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In this circuit:

The diode is an ideal diode

The diode is on since V(anode) > V(cathode) and then we replace it by a short circuit.

Now, how to calculate Vo when two sources are existed?

I = (10 + 2) / (2k + 4.7k) = 1.79 mA.

I tried using KVL:

-10 + 2k (1.79 mA) + Vo - (-2) = 0

=> Vo = 4.42

Or:

Vo = 4700 * (1.79 mA) = 8.413

Which answer is correct? and why the other one is not correct?

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  • \$\begingroup\$ Imagine what would happen if you added 2 volts to all the voltages - it would become a straight forward potential divider with 12 volts at the input and the 4k7 to 0V - work this out then subtract 2 volts. \$\endgroup\$ – Andy aka Oct 18 '14 at 9:47
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Neither answer is correct, but you mostly had the right idea.

You correctly calculated the current through the whole circuit: \$\dfrac{10 - (-2)}{2000 + 4700} = 1.791 mA\$

The next step you could take would be to calculate the voltage drop across each resistor:

  • Across the 2K resistor: 2000 * 0.001791 = 3.58 V
  • Across the 4.7K resistor: 4700 * 0.001791 = 8.42 V

To find the voltage at Vo, start at either end and add or subtract the voltage drop across the resistor.

  • If we start at the 10V end: Vo = 10 - 3.58 = 6.42 V
  • If we start at the -2V end: Vo = -2 + 8.42 = 6.42 V
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  • \$\begingroup\$ First step is fine, but second step is simply 4.7E3*I-2 \$\endgroup\$ – lucas92 Dec 15 '17 at 18:00
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You don't "ignore" the other source(s) to use superposition, you connect them to ground.

So the the voltage from the 10V source is 10V * (4.7/(4.7+2)), and the voltage from the -2V source is -2 * (2/(4.7+2)).

You can also use a quick and general method that works for any number of resistors and corresponding sources:

Vo = (V1/R1 + V2/R1 + ... Vn/Rn) * (R1 || R2 || ... || Rn)

Where || represents the parallel resistance 1/(1/R1 + 1/R2 + ... + 1/Rn)

So in this case, Vo = (10/2 - 2/4.7)(4.7 || 2)

Either method should give the same result if you don't make any errors.

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The diode is ideal and forward biased therefore it becomes a short.

The current in the resistors is \$\dfrac{12V}{4k7+2k0}\$ = 1.791mA (which you have already calculated)

The voltage across the 4k7 is 1.791mA * 4k7 = 8.418 volts.

But the bottom of the 4k7 resistor is 2 volts lower hence the output voltage is 6.418 volts.

Neither of your answers are correct.

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  • \$\begingroup\$ Thanks but why can't I consider Vo as the voltage across the 4.7k resistor? \$\endgroup\$ – ammar Oct 18 '14 at 10:06
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    \$\begingroup\$ Because all voltage measurements must have a reference point and that reference point is 0V and not -2V. Vo is relative to 0V - just because 0V isn't explicitly shown in the schematic it doesn't mean it doesn't exist. \$\endgroup\$ – Andy aka Oct 18 '14 at 10:16
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First task at hand is to identify whether the diode is forward or reverse biased. You can think of it this way, imagine the diode is reverse biased, in that case the anode receives a 10v and the cathode receives a -2, which means your assumption of the diode being reverse biased is wrong and it should be forward biased.

Now that we know the diode is conducting, apply a simple nodal equation at point of output,

(Vo-10)/2k + (Vo+2)/4.7k =0;

Solve for Vo, which will be approx 6.42V.

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Since you have more than one source of voltage in the circuit, why not use the Superposition Theorem (http://en.wikipedia.org/wiki/Superposition_theorem). Put the -2V source to zero and apply the voltage divider rule to find: Vo(-2V grounded) = 10V x 4.7k/(4.7k+2k) = 7.0 V. Next, put the +10V source to zero and apply the voltage divider rule to find Vo(+10V grounded) = -2V x 2k/(4.7k+2K) = - 0.6V. Sum these two results to get the actual Vo: Vo = Vo(-2V grounded) + Vo (+10V grounded) = 7.0 - 0.6 = 6.4V You only need to use 2 digits of precision because that's all you have to start with in your circuit data. Note: The concept of grounding each independent voltage source is for analytical purposes only. You wouldn't do it in practice. In practice, you would measure Vo with a voltmeter relative to ground (or the [-] terminal of the +10V supply OR the [+] terminal of the -2V supply). Notice this solution doesn't require the need to calculate circuit current.

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enter image description here

Apply KVL only on closing loop circuit.

-10 + 2k (1.79m) + V1 - 2 = 0

V1 = 8.42 Volt

Or

V1 = 4k (1.79m) = 8.41 Volt

Note that Vo is Voltage between 4.7k resistor and 2 Volt DC supply.

Hence, Vo = V1 - 2 = 6.42 Volt

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