Z
Z
zisp2022-02-14 19:55:48
Electronics
zisp, 2022-02-14 19:55:48

The motion sensor at the output gives 3.3V, how to raise the output to 5V?

Hello.
The SR-505 motion sensor works from 5V to 20V, I supply 9V to it, but in any case it gives 3.3V at the output, and I would like either the same input voltage or 5V. What are the simplest options for converters? There are many capacitors, transistors are in short supply, one DD127. There is also a powerbank module, to which you supply 3.7, and it gives 5V at the output. I tried it with a powerbank module, but it does not output anything at the output (0V, or 0.10-0.20V. In general, it does not work).

Answer the question

In order to leave comments, you need to log in

3 answer(s)
A
Armenian Radio, 2022-02-14
@zisp

Apply this signal to the base of the NPN transistor C3779 through a resistor of approximately 150 to 310 ohms, building a simple transistor switch. You definitely need a diode connected in anti-parallel to the relay winding, otherwise the transistor will turn out to be disposable.
This is called - a key on a bipolar transistor - it describes how to accurately calculate the resistance.

P
pfg21, 2022-02-14
@pfg21

If we exclude that dd127 is a high-voltage power transistor. Those. somewhat not for low-voltage signaling systems :)
Then it is quite possible to use this transistor in common base mode.

V
Viktor, 2022-02-14
@nehrung

The easiest way is to use a field effect transistor. You need to turn it on according to this simple circuit, which will fit almost any medium or high power field effect transistor with an N-channel (it is unlikely that you will come across those among them that cannot withstand 9 volts or do not provide the current your relay needs).
i?id=81fec0c84bf617ec662053b6b4ac5796-5660528-images-thumbs&n=13&exp=1
More precisely, even more simply, since you need to apply your output 3.3 volts directly to the gate of the transistor, so all these input resistors and a capacitor are not needed (however, you can leave a resistor between the gate and source with a value of 10 kilohm). And of course, you take the relay for 6 ... 9 volts, and at the point where 12 is written, you supply your 9 volt power supply. The diode parallel to the relay winding is also almost any rectifier.
I wrote "almost any" here - this is not good for a beginner, so if you have any difficulties with the choice, write in the comments what you have found, and I will tell you if it is suitable.

Didn't find what you were looking for?

Ask your question

Ask a Question

731 491 924 answers to any question