High/Low Logic Probe Circuit

[Kalagaraz]

Ok so my book says when switch is high, the low led won't light. But why not? Doesn't it have a straight path from +5v to ground if it goes straight down on right side?

Same thing when it's LOW. Why can't the +5v just flow straight down at all times?

 

[john blue]

When the switch is high, OUTPUT Y will be at +5V. In order for the LED to light up, it needs to have pottential difference between the anode and cathode. Thus, only HIGH indicator lights up.

 

[Kalagaraz]

When the switch is high, OUTPUT Y will be at +5V. In order for the LED to light up, it needs to have pottential difference between the anode and cathode. Thus, only HIGH indicator lights up.

Ok so current won't flow if there is a positive voltage below it, even if there is 0 voltage below that? I was imaging it in my head like there would be two flows of current like water flowing, one flowing from battery on switch side to bottom right ground, and one flowing from top 5v to bottom ground and they just joining like 2 rivers... Guess electricity does not work that way though

 

[john blue]

Yup. Let's assume +5V to the output Y which is also 5V even though you have GND below, current for that section would not flow. Water only flows from higher potential to lower potential. If the potential is the same, then water would not flow right? ~~

 

[Kalagaraz]

Yup. Let's assume +5V to the output Y which is also 5V even though you have GND below, current for that section would not flow. Water only flows from higher potential to lower potential. If the potential is the same, then water would not flow right? ~~

Makes since, but I'm still confused sorry . How exactly do you determine what the voltage at point Y is? I'm assuming that black dot just means those wires are connecting and not just over one another. I mean, why does the +5v from battery take precedence over the +5v on top? Couldn't I just as easily say that +5v on left doesn't flow because there is already +5v coming into that point from the top?

Again sorry, I'm not questioning whether your right or wrong, your obviously right. I'm just not understanding it .

 

[john blue]

hahaha.... it's alright.. The dots means solid connection. Thus, there will be 5V at output Y. Nope. you can't say that. You need to determine the voltages at the nodes before saying that whether the voltage is flowing or not .. ^^

 

[Kalagaraz]

hahaha.... it's alright.. The dots means solid connection. Thus, there will be 5V at output Y. Nope. you can't say that. You need to determine the voltages at the nodes before saying that whether the voltage is flowing or not .. ^^

So how do you determine what the voltage at the node is? I'm assuming node voltage analysis, but I can't understand that either I'm hopeless. Like what defines a "node" I use to think it was any time wires connected, but that's not the case. I can do this (How To Quickly Determine or Locate a Node In Node Voltage Analysis (NVA) | KernelTrap) in my head and figure out how many nodes there are, but I'm not exactly sure where to put the dot to measure from...

 

[john blue]

If you need to measure the voltage at one node, you just need to measure that node with respect to ground. Since in the case here, if you try to measure output Y with respect to ground, you will see that you are actually measuring the voltage of the battery, which is 5V.

 

[Zanderist]

Yup. Let's assume +5V to the output Y which is also 5V even though you have GND below, current for that section would not flow. Water only flows from higher potential to lower potential. If the potential is the same, then water would not flow right? ~~

I was going to say something like this, but you said it best.

 

[permute]

starting from the left.

if the switch is set to "high" then Y =5V. The "low indicator" and 390ohm resistance in series with "low indicator" will have 0V across them. After all, the voltage from the diode+resistor pair is 5V - Y. if Y = 5V, then this is 0V.

At the same time, the voltage across the "high indicator + resistor" is 5V. this allows current to flow and the "high indicator" will light up.

flip the switch and the "low indicator + resistor" sees 5V, and the "high indicator + resistor" sees 0V.

 

[oh_jani]

super prob

to see whole project of super prob
see the web site
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