Simulate FST3126 4x bus switch in multisim

[FrankJ777]

I'm need to simulate a circuit with a FST3126 4x bus switch in Multisim. I was planning on using the component wizard to make an FST3126 as it's not one of there components, but I'm not sure how it works. Looking at the data sheet logic diagram I though the component acted like 4 switched inputs on the A side to outputs on the B side. I assumed that the switches were closed by asserting a high on OE1-4, and that closing the switch would essentially create a "short" from the A side to the B side. Reading another post though, it indicated that the outputs of the FST3126 could only be between 0V to VCC. So this would not seem like simple switched, but maybe FETs?? I'm not sure what the logic diagram is supposed to indicate. I'm hoping someone can enlighten me. I've attached part of the data sheet, and you can find the full datasheet here: http://http://www.fairchildsemi.com/ds/FS/FST3126.pdf

Thanks

 

[DragonPetter]

What is meant by a simple switch?

 

[FrankJ777]

Sorry.
By simple switch a meant that when open there is infinite resistance between the A side inputs and B side outputs; when the switch is closed there is a short circuit between the A side inputs and B sides outputs.
Also I meant that the OE leads open or close the switch completely, instead varying the resistance of the paths between the inputs and outputs, or amplifying the inputs.

 

[DragonPetter]

Those are FETs in the block diagram.

 

[FrankJ777]

So would they be N channel or P channel? Also how can you determine which leads are the source and which are the drain?

 

[gnurf]

I'm not sure where your confusion lies. From the datasheet:

When OE is HIGH, the switch is ON and Port A is connected to Port B. When OE is LOW, the switch is OPEN and a high-impedance state exists between the two ports.

I don't know Multisim, but for simulation purposes you can just treat them as either on or off.

 

[FrankJ777]

The confusion is because in another post someone said the output at the B side can only be in the range of 0V to VCC. This does not seem like Port A is connected to port B when the switch is closed. For instance what happens when port A is negative?
Also the confusion is because they are using a symbol that looks kind of like a FET. I would think that if functionally port A is connected to port B when OE is high, they would have used for symbols that look like switches instead of FETs.

 

[gnurf]

The confusion is because in another post someone said the output at the B side can only be in the range of 0V to VCC. This does not seem like Port A is connected to port B when the switch is closed. For instance what happens when port A is negative?
Also the confusion is because they are using a symbol that looks kind of like a FET. I would think that if functionally port A is connected to port B when OE is high, they would have used for symbols that look like switches instead of FETs.

It's funny because I can't think of a better symbol for a switch than a FET. But you're right, it's not an ideal switch. If you look at the datasheet (again), the ON-resistance of the FETs are 4Ω and they recommend input/output voltages between 0V and 5V. If you can't live with that you need to find another chip.