0
\$\begingroup\$

Why does adding a linear regulator to this circuit cause the input voltage (V_c) to drop?

Power circuit schematic

  • The 9V input coming in is regulated. The voltage measured at the branch between D5 and R22 is ~8.7 V, so it should not be voltage sag due to the load of the regulator.
  • If I remove REG1 from the circuit, V_c reads ~8.7 V.
  • With REG1 present, V_c reads ~6.7 V. The drop in voltage is across R22.

The only other place in the circuit where V_c is used directly is to power a 2N7000 MOSFET used as an input buffer.

input buffer circuit

My best guess is that REG1 is somehow creating a voltage divider with R22, but I did not think this could happen. I don't believe my PCB has any shorts.

Note: I'm actually using a 78L05 regulator. I've tried two different ones (this and this). The full circuit diagram can be found here (on page 4).

Thanks.

\$\endgroup\$
  • \$\begingroup\$ Check TO92 pins 123 = Out-gnd-In \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 5 at 3:52
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 REG1 (TO92) pinout is correct. V_c on pin 3, and 5 V out on pin 1. \$\endgroup\$ – KLing Jan 5 at 4:16
  • \$\begingroup\$ Load/Source impedance ratio acts as a voltage divider. Find out where the fault lies. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 5 at 4:18
  • 1
    \$\begingroup\$ Hi, I am not entirely clear on the purpose of having R22 as simply dissipates power and makes all voltages after R22 load dependant. Secondly, Is Vb a voltage reference? if yes, it would be better to create a reference from after your linear regulator. Thirdly, if you see sudden voltage drop without any load, it could be a faulty regulator, if not incorrect pin usage. The use of 2N7000 is not clear aswell in terms of what is outcome you desire from use of 2n7000 circuit? \$\endgroup\$ – Nouman Qaiser Jan 5 at 5:31
  • \$\begingroup\$ I agree but L05 rated for 100mA means 3.3V drop max If more its a faulty regulator and should be hot \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 5 at 6:44
1
\$\begingroup\$

The regulator needs to have a capacitor, around 0.33uF nearby between input and ground. Also a small capacitor 0.1uF between output and ground is good practice. The regulator could be unstable without it, and without a little bit of load on the output. That may explain the current draw, and resulting voltage drop across R22. Speaking of R22, normally it is a bad idea to have a high resistance on the power feed to a regulator. Also, it is worth noting that the regulator will only work if the voltage drop from input to output is at least 2 volts. So if you need 5 V out, the input pin on the regulator must be at least 7 V.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.