help with circuit flow directions vdd vss vout

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duxbuz

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Hi


I am reading this article :
https://www.pcbdesign007.com/pages/columns.cgi?clmid=17&artid=20146

and I am having trouble understanding the current flow between Vdd, Vss and Vout in the circuit diagram in figure 2

The author states:

I thought that the current would be flowing from Vdd to Vout (or Vdd to Vss if switch closed) but when I read the article it seems to be flowing from Vout to Vdd?

This sounds like its going backwards, unless Vout is more positive than Vdd??


Is it something to do with electrons flowing neg to pos?

If this is the case then being more neg vout would flow to vdd (pos), and then if switch is closed vout would then be more pos than vss(ground) so errr.... then what???

I know this maybe trivial but could someone help me understand how current flows in this circuit?

I realise that I do not have much of a grasp of current flow, so any help is good.

Thanks

p.s. If anyone understands why I don't understand!
 

I think you are complicating the operation too much. At the bottom end of the resistor, imagine an arrow, this is the current being "pushed down" by Vdd (the most positive voltage) . If the switch is closed, the current flows through the switch down to Vss. If the switch is open the current would like to flow out through the external load and finally back to Vss. If it does or not depends on the external load and what its connection is to Vss. If its connected to a voltage equal to Vdd, then there is no current. If the external voltage is greater then Vdd, then that voltage supplies the current backwards to Vdd.
Frank
 

It doesn't matter which way electrons flow. VDD can be positive, negative or an AC audio tone instead of a DC voltage.
The resistor connects the output to VDD when the switch is OPEN (off).
The switch connects the output to VSS when the switch is CLOSED (on).
 

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