Something wrong

• posted

Hi, Based on this circuit, I have some strange results from simulations.

5V ___ | | \ / 2K \ | |--------------------------o C | | | | \ | / 2K | \ | | | | | | | o o A B

---------------------------- A B C 0v 0v 0v 5v 0v 0v x 5v 5v (something wrong here)

---------------------------- I think that when point B is 5V when point A either 5 or 0v, point C should not get 5v as well. But I just cant figure out why the simulator behave like this. I am using Electronics Workbench 5.12. Thanks in advance

• posted

Aren't B and C the same node?

John

• posted

But with point A equal to 0v. The node between 2 resistors should have

2.5v according to the voltage divider rule. So this is something that I am not so sure with the 'mix' of 2.5v and 5v (point B). Is point C still 5v under this circumstances ?

• posted

It would if nothing else is connected to it. But when you apply a hard voltage source to the junction, the source wins.

John

• posted

This is correct, as long as *no voltage* is being applied at point B. The resistance between B and C is 0 ohms, so Vc will always equal Vb.

The voltage chart is correct for the circuit as drawn.

If you connect B to 5v, you're putting a 0 (that's zero) ohm resistance in parallel with the upper 2k resistor. What does your 'voltage divider rule' say about this?

If you connect B to A (0v), you're putting a 0 (that's zero) ohm resistance in parallel with the *lower* 2k resistor. What does your 'voltage divider rule' say about this?

• posted

Do you have a connection to ground in your circuit? Electronics Workbench (Multisim) and similar simulators need one, in order to calculate voltages.

• posted

If the ASCII drawing is correct, B and C will always be the same; they are directly connected.

```--
John Miller
• posted

eh? Point B and C are connected! why shouldn't they be the same? just think about that. if Point B is applied voltage from a source its ovbious that Point C will look the same.

W> Hi,

• posted

Think that I am confused by the word 'input' and 'output'. FYI, A & B label as input points and C point is output. And I have been given the waveforms of C in the middle of 0v to 5v(half Vcc) for A=0v and B=5v. Yup, u guys are correct since output C is followed input of point B. Thanks.

• posted

When you say "0v" do you mean putting the point to ground or letting it floating. There's a big difference between the two options.

If point A is floating (not connected to anything...) you have 0v connected to this point BUT there is no current flowing on the branch... so you can remove the entire line from point A to the next node (with the 2k resistor) since this part of the circuit is doing nothing.

If A is connected to Ground (zero volt), current can flow between point A and the rest of the circuit. You can then apply voltage dividing and current dividing rules.

Also, when you say " [...] pint A equal 0v [...]" is it a measured voltage or an applied voltage ? You have to do the distinction between these notions for us (and you) to understand what you are doing.

I think that you are on a good way to succeed your simulation, you're probably not doing exactly what you want to do. Be shure that, in all time, all your points are connected (either to ground, source or instruments).

message news:...

If point B = 5v then point C = 5v (same node), the upper resistor is shortcutted (no voltage difference on it). You can then ignore this branch.

If point A = 5v and point B = 5v then point C = 5v and no current is flowing on the circuit. If point A = gnd and point B = 5v then C = 5V and a current is inducted in the 2K resistor (I let you do the math... I = V/R).

For clarity, be shure to define the conditions on all nodes an any time.

Regards.

Jean-François Martel

• posted

Putting the point to ground.

Mind to explain what is measured voltage and applied voltage? What's the different ?

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