorado
    Commented May 21, 2019 at 9:11
  • \$\begingroup\$ LM338K is a 5A output IC And why am I not getting required Voltage if the ckt is fine? \$\endgroup\$
    – El_Dorado
    Commented May 21, 2019 at 9:16
  • \$\begingroup\$ If I don't add Load resistors I can't draw 5.1 V right? And can't use it for my application.I will eventually get Input Voltage of IC only \$\endgroup\$
    – El_Dorado
    Commented May 21, 2019 at 9:29
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    \$\begingroup\$ From your comments I see that you are still missing the point and do not understand how things work. I can't draw 5.1 V right? Without load resistors you should see 5.1 V if the resistor divider is OK. Currents are "drawn", voltages are not. I will eventually get Input Voltage of IC only No, that's not how voltage regulators work. There is always a small load: the voltage divider. You really need to educate yourself more on this or just use ready made modules like: aliexpress.com/w/wholesale-lm2596-module.html \$\endgroup\$
    – Bimpelrekkie
    Commented May 21, 2019 at 9:45
  • \$\begingroup\$ I was just confirming Sir. I am not getting 5.1 Voltage as an output. Please, can you check it with the datasheet ckt reference I have provided above, what are my mistakes and all? \$\endgroup\$
    – El_Dorado
    Commented May 21, 2019 at 12:42
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For a 1.7A load you are dumping heat with the voltage * current drop that tries to heat up U2 to 434'C before OTP cuts in.

Not a great way to provide power, but the lossy way is to use a power resistor or light bulb up front.

Next time try to match your supply to the load, rather than drop down to 1/3rd the input for high current.

enter image description here

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. Any conclusions reached should be edited back into the question and/or any answer(s). \$\endgroup\$
    – Dave Tweed
    Commented May 31, 2019 at 18:45

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