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[coolerro]

i want to manually connect power pin (ex. CMOS 4060). VDD is pin 16 and Vss is pin 8, but they are not displayed. how can i do it? for other IC like 4066 it is possible...

[NatashaB]

Hi,

In the case of the 4060, the power pins are hidden on the symbol.

To connect the power pins, just place down a VDD/VSS net associated with the desired voltage.  One way to do this is to place down the VDD/VSS CMOS supplies from the Sources group (Power Sources family).  This will form a virtual connection from the net to the hidden power supply pins.

Note: if you want to verify that hidden pins exist on a component, double-click on it and go to the Pins tab.  This will show you all visible and hidden pins that exist on the component.

Hope that helps.

[coolerro]

i want to place few CMOS with different voltage supply. when you place a VDD supply, the voltage will be the same for all IC. this is the problem....

[clansing]

For the hidden pins, if you don't want connect them to their default nets (VDD for example in your case) you can edit them in the table of pins in the component properties.  So if you want one of them to connect to VDD2, you could place another digital power, name it VDD2, then go to the component properties, to the pins tab, select the net name VDD and enter VDD2 instead.  The hidden pins are the only pins in the table you can change in this way.

Alternatively, if you have/use the component spreadsheet, there are columns for the different hidden digital power connections for each component. That entry in the spreadsheet will give you a dropdown from which you can select any power net you have in the circuit, so you could just place another digital power and then select it for that component in the spreadsheet (you can also multi-select with the spreadsheet, so if you want to use VDD2 for a few different components you can select them all and change them all to VDD2 at once).

Hope that helps,

[howdyrichard]

My application (with a 4060 & a 4093) requires VDD to be equal to a source when it is turned on.  In this case, when a phone is picked up and the source is regulated with a 3.3V zener to GND.  It cannot have a continuous source for the "switching" part of the application (off-hook, on-hook).

How would I do that?

Howdyrichard

[_user32]

howdyrichard,

From a symbol/footprint standpoint this is quite easy.   First place the existing 4060 part (I used the one from the CMOS_5V family).   Right-click and choose.'Edit component in DB'.   On the Symbol tab copy one of the pins and paste 1 on top and paste another on bottom.  Change the name in spreadsheet view to VSS and GND (or VDD) respectively.   Exit (Save Changes) and once in the Symbol tab change the number of pins from 18 back to 16 (the hiddens VSS/GND pins will go away).  You'll also need to hit the Copy to button and change the DIN symbol as well to match...

You'll next need to go to the Footprint tab and correctly map the footprint pins for the new VSS (pin 16) and GND (pin 😎 pins you created.  Save this back to the USER or CORP database and from a symbol and footprint standpoint you should be good (verify in the layout to make sure).

However if you want to emulate the effects of turning off the 3V source regulated by the zener for simulation, you may have some additional difficulty.  You'll note that when you click the 'Edit model' button that the VSS and GND are not officially assigned in the model since this is a d_chip (XSPICE) model which means that the pin nodes are modeled slightly differently than the a normal Analog SPICE model.   VSS and GND are actually modeled at the pin level.   You'll need to go back and edit the part and on the Pin Parameters tab verify that VSS is assigned to VSS and GND is assigned to VDD.

Try this and see if this works.

Regards,

Pat

[howdyrichard]

Thank you!  I will work with this and let the forum know what happens.

[ECEkid]

There is an issue with Multisim spice model.  For example, I change the hidden pin VCC and GND of the device 74F14 from the default 5V and ground to -3.3V and -8,3V.  The output swing of the device suppose to follow the new VCC and ground, but I still observe the output swing is from 0V to 5V.  See the attached image.

[Tayyab_R]

Hi ECEkid,

I made a test circuit and it seems to be simulating correctly. It is using the -8.3 to -3.3V. You can take a look at attahced circuit.

Hope this helps.

[ECEkid]

Thanks,

D. Nguyen