CPC1002N - Using as Relay - Page 1
Forum section: electronics / beginners

author: Yaroooo
author_rank: Regular Contributor
timestamp: April 11, 2019, 04:23:26 pm
content: 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 CPC1002Nhttp://www.ixysic.com/home/pdfs.nsf/www/CPC1002N.pdf/$file/CPC1002N.pdfIn 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?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 12, 2019, 04:48:59 am
content: 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

author: free_electron
author_rank: Super Contributor
timestamp: April 12, 2019, 04:53:59 am
content: get the variants that use two back to back mosfets. polarity is no longer a problem then

author: Yaroooo
author_rank: Regular Contributor
timestamp: April 12, 2019, 12:37:15 pm
content: Quote from: T3sl4co1l on April 12, 2019, 04:48:59 amI 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.

author: Zero999
author_rank: Super Contributor
timestamp: April 12, 2019, 01:18:10 pm
content: 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-bIn 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 12, 2019, 01:25:40 pm
content: What diode are you referring to? The LED?They're not electrically connected, internally, so it doesn't matter.Electrically speaking, most passive loads don't care if they are common-ground or common-supply. There may be other considerations, like safety: you might not want active voltage on a connector or terminal block that exposes the connections.Tim

author: Yaroooo
author_rank: Regular Contributor
timestamp: April 15, 2019, 08:52:32 am
content: Quote from: Zero999 on April 12, 2019, 01:18:10 pmOk, this what I was looking for.Quote from: T3sl4co1l on April 12, 2019, 01:25:40 pmYes, to the emitter side. The LED that "simulates" gate pin of the MOSFET.I've got answer on both questions. Opto may not share ground of the MOSFET in isolation application. And Vgs, in the case of opto, is not applied by driver between MOSFET source and gate so if a load is connected to Source MOSFET high side considerations doesn't applies.