Just forgetting to turn off the multimeter several times before the long holiday weekend, I found that the batteries were completely discharged. I decided to make a simple switch off delay relay for 4-6 minutes. There are many circuits in the public domain, from the simplest on one field-effect transistor
So on the microcontroller
But I am not friends with the latter – this time.
My working multimeter is powered from 3.2 to 4.5 V – that's two …
And I would like the current not to be consumed in standby mode – that's three! < br> At first I assembled on transistors:
Works well, but standby current is 1.5mA. I think that this is quite a lot and gave up on this scheme.
Here is a very good scheme
But my multimeter is powered from 4.5 to 3.2 V. You can put transistors with a logic open level (full channel opening at 5 V). But at a supply voltage of 3.1 V, the multimeter starts to lie madly in the big direction and, moreover, very much! And the transistors in most simple circuits close linearly and the moment may coincide when I will measure the voltage and on the multimeter at that moment there will be just 3.1 V … There is of course no indication of the battery discharge in this miracle of measuring technology!
I remembered the previously assembled fan off delay relay and decided to make a similar relay for a multimeter based on it.
After some time spent with a soldering iron, it turned out like this
Powered by today, the shutdown delay time is about 4-5 minutes.
To turn it on, I used the HOLD button, I still don't use it. But you can attach any miniature button with normally open contacts.
The principle of operation is quite simple:
When you press the crop S1, a negative voltage is applied to the base of the upper transistor BC857, it opens and supplies power to the time relay assembled on the NE7555 timer. Through the upper resistor of 2.7 MoM, a 100 uF capacitor connected to the 2 and 6 legs of the 7555 timer begins to charge. While the capacitor is discharged and the voltage at 2 and 6 legs is less than 2/3 of the timer supply voltage, there is a positive voltage at the 3rd output of the NE7555, which opens the lower transistor BC847. Which, in turn, duplicates the pressed contacts of the S1 button and keeps the upper transistor BC857 open. If you release the S1 button, then its contacts will be duplicated by the open transition of the K-E transistor BC847 and the circuit will continue to work.
As the 100 μF capacitor is charged, the voltage at pins 2 and 6 of the 7555 timer begins to rise, and as soon as it reaches 2/3 of the supply voltage (in this case, about 3 V), the timer switches to off. On its 3rd pin, the voltage will disappear, the lower transistor BC 847 will close and close the upper transistor BC857, which will turn off the timer power. The power supply to the multimeter will break. Perhaps some tenths of microAmperes will pass through the closed transition, but this is commensurate with the self-discharge current of the batteries and can be neglected. This is my opinion and I could be wrong.
The next time you press the S1 button, the cycle will repeat.
I drew the board for SMD components
So: I draw, iron, baiting, soldering connecting ….
And after setting for a convenient delay time, I attach it to the case.
Details are not critical, the main thing is that they fit in type and face value.
For more details on the applied details and settings, see archive at Yandex disk.