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I designed a simple LDO circuit using LM1084 (circuit A as in the picture) to power a small MCU board (circuit B) and the two circuits are connected with a 40cm wire because of installation constraint. enter image description here

When the circuit B is not connected, circuit A can output 7V as expected. However, after circuit B is connected, the circuit A output dropped to about 2.4V.

I understand that R45 should be as close to ADJ pin as possible, and the PCB picture below shows the actual layout. Is this considered close enough? enter image description here

I understand that the wire resistance could deteriorate the output as explained in the LM1084 datasheet, but it seems that would happens only when R45 is close to the load side, which I thought was not the case here.

So would anyone please advise how the voltage drop happened and how I can solve the problem?

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    \$\begingroup\$ How much load is on the output of circuit B? How much current is the 12 V supply capable of providing? \$\endgroup\$
    – The Photon
    Sep 14, 2017 at 3:58
  • \$\begingroup\$ As shown in the picture, circuit A load current to circuit B is about 35mA. The 12V supply is a DC power supply that has adequate output capability. \$\endgroup\$ Sep 14, 2017 at 4:33
  • \$\begingroup\$ I hope that isn't a wirewound pot \$\endgroup\$ Sep 14, 2017 at 4:39
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    \$\begingroup\$ Do you have an oscilloscope to check if either regulator is oscillating? Oscillation would likely lead to inaccurate measurements from your DMM, and/or cause the LM1084 to go into thermal shutdown. \$\endgroup\$
    – The Photon
    Sep 14, 2017 at 4:46
  • \$\begingroup\$ The problem was fixed by changing the resistors as AnalogKid suggested. Thanks for comments anyway. \$\endgroup\$ Sep 15, 2017 at 0:56

2 Answers 2

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The datasheet strongly recommends that R45 in your schematic be 121 ohms and not variable. For a 7 V output, R46 would be about 550 ohms. A 1K pot would center up nicely.

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    \$\begingroup\$ I thought the value 121 ohms was only for illustration purpose... Anyway, i will change the circuit accordingly to verify. Thanks for the answer. \$\endgroup\$ Sep 14, 2017 at 5:37
  • \$\begingroup\$ Yeah, it worked after changing the resistors. Thanks a lot. \$\endgroup\$ Sep 14, 2017 at 6:13
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The adjust terminal can be bypassed to ground with a capacitor (Cadj). The impedance of the Cadj should be equal to or less than R1 at the desired ripple frequency. This bypass capacitor prevents ripple from being amplified as the output voltage is increased. 1/(2 × fRIPPLE*Cadj) < R1

Try 0.1uF from Vadj to 0V, then adjust C down to make it only 10x bigger than the minimum needed to make it stable.

With R45=330 ohms and having ~1.25V across it , will conduct 3.78 mA , Pot R46 needs to have wiper pin 2 shorted to pin 1 for CW turn-increasing voltage with an R46 value of (7V/1.25)-1)*330=1518 ohms.

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  • \$\begingroup\$ Thanks for your answer, and probably a simpler way is to try 121 ohms for R45. \$\endgroup\$ Sep 14, 2017 at 5:38

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