Circuit with two voltage sources, finding voltage at nodes

[vizakenjack]

Okay, so why is it that V_c assumed to be 12v - 0.7v (D₁), how about contribution from the 9v battery?? Why is this not affecting the voltage at node C?

Why V_c isn't 9v - 0.7v (D₂) = 8.3v?

 

[Baluncore]

A forward biassed diode drops only the voltage it takes to turn the diode on, say about 0.7 volt.
A reverse biassed diode drops whatever voltage is available, no current flows.
As no current flows through a reverse biassed diode, it effectively disconnects part of the circuit, a bit like an infinite resistor or insulator.

 

[COWilliam]

I would assume it has to do with how it is wired in parallel. In section B, where you have the 12v disconnected you do have 8.3v, but otherwise your stronger 12 volt power source is just discharging into the smaller one and providing a 11.3v charge to C. If you had it wired in series then you would be getting a contribution from the 9v, but otherwise it does not have an impact on the voltage at node C.

 

[vizakenjack]

A reverse biassed diode drops whatever voltage is available, no current flows.
As no current flows through a reverse biassed diode, it effectively disconnects part of the circuit, a bit like an infinite resistor or insulator.

Even if no current flows, voltage would still be there, no? Infinite resistor would have a voltage across it...

 

[COWilliam]

Even if no current flows, voltage would still be there, no? Infinite resistor would have a voltage across it...

I believe so, seeing as how potential difference does not actually require current to flow, since electrical fields can exist in a vacuum. Voltage just represents stored energy that could be put to work if it was released.

 

[vizakenjack]

I believe so

So if voltage across open (V_D2) is -2.3V (9-11.3), how come it's not affecting the voltage at V_c

 

[vizakenjack]

Here's another one that perplexes my mind:

So why is V_o = V_i??

If D1 is off, it's basically an open, right? Like this:

So Vo should be affected by V_ref then, no? If you do KVL for V_o

Vi + V_Rs - Vref = 0 ... have I got that right?

 

[COWilliam]

That is kinda confusing to me as well. An open circuit means the wires are cut off so there will be no current flow, but there is voltage, whereas a closed circuit means the wires are connected so there will be flow of current, but there is no voltage... I think I get what you are saying, even if there is no current there should still be voltage that should have an impact on the system, but I am not sure what the answer is.

 

[Baluncore]

You are confusing this thread with unnecessary complexity. Your perplexing observations have no question and do not help sort out misconceptions in the OP.

Re: the OP. If a diode is reverse biased it does not conduct and so behaves like an infinite resistance. There is zero current but there is still a voltage gradient and field. The circuit node that the turned off diode is connected to has a very low resistance which sets the current and so sets the voltage gradient across the reverse biased diode.

 

[vizakenjack]

There is zero current but there is still a voltage gradient and field. The circuit node that the turned off diode is connected to has a very low resistance which sets the current and so sets the voltage gradient across the reverse biased diode.

V_c is only affected by D₁ but not D₂, why? There's voltage across both diodes... Even though D₂ is off.

 

[Baluncore]

Vc is only affected by D1 but not D2, why? There's voltage across both diodes... Even though D2 is off.

The voltages are determined by the currents flowing through the components.
If a switch is open, no current flows, the voltage across the switch is irrelevant and the open switch can be removed from the circuit diagram without changing the analysis.
If a diode is reverse biassed, no current flows, the voltage across the diode is irrelevant and the reverse biassed diode can be removed from the circuit diagram without changing the analysis.
How can the voltage at node B effect the voltage at node C if D2 is effectively removed from the circuit.

 

[vizakenjack]

The voltages are determined by the currents flowing through the components.
If a diode is reverse biassed, no current flows, the voltage across the diode is irrelevant and the reverse biassed diode can be removed from the circuit diagram without changing the analysis.

Wait a sec, so for negative clipper in post #7
When D₁ is on it's practically a short, right? Like this:

So why is V_o = -Vd + Vref. Wouldn't V_i affect it too?

 

[Baluncore]

So why is Vo = -Vd + Vref. Wouldn't Vi affect it too?

The minimum output voltage, Vo, will be Vref - Vd.
If the input voltage is less than Vref-Vd, the difference voltage ( Vref-Vd - Vs ), appears across Rs and a current flows back to Vi from Vref.
The output remains supported at Vref-Vd as the Vi component appears across Rs, not at Vo, which is controlled by Vref-Vd.

 

[vizakenjack]

The output remains supported at Vref-Vd as the Vi component appears across Rs, not at Vo, which is controlled by Vref-Vd.

So in this simple circuit, V_o = 3v?

 

[Baluncore]

So in this simple circuit, Vo = 3v?

Vo must be the 3V battery voltage because that is what the output is connected to. How could it be different ?

5V, the voltage source on the left has zero output resistance. The battery is ideally also a fixed voltage source.
The battery is being charged by the 5V source through R1. Charge current = (5V - 3V) / R1

 

[vizakenjack]

The battery is being charged by the 5V source through R1. Charge current = (5V - 3V) / R1

so if R₁ = 1k ohm, I_R would be = (5v-3v)/1000 = 2 mA, current across R₁.
Voltage across R₁ is 5V-3V = 2V
But at this point here:

Isn't the voltage also 5V-3V=2V?
See, I thought that voltage at that point would be the same as V_o, but apparently not ...

 

[Baluncore]

The current flows “through” R1.
The voltage appears “across” R1.

See, I thought that voltage at that point would be the same as Vo, but apparently not ...

Why not.
All voltages and heights are measured relative to local ground unless otherwise specified.

If you fall from the roof of a 5 story building, onto the roof of a 3 story solid building, you have only fallen 2 stories.
The height of your fall, or drop, may have been only 2, but you are still 3 stories above ground.
The height of the building you fell onto did not change.

Electrons fall from one voltage to another, when they do fall, they release potential energy.
That energy can be released in different forms. In a resistor it appears as heat.

 

[vizakenjack]

Why not.

Well then V_o should be 5V-3V = 2V not 3V. Don't you see the contradiction?If I were to guess for an explanation, I'd say that for the voltage where the arrow is (in the figure of my previous post), potential difference is taken between two voltage sources. Where is for V_o Potential difference is measured between one voltage source only, which is 3V...

 

[Baluncore]

Well then Vo should be 5V-3V = 2V not 3V. Don't you see the contradiction?

I do not follow your logic and cannot see why you have that opinion.
You appear to be confusing Vo = the output voltage, with the voltage drop across R1.

Vo is the voltage between ground and the arrow. It is also the 3V battery voltage.
The voltage difference across R1 is 5V – 3V = 2V.

Where is for Vo Potential difference is measured between one voltage source only, which is 3V...

Measured relative to what. You show Vo is clearly being measured relative to ground.
There is no such thing as an unreferenced voltage. Ground is the default reference.

 

[sophiecentaur]

so if R₁ = 1k ohm, I_R would be = (5v-3v)/1000 = 2 mA, current across R₁.
Voltage across R₁ is 5V-3V = 2V
But at this point here:

Isn't the voltage also 5V-3V=2V?
See, I thought that voltage at that point would be the same as V_o, but apparently not ...

The definition of an ideal battery is one that maintains its voltage, whatever current is taken from it. Your arrow (Vo) must be at 3V. Quite incidentally, current will flow through R₁, given by 2/R₁.
In the analysis of 'ideal' circuits, you have to stick rigidly to the 'rules' that apply. It's no use having a feeling about the way things should be. It's very easy to have a wrong feeling. We have all been there and kicked ourself afterwards.