# 7805 Voltage Regulator powering astable 555 Timer yields LOW voltage as input to timer circuit

Hello stack exchange,

So, I have two batteries powering a circuit and I want to only use one. 12.7V comes from a lead acid battery, and the smaller one is a stack of 4xAA that yield around 6.4V. The ~12.7V powers a fan, the ~6.4V is enough to power the logic.

There's a 555 timer with an astable configuration and diode parallel to R2. Another diode is in series from R2 to ground. This set up allows a full duty cycle form ~0% to ~100% to exist without being limited to only 50%-100% or 0%-50%. R2 is a pot. and rotating it changes the 555's duty cycle.

The KA7805 voltage regulator works fine by itself as well. Here are measurements taken from it without connecting the output to any load.

Both circuits work as intended when isolated. A problem occurs when they're joined together with common grounds, and the 7805's 3rd pin output becomes the 555's voltage input. At this point, the voltage from ground to the 7805's 3rd pin reads below 2V. This value isn't high enough to trigger the voltage threshold of HIGH on the 555 and so it remains forever unpowered.

*at this point, I'm not even using the AA batteries, just the 12.7V. *A DC-DC converter would be more ideal, but a Recom Power R-785.0-1.0's recommended application is similar to the 7805's. Using those could yield a similar problem.

The astable 555 set up looks similar to this but with an additional diode in series below R2. (Ra = R1, Rb = R2)

The 7805 circuit looks like this.

So, to recap: the 555 works by itself. The 7805 works by itself. Together they do not. My assumption is that somehow, between R1 on the 555 circuit and the capacitor in series with R2 (all going form V to grd) the 7805 interprets that as being in parallel to its pin 3 and its pin 2. Somehow, this set up lowers the voltage difference between those pins.

**Also, as the value of R2 (the pot) changes, the voltage across the 7805's 3rd and 2nd pin is changed.

The 7805 datasheet show that applications like this are 'intended'.

And that's a great function to have with a 7805 regulator. The ability to change the the voltage difference between pins 3 and 2. However, this is interfering with keeping the 555 circuit powered. How can someone connect the 7805 to the 555 such that the input voltage remains high?

Thank you!

• Show a schematic of the whole setup, small parts and guesses how you connected and modified them isn't going to lead anywhere. – PlasmaHH Feb 6 '18 at 10:52
• Simplify the question and add a schematic. We don't need to know what type of batteries you aren't using if you aren't using them! e don't know what vinToGrd is or what sources it. BUT when it's 6V it's too low to drive a 7805. Even your example 7805 circuit says "+ 9 to 15V" on its input - get the basics like that right, and come back to us with a focused question if there still is one. – Brian Drummond Feb 6 '18 at 11:09

## 1 Answer

You cannot power a 7805 from 6.4 Volts and expect to get 5V out of it.

The KA7805 has a dropout voltage of 2V. 6.4V minus 2V is already less than 5V.

You need to either power the 7805 from your 12V source (and properly heatsink it because it will waste a lot of power as heat) or you need to use a low drop out regulator.

Or, you could use a switching regulator from either the 6.4V or the 12V source. In either case it will waste less power than the 7805.