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I'm currently working with a simple 555 variable frecuency aestable circuit, that switches a relay at the desired frecuency. Here is the schematic so you can understand my issue better: Circuit schematic The circuit is quite simple. J1 should have a 12v battery connected to it, and depending on the state of the SW1, power can go to the 7812 or directly to the relay. This is because when the switching has to stop, the relay has to stay energized, and I figured that by "unplugging" the 7812 and the 555 I would be able to "reduce" energy waste. Now, the rest of the circuit is self explainatory: The 555 switches on or off the PNP transistor at the desired frecuency (selected by the potentiometer) and as a bonus feature, an LED lights up when the relay is de-energized.

The issue is the next: Whenever SW1 is "OFF", the LED should stay static (on or off, it dosent matter), but instead, it keeps oscilating. I can still tune the frecuency using the potentiometer, and as soon as i switch SW1 to the "ON" position, everything goes back to normal working order.

I've tried to figure out by my own why could it be. I started looking to the LED, and noticed that maybe, by having the LED connected directly to the battery instead of the 12v regulated (to reduce wire count on the connector) may be having some effect on the 555 circuit (like a reverse power or something). What was a bit more concerning is that i took a look at the "12v" line, and saw, with my multimeter, around 12v, but with some fluctuation. When i cranked up the frecuency to the max (30 something Hz) the voltage got more stable, and lower. Its there, that i decided out of curiosity, to measure the voltage at the input of the 7812, and saw around 5v, which for me, makes absolutely no sense.

Now, I would like to make "reverse power" go away, but I dont know what to do/add/cut(yes, i already did the pcbs). I attached a video of the circuit working and reproducing the issue. Video Demonstration

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  • \$\begingroup\$ N.O. and N.C. (normally open and normally closed) are descriptions of the contacts not descriptions of relay state. Use energized and de-energized if that's what you mean. The transistor is PNP, not NPN. It's not clear what you expect to happen when the switch is in the 'off' position, why are you even using the back contact on the switch? If you interrupt the power to the whole circuit (by only using two connections to the switch), how is that different from what you want? \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 11, 2023 at 16:21
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    \$\begingroup\$ When SW1 is on the "off" position, the relay should be energized and the 555 and 7812 should be de-energized, meaning that the LED should stay off. When SW1 is in the "on" position the 555 and 7812 should be energized, and the PNP should drive the relay using the output from pin 3 of the 555. When SW1 is on the "off" position, i need the relay to be energized. Thats why i use the three connections of the switch \$\endgroup\$
    – fpp
    Commented Jul 11, 2023 at 16:36

3 Answers 3

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Yes, that is normal. When a chip is without a supply, there generally must not be any voltages present on any input or output pins, or the chip internal structures can get forward biased and leak current. Inside, it's just a bunch of transistors and resistors.

So currently, if the 555 has no supply, current will leak via Q1 and R3 into chip output. It may rise the chip supply so it starts operating in some way, correctly or incorrectly. While current leaks via Q1 it gets turned on and drives the relay. Another leakage path is via LED to 555 output.

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  • \$\begingroup\$ Thanks for making me notice that, but i have a question. Is there a way to know how much current can the LED and Q1 leak? That way i can know from where to start modifying stuff (one thing for sure is to replace the BJT for a MOSFET) \$\endgroup\$
    – fpp
    Commented Jul 11, 2023 at 18:05
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    \$\begingroup\$ @fpp Just measure it with a multimeter. If you can't measure current, measure voltage over resistors and use Ohm's law. But the point is not how much it leaks. Less leakage means just slower rise of voltage and it won't remove the problem, although very low leakage levels might not charge up to high enough voltage to start it up. Generally, any leakage is bad and it should be zero. Don't add leakage protection to a "bad" circuit, design a better circuit with no leakage to begin with. \$\endgroup\$
    – Justme
    Commented Jul 11, 2023 at 18:28
  • \$\begingroup\$ With my multimeter on current mode i measured around 3mA from the base of Q1 before the 1K resistor connected to pin 3. Doing the same from the cathode of the led to before the 1K resistor i got around 100uA. I dont think those measurments are good, but i can defenetly see a difference in intensity of the LED when unplugging the base resistor \$\endgroup\$
    – fpp
    Commented Jul 11, 2023 at 19:37
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I think you can fix this simply by replacing the 2N2907 with a P-channel MOSFET

enter image description here

The pinout is even compatible.

You will also have to remove the 'bonus' LED. It can be wired across the relay coil instead (with a suitable series resistor) if you really need it.

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  • \$\begingroup\$ Ok, seems easy enough and I wont have to remake the pcb. From what @justme said, replacing the BJT with a MOSFET would solve half of the problem by getting rid of the base current leak, but what if I cant get rid of the LED? Because, i want to keep it, but also, I dont want to add another pin to the connector to get 12v and GND. Can i just move the cathode of the LED to the mosfet drain with a pull-down resistor? Would that still affect the 555 and leak enough current to power it up? \$\endgroup\$
    – fpp
    Commented Jul 11, 2023 at 18:03
  • \$\begingroup\$ Maybe you can move one end of R4 to GND then the LED between 4 and 1. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 11, 2023 at 18:40
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After a lot of thinking, i noticed that i could've just wired things a bit different to get the same functionality and with no current leaks. Instead of connecting my load between the common and N.O. pin of the relay, I instead decided to connect it to the common and N.C. pin. That way, when the circuit is off (either no power or SW1 on the off position) the path for current will remain closed, so i can get rid of that +BATT voltage when SW1 was on the off position (Wich was one of the biggest current leaks). With that modification, i guess that Q1 stopped leaking current not oly to the base but also to the emmiter, wich i guess was the reason (that i didnt see) of why the 555 was always powered. Now the LED remains OFF when SW1 is on the off position. Only downside is a half second delay from when i change SW1 to the on position until the first pulse appears

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