# input voltage drop after connect to voltage regulator LM7805cv

So this is an inductive sensor, it has 3 wires ( black, brown, blue). According to the manual the brown is for VCC input (10-30v), blue is GROUND, black is signal(OUTPUT). This sensor is labeled as NPN NO (normally open).

First i did some test using 12V power source and from the output signal i got roughly about 12V.( Also if i use 9V vcc, i got output around 9V) So no problem so far.

Now i want to regulate that ouput signal from 12v to 5V. So i'm thinking to use LM7805cv voltage regulator to convert the output signal from 12V to 5V, nothing fancy ... But then here come the problem. As soon as i connect output signal from sensor (12V) to lm7805 input, the output that coming out from lm7805 is about 2-3V ( it's supposed to be 5V). So then i check the output voltage from sensor ( which is connected to the input of LM7805cv) and surprised that i got about 4-5v, it's supposed to be 11-12v right ?

If i disconnect the lm7805, the output from sensor back to 12V, but if connect to lm7805, that output from sensor drop to 4-5V.

I also check that there's nothing wrong with this lm7805. As i tried to connect it to 12v or 9v directly (without sensor) and i got 5V output.

Anything wrong? is it normal i got voltage drop that big? i'm not an electrical guy, i just know a bit about simple wiring. So anyone can explain this?

Using a voltage power regulator is not a right idea for voltage translation. According to the device internal diagram, it has effectively 47k of output impedance when it is OFF (output is HIGH), and 10-12 Ohms when active ON. When you load the 47k output with LM7805, it sucks all available current and voltage drops low.

The simplest way to get 5V pulses out of this device is to simply use ~33k resistor to ground (assuming 12V power supply). If you need more powerful signal, you can use, say, 2k resistor pull-up to 5V power rail of your MCU or whatever you are using to acquire the switch information.

• Hi Ali, thanks for the reply, i will try your suggestion and see if i can work it out. thanks. – andio Oct 20 '16 at 6:23

Looking at the schematic of the sensor on the website you linked to, the output is an NPN transitor, emitter grounded, and as 47K pull-up resistor to Vcc. You can't draw much current from that output without the voltage dropping due to the current through the 47K resistor.

@Peter has the cause- to solve it you need only add a pull-up resistor on the output to your 12V source. The 7805 needs about 5mA internally and you have to account for whatever load you intend to put in the regulator output. If I assume that load is say 1mA then the resistor should be:

$R_P = \frac{12V - 7.5V}{5mA + 1mA} = 750\Omega$

The minimum value is about 60 ohms because the switch can handle 200mA, but too low a value wastes power.