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schematic

simulate this circuit – Schematic created using CircuitLab

If V2 is NOT present(VB=GND) i can solve it easily using voltage divider formula: I know the steps:

  1. I = V1/(R1+R2)
  2. Vout = I * R2 =>(i have 'I' from step 1,R2 is known)=> Vout = (V1*R2)/(R1+R2)

Questions:

  • A) I would like to request some help on how to solve it when v2 = 3.3 volts is present. Please do not skip steps if you answer it.

    B) Vout is calculated with reference to GND. Right? If yes, that means the voltage drop at R2 equals to Vout = Vout - GND = Vout - 0 = Vout assuming GND as reference point?

    C) Is it correct to say the voltage drop at R2 = Vout - VB? (So in this case i do not take GND as a reference point).

    D) So there are 2 different ways to describe voltage drop? Like case B) and case C).Am i correct?

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  • \$\begingroup\$ Since GND is just a reference point why don't you: 1) connect the GND at VB, Do note that the GND connected to V1 will than change, it will be at - 3.3 V it is easier to replave V1's GND with a wire to VC 2) solve the voltage divider 3) write all the voltages across the components in the schematic 4) move back the GND to VC. 5) Determine Vout. \$\endgroup\$ Jan 14, 2018 at 15:52
  • \$\begingroup\$ The superposition is your friend Vo1 = 5*667/(200+667) = 3.8465V and Vo2 = 3.3*200/(667+200) = 0.76124V therefore Vout = Vo1+Vo2 = 4.6V \$\endgroup\$
    – G36
    Jan 14, 2018 at 15:54

4 Answers 4

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The key is voltage ACROSS the resistive chain

V = V1 - V2

Then it is a simple voltage divider to work out the voltage across R2

Vr2 = V(R2/(R1+R2))

Finally the output is raised by V2

Vout = Vr2 + V2

Thus. Vout = (V1-V2)(R2/(R1+R2))+V2

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Answers:

A) I would like to request some help on how to solve it when v2 = 3.3 volts is present. Please do not skip steps if you answer it.

You need to use the full equation here.

\$V_{OUT} = V_B + (V_A - V_B) * R_2/(R_1 + R_2)\$

which in your example is

\$V_{OUT} = 3.3 + (5 - 3.3) * 667/(200 + 667) = 3.3 + 1.7 * 667 / 867 = 4.61V\$

B) Vout is calculated with reference to GND. Right? If yes, that means the voltage drop at R2 equals to Vout = Vout - GND = Vout - 0 = Vout assuming GND as reference point?

Ok that is a confusing sentence, but essentially correct.

C) Is it correct to say the voltage drop at R2 = Vout - VB? (So in this case i do not take GND as a reference point).

No, the voltage drop ACROSS R2 is Vout-VB.

D) So there are 2 different ways to describe voltage drop? Like case B) and case C).Am i correct?

Since you got off course with C, then this assumption is also incorrect.

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    \$\begingroup\$ Or Vout = Va - I*R1 = Va - (Va - Vb)*R1/(R1+R2) = 5V - (5V - 3.3V)*200/(200+667) = 4.60784314V \$\endgroup\$
    – G36
    Jan 14, 2018 at 16:13
  • \$\begingroup\$ @G36 indeed.. though let's not confuse the OP more ;D \$\endgroup\$
    – Trevor_G
    Jan 14, 2018 at 16:24
  • \$\begingroup\$ i would accept your response but you magically popped out vout equation without showing how to do it. Question A) was my most important of all. Anyway, overly it is an almost comprehensive and good answer, thanks.. \$\endgroup\$ Jan 14, 2018 at 22:03
  • \$\begingroup\$ I have accepted somebody's else response and i just noticed @G36 comment right here. Thus, if i combine G36's comment and my 'accepted answer' i found a 'good' detailed solution: Vout - I* R2=Vb => Vout = I* R2 + Vb. Where I = (Va-Vb)/(R1+R2) \$\endgroup\$ Jan 14, 2018 at 22:25
  • \$\begingroup\$ @DontAskTheEye it's all the same just written different ways. As long as you have your solution, all is well. \$\endgroup\$
    – Trevor_G
    Jan 14, 2018 at 23:34
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The current I is (5 volts - 3.3 volts)/(200 ohms + 667 ohms) = 1.9608 mA.

That current multiplied by R2 (667 ohms) gives you the voltage that Vout is above 3.3 volts i.e. 1.3078 volts hence, Vout = 4.6078 volts.

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You could also solve this by producing 2 current equations:

  1. I = (VA - VB)/ (R1 + R2)
  2. I = (Vout - VB) / (R2)

set them equal... algebra...

Vout = (V1R2 + V2R1)/(R1 + R2)

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  • \$\begingroup\$ In the real world the 3V3 would have to sink some current, if it cannot its output will become 5V as there is no load. \$\endgroup\$
    – Gil
    Nov 17, 2022 at 19:40

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