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In PSoC Creator, there is a resistive pull up and resistive pull down. From the documentation, the initial drive of pull up is logic 1 when for pull down is logic 0.

Pull Up:

When the DR=1, the upside transistor will switch on, whereas the downside will switch off. Then, what happens in the buffer?

Pull Down:

When the DR=0, the upside transistor will switch off, whereas the downside will switch on. Then, what happens in the buffer?

Actually, what is the effect of the pin that is connected to the buffer?

From cypress documentation

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The purpose of the buffer is to reflect the value of the Pin at all times. For this circuit, all you have to ask yourself is 'what is the Pin doing?'.

Digital buffers are often used to increase the sourcing current capability of the node. For instance, if the resistor is in the path, then the Pin can't source very much current. The buffer takes care of that part only while accurately reflecting the state of Pin.

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  • \$\begingroup\$ So, if the pin is push button and I push the button, then the buffer act as switch ON. Is it right? Then, what happens with the current flow(in resistive pull up) \$\endgroup\$ – ardiantovn Oct 23 '17 at 1:45
  • \$\begingroup\$ Depends on how the button is configured. If the button is configured to pull the Pin up when it is pressed, then the output of the buffer will be high when the pin is pressed. The buffer always reflects the pin, so make the pin do what you want it to do. \$\endgroup\$ – slightlynybbled Oct 23 '17 at 1:47
  • \$\begingroup\$ Actually, if I set the initial drive = 1 in PSoC Creator, do I set DR / PS to have initial value logic = 1 or set Pin value to have initial value of logic 1? \$\endgroup\$ – ardiantovn Oct 23 '17 at 1:57

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