# Why Does the Input Voltage Drop When a Linear Regulator is Added as Load

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

• 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.

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.

• Check TO92 pins 123 = Out-gnd-In Jan 5, 2020 at 3:52
• @TonyStewartSunnyskyguyEE75 REG1 (TO92) pinout is correct. V_c on pin 3, and 5 V out on pin 1. Jan 5, 2020 at 4:16
• Load/Source impedance ratio acts as a voltage divider. Find out where the fault lies. Jan 5, 2020 at 4:18
• 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? Jan 5, 2020 at 5:31
• I agree but L05 rated for 100mA means 3.3V drop max If more its a faulty regulator and should be hot Jan 5, 2020 at 6:44