The question is in title. Why there is no ground pin on the 12V RGB header? The 5V ARG header which uses only 3 pins has a ground.
How does it work without a ground?
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4\$\begingroup\$ Who says you have to drive the RGB lines high? ;) Drive them low and you have your ground. \$\endgroup\$– MichaelCommented Apr 14, 2023 at 19:57
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4 Answers
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This is how it looks on the driver end. Whatever channel you want to drive, the R, G and/or B becomes ground with the help of a transistor, completing the circuit and allowing current to flow.
simulate this circuit – Schematic created using CircuitLab
EDIT: Added colors for clarity.
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3\$\begingroup\$ Good answer. Note that this configuration is an example of low-side switching, where the switching device (transistors in this case) are in the return, or low-side path of the power flow. \$\endgroup\$– SteveShCommented Apr 14, 2023 at 13:29
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1\$\begingroup\$ @SteveSh Unless someone invents a new type of transistor with majority carrier yet controlled by negative gate voltage (depletion MOSFET comes close if they would be off with 0 V Vgs), we will have low side switching for the foreseeable future. \$\endgroup\$– winnyCommented Apr 14, 2023 at 13:46
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\$\begingroup\$ @winny Isn't that just a P-channel FET? \$\endgroup\$– HearthCommented Apr 14, 2023 at 21:36
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2\$\begingroup\$ @winny No, p-channel devices are still majority carrier devices. The fact that holes have lower mobility in silicon doesn't make them minority carriers--it's the doping that determines that, and the doping in p-channel devices makes holes the majority carrier. Just like the doping in n-channel devices makes electrons the majority carrier there. \$\endgroup\$– HearthCommented Apr 15, 2023 at 2:45
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2\$\begingroup\$ @winny Yes. Minority carriers are only involved in conduction in any significant way in BJTs, diodes, and thyristors; MOSFETs, JFETs, and HEMTs are majority-carrier devices. \$\endgroup\$– HearthCommented Apr 15, 2023 at 11:36
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But there is a ground. Or actually, three grounds.
The three RGB wires are the grounds for the three LED colours.
The three colours just share a single +12V supply wire.
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\$\begingroup\$ The RGB wires are the cathodes. They're grounded through the controller. (I know you know. You just didn't explain it well.) \$\endgroup\$ Commented Apr 16, 2023 at 20:03
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The anodes of all LEDs are connected to the 5V/12V. The cathodes of the LEDs go to the R, G, and B pins of the connector. The controlling device then connects the pins to ground to make the respective color light up.
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\$\begingroup\$ The 5 V version is the addressable type. The connector shows +5, D, nc, GND. \$\endgroup\$ Commented Apr 16, 2023 at 20:13
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simulate this circuit – Schematic created using CircuitLab
Figure 1. A manually switched controller for an RGB LED strip.
Closing the R, G or B switches will ground the last cathode of each string of R, G or B LEDs and cause them to light up. The switches can be replaced by transistors and these can be controlled by a microcontroller.
How does a 4 pin RGB connector work without ground?
The ground is provided by the control circuit.
answered Apr 16, 2023 at 20:12