Camera Alarm output 12VDC, 30 ma

[jalvarez]

Hi! Im new here! I have a hikvision camera model DS-2CD2163G2-IS and it has a output alarm of max 12VDC and 30 mA, how can i supply these amps knowing that the adapters of today be of 12V and 1 A min. ? I know that i must to have an relay to connect with a siren or strobo but how i can supply the 30 mA?

 

[Mark_M]

Amperage/Amp/A is a unit of current measurement.

When a power supply says it can provide '1A' it means this is the maximum amount of current it can support before failing.
It is like a weight limit, a maximum size bag on an airplane can be 20Kg. The air plane storage can support up to 20Kg but the bag could weigh less and there would be no issues.

The Hikvision camera can support a maximum of 30mA of current through the relay. This means the device you connect to the relay is a maximum of 30mA. But the power supply does not need to be less than 30mA.
The current is not forced by the power supply, it is just a maximum it can support.
You could connect a 12v 200A power supply and the camera would not become damaged, unless your load (siren) is more than 30mA.

 

[Alaska Country]

This is a crude drawing of a circuit could could work for your application.



It uses a SSR (solid state relay) that handles the AC load of the alarm device. This also assumes that the alarm is an AC line voltage device. The camera relay is also assumed to be a NO (normally open) device and only completes the circuit when it is triggered.

The 9 volt DC battery would be a temporary testing source for camera relay power. However, if the camera outputs 12 VDC at a max load of 30 ma, then the camera can be directly connected to the SSR without the battery power source.

To check, use your DMM or DVM to see if there is any DC voltage on those two output connections when triggered. Some cameras may provide an output voltage while others may provide a dry relay contact.

 

[jalvarez]

Amperage/Amp/A is a unit of current measurement.

When a power supply says it can provide '1A' it means this is the maximum amount of current it can support before failing.
It is like a weight limit, a maximum size bag on an airplane can be 20Kg. The air plane storage can support up to 20Kg but the bag could weigh less and there would be no issues.

The Hikvision camera can support a maximum of 30mA of current through the relay. This means the device you connect to the relay is a maximum of 30mA. But the power supply does not need to be less than 30mA.
The current is not forced by the power supply, it is just a maximum it can support.
You could connect a 12v 200A power supply and the camera would not become damaged, unless your load (siren) is more than 30mA.

Great explanation my friend! And very clear, I will put it into practice. Thank you so much!

 

[jalvarez]

This is a crude drawing of a circuit could could work for your application.

View attachment 188054

It uses a SSR (solid state relay) that handles the AC load of the alarm device. This also assumes that the alarm is an AC line voltage device. The camera relay is also assumed to be a NO (normally open) device and only completes the circuit when it is triggered.

The 9 volt DC battery would be a temporary testing source for camera relay power. However, if the camera outputs 12 VDC at a max load of 30 ma, then the camera can be directly connected to the SSR without the battery power source.

To check, use your DMM or DVM to see if there is any DC voltage on those two output connections when triggered. Some cameras may provide an output voltage while others may provide a dry relay contact.

Thank you my friend! The diagram is very useful. In fact that was the idea, all that remains it's to confirm if the voltage is provided by the camera.