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I created a circuit running on 5V DC, with an ATMEGA328P chip on it, to turn a 2-channel relay module on and off and receive signals with a 433MHz receiver.

As a power source I use a 9V battery which is connected to a 7805 voltage regulator to get 5V.

To have a more reliable power source I want to use a 9V power supply, instead of the battery. So I cut the wires of a 9V power supply (I bought for my Arduino), I connected them to the voltage regulator and plugged it in.

But when I tried to send a message over my 433MHz transmitter but the relay did not turn on. Then, I unplugged the power supply and connected the battery to the voltage regulator resent the signal and the relay turned on. I used a multimeter to make sure the power supply was working and the voltage and amperage where pretty much the same (arround 9V and 2.5 amps).

Why does not the power supply work?

Power supply label

AC/DC ADAPTER
Model No: MTR-910
Input: 100-240V ~ 50/60Hz 0.2A
Output: 9V DC 1A

Remeber that it worked with the battery, it doesn't work with the power supply.

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    \$\begingroup\$ My guess is that the current required by your relay is more than the power supply can supply. Do you have a datasheet for your relay? \$\endgroup\$ Jun 15 at 17:54
  • \$\begingroup\$ No but the relay is has printed on it the text SRD-05VDC-SL-C. \$\endgroup\$
    – manarinian
    Jun 15 at 17:57
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    \$\begingroup\$ After connecting your power supply to the 7805, perform a load test of the 7805's output. That is, connect a load which will draw 100, 200, 300mA from the 7805 (like resistors) and make sure the output of the 7805 remains at 5V. \$\endgroup\$
    – ErikR
    Jun 15 at 17:58
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    \$\begingroup\$ Ammeter is a meter that measures current. If your multimeter has current measuring ranges then those are ammeters. \$\endgroup\$
    – Transistor
    Jun 15 at 18:04
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    \$\begingroup\$ (1) Did you add the decoupling capacitors to the input and output of the 7805 as shown in the datasheets? (2) If not, why not? What might happen if you leave them out? \$\endgroup\$
    – Transistor
    Jun 15 at 18:07
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I will take a SWAG and assume it does not work because the 7805 is oscillating at a high frequency. I would suggest reading the data sheet for the 7805 you have and follow the recommendations it gives for the input and output. Not all 7805s are the same. You can also post a schematic of your circuit, pictures and frizzy things do not count. You probably will need a heat sink for the 7805. For the cost of the heat sink you can purchase a buck converter that will do the job probably for less money. Let us know how you make out.

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  • \$\begingroup\$ It worked with the battery, it doesn't work with the power supply. \$\endgroup\$
    – manarinian
    Jun 15 at 21:25
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Without any datasheets or schematics for your circuit, I can only make guesses. But first I have a couple questions. When you measured your power supply did you connect directly to the wires for a current measurement, with no other circuitry, i.e resistors? Did you check the power supply with it plugged into your circuit? Did you measure the current and voltage concurrently while in circuit?

Based on the label of your power supply and the fact that you measured 2.5 amps (on a power supply rated for 1A), my guess is that your power supply is dropping it's voltage to make up for the excessive current being drawn. Most likely the voltage drops below the drop-out voltage of your regulator resulting in a regulator with no output. Once the regulator shuts off, the voltage rises and the cycle begins again, essentially feeding your circuit with 5V AC.

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