CD4066B CMOS Quad Bilateral Switch question

Look at the On-State resistance data.

Typical resistance at room temperature is about 500 ohms and could be as high as 1K.

This means if you run 20 mA through the switch the voltage will typically drop 10 volts.

They are not very good at switching power. You can switch several in parallel to reduce the resistance, but it still won't be very good.

High power analog switches are available.

Thanks for the reply.

My problem is that there are extremely limited weight and size limitations for switching power in this project. As well there is a need to switch around 10 channels - hence i'm looking for a chip that weight ~1 gram, which can handle several channels, rather than using 10 times a solid state or a mechanical rely, which add up to more than 50 grams and 3 match boxes volume, which is a lot in this case. 

Do you know of a small chip that does what the CD4066B CMOS Quad Bilateral Switch does and has the same small chip size - but rather can handle around 1-2 Amps of current?

If you really need the isolation, you're probably back to a solid-state relay.
Something like this will give you 4kV of isolation and switch a couple of amps

http://uk.farnell.com/infineon/pvn012spbf/relay-photomos-spst-no/dp/1608203

What are you switching - can the things you are switching share a common ground with the Arduino ?

MK

No they can not share a common ground with the Arduino.

Jon, this relay chip looks interesting weight and size wise.

Could it be switched easily from an Arduino?

How do i handle its legs pin out? what legs are for the main voltage/current and what are for the logic voltage/current? can it handle an Arduino 5V 20mAh logic signal? why does it have 6 legs?  

Datasheet

http://www.farnell.com/datasheets/690678.pdf?_ga=1.212576983.298976934.1443043972

See bottom of page 2 for connection diagrams.

Input is an LED. Needs 25mA. If it's a hobby project, the Arduino will probably cope, particularly if you don't have all 10 on at once. If it's a product use transistors like you would with a relay coil.

In the following page, it says 'Voltage - input: 1.2V DC' does it mean that i can't use the Arduino 5V logic output directly and need to step it down to 1.2V?

Here it is: PVN012SPBF Infineon Technologies | Relays | DigiKey 

It's an LED in the input, you put a series resistor to limit the current. The Vf of the LED will be 1.2V so you take your output voltage from your Arduino pin, subtract Vf and then select your resistor to give the appropriate current using ohms law. Looking that the IRF datasheet it'll allow 2.5A output current with a 5mA LED current so 5V - 1.2V = 3.8V. R = V/I so 3.8 / 0.005 = 760R. Choose the nearest standard value to that and it should be fine driving it from the Arduino output pin. You might want to check the actual output voltage you are getting in case it's lower than 5V and adjust your resistor accordingly.

Best Regards,

Rachael

Oh, where did I get 25mA from? I know where it was - it was the Farnell product page:

http://uk.farnell.com/infineon/pvn012pbf/relay-photomos-spst-no/dp/1608202

Sorry about that - next time I'll read the datasheet.

Depending on what the load is, you might want some protection on the output.

Jon Clift wrote:

 

Oh, where did I get 25mA from? I know where it was - it was the Farnell product page:

 

http://uk.farnell.com/infineon/pvn012pbf/relay-photomos-spst-no/dp/1608202

 

Sorry about that - next time I'll read the datasheet.

It's the max control current given on page 2 of the datasheet so you weren't wrong. If you need 4.5A at the output then you'll need to go to 25mA

Jon Clift wrote:

 

Depending on what the load is, you might want some protection on the output.

Yes, that's a very good point.

Nice choice Jon,

The device is fully on with just 5 mA through the LED so it can easily be driven by an arduino with an appropriate resistor as indicated by rachaelp.

This gets you 2.5 A ac when connected like this:

There would be a voltage drop of 0.25V with a 2.5A load.

5 mA input allows a 4.5 A unipolar output when both FETs are in parallel, like this:

Nice choice Jon,

The device is fully on with just 5 mA through the LED so it can easily be driven by an arduino with an appropriate resistor as indicated by rachaelp.

This gets you 2.5 A ac when connected like this:

There would be a voltage drop of 0.25V with a 2.5A load.

5 mA input allows a 4.5 A unipolar output when both FETs are in parallel, like this:

Yes, it has the virtue of simplicity, but with 10 outputs it's an expensive solution, particularly if you don't actually need the isolation. I suspect Michael's approach might be better if we knew the problem we were solving.

Even with isolation, it would probably be cheaper to use quad optos and separate MOSFETS.