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CPC1002N - Using as Relay
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Yaroooo:
After a research on MOSFET use, I've found that I'm unable to connect load on Source in N-type without additional consideration.

About this component CPC1002N http://www.ixysic.com/home/pdfs.nsf/www/CPC1002N.pdf/$file/CPC1002N.pdf

In this case I've a MOSFET N-Type driven by a LED.

This component is described as Relay on datasheet, in this case should I take in consideration that load still need be on Drain becouse there's a MOSFET inside?

Or can I put load where I prefer since Vgs is given by an optocoupler?
T3sl4co1l:
D/S doesn't matter, polarity matters.  D must be positive, S negative.  Whether it's placed "above" or "below" the load doesn't matter. :)

I don't very much like that they don't discuss the DC polarity of this part.  The only signs indicating it, are the few words in the description, and the symbol.  There is no specification for the reverse diode drop, if one is present (which there should be, if the diagram is correct).  Very easy to overlook, and easy to goof up in the design process!

Tim
free_electron:
get the variants that use two back to back mosfets. polarity is no longer a problem then
Yaroooo:

--- Quote from: T3sl4co1l on April 12, 2019, 04:48:59 am ---D/S doesn't matter, polarity matters.  D must be positive, S negative.  Whether it's placed "above" or "below" the load doesn't matter. :)

I don't very much like that they don't discuss the DC polarity of this part.  The only signs indicating it, are the few words in the description, and the symbol.  There is no specification for the reverse diode drop, if one is present (which there should be, if the diagram is correct).  Very easy to overlook, and easy to goof up in the design process!

Tim

--- End quote ---

I can respect polarity, but the fact that a MOSFET is inside make me think about S and D load position.

First thing is, if grounds are shared, so diode (-) is connected to MOSFET (-), in this case I should consider voltage drop on Vgs and place the load on Drain side?

Do you think there're construction particularities that MOSFET (-) should be source?

An additional question. In an Optoisolated circuit, ground of the load should or not be connected to controller ground? Maybe I'm doing wrong so GND shouldn't be shared becouse we're loosing physical isolation of the two circuits.

I'm always talking in DC-DC connections.
Zero999:
No, the diodes anode is connected to the source and the cathode to the drain. If you connect it up backwards, current will simply flow through the diode, whether the MOSFET is on or off. The body diode is present in all MOSFETs and this SSR is no different.
https://www.google.com/search?q=MOSFET+body+diode&ie=utf-8&oe=utf-8&client=firefox-b

In an opto-isolated circuit, it doesn't matter whether the driver's ground is connected to the load's or not. If you want isolation then don't connect them. If you don't want isolation, then you can connect them, but the there are probably cheaper alternatives to a solid state relay.

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