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This ckt comes as a question in our previous year exam. what will be current " I " and voltage drop " Vo " . Generally voltage drop across any diode around 0.7 V (which is used for other problems).

If I guess initially \$ V_o \$ is zero then D1 become ON and voltage at \$ V_o \$ become 3-0.7 = 2.3V , which cause D3 to be ON . If D3 is ON then voltage at \$ V_o \$ should become 1+0.7=1.7V. So there is no way that diode voltage drop remain at 0.7V .

How could I solve this kind of problem? How could I decide which will be the voltage drop across the diode in this kind of situation?

It'll be very helpful If someone explain it also. Thnx in advance .

  • 1
    \$\begingroup\$ Since this is an exercise, it would be really appreciated if you could explain what you understand of the problem and how you would go to solve it. The risk is it being closed as "Too Broad" \$\endgroup\$ – clabacchio Apr 17 '14 at 9:13
  • \$\begingroup\$ Hint - D2 can be removed. \$\endgroup\$ – Andy aka Apr 17 '14 at 9:48
  • \$\begingroup\$ Vo, in the location it's marked, will be zero volts, unless the wire is quite thin :-) And my vote for RL is 0.8mA, if the diodes are matched. Which should not be depended upon in any real circuit. \$\endgroup\$ – gwideman Apr 17 '14 at 10:00
  • \$\begingroup\$ What sort of model are you using for a diode when analyzing this problem? That information should have been part of the problem definition. Without specifying parameters for the diodes this would appear to be a terrible exam problem. \$\endgroup\$ – Joe Hass Apr 17 '14 at 10:49
  • \$\begingroup\$ @joe ;to solve this kind of problems theoretically constant voltage drop model of diode is used and the drop is 0.7V . There is no model number but just this ckt diagram. \$\endgroup\$ – Anklon Apr 17 '14 at 11:02

This problem can be answered fairly easily, if you use a diode model that's a bit more than the 0.7V/off model.

I suggest finding the current through the diodes without the resistor load (1V across each diode) using the Shockley diode equation. That will give you the diode current in terms of \$I_S\$ and \$V_T\$ and n.

The (unloaded) output voltage \$V_O\$ will obviously be 2V, by symmetry.

Then use small signal analysis to find the output resistance of the diodes in terms of n, Vt and current.

You should then be able to write down a closed-form expression for the output voltage in terms of \$R_L\$, \$I_S\$, \$V_T\$ and n.


When I learned this sort of problem, we were taught to make assumptions about each respective diode's state, and then use the process of elimination to verify our assumptions and deduce a solution.

  1. Assume all diodes are on, and that they follow the 0.7 voltage drop model. (The 0.7V drop model states that in order for a diode to be conducting, the voltage drop across its terminals must be greater than, or equal to, 0.7V)

    • For D3 to be active, Vo must be >1.7V

    • For D2 to be active, Vo must be >2.7V

    • For D1 to be active, Vo must be <2.3V

  2. Analyze the assumptions

    • Lets look at D2. Vo will see 2.7V only if the voltage somehow comes from the 3V source connected to D3 (The 3V source is the only one in the circuit high enough to create a 2.7V node). If the 3V source passes through D1 to reach Vo, the highest voltage it can create on Vo is 2.3V due to the voltage drop of 0.7V across D1. (2.3V < 2.7V) We can conclude that D2 will not be conducting.
    • Now lets look at D1 and D3. For D1 and D3 to be active, Vo has to fall within (1.7 < Vo < 2.3). With D2 eliminated, the only place Vo can potentially pull this kind of voltage from is again from the 3V source. If Vo is to see the 3V source, it will only happen after a voltage drop through D1 > 0.7V. So we now know that Vo<2.3V ... but by how much?

      Lets imagine that the resistor temporarily doesn't exist. Lets also look at the effective resistance of a diode. The effective resistance of a diode is given by the inverse of the slope of the diode's I-V characteristic.

      I-V characteristic from www.allaboutcircuits.com/vol_3/chpt_3/1.html

      With the resistor in the problem ignored, we have two diodes connected in series passing current. If they are passing an equivalent current, they will be equivalent effective-resistors. But notice from the graph, as the current that one passes increases, the change in effective resistance is marginal. So even if each diode is passing a slightly different current, their effective-resistance is approximately equivalent as long as the current is large enough. Note that the effective-resistance of the diode is VERY small for such large currents (this is important for later). If the diodes are equivalent effective-resistors, then the voltage division formula allows us to conclude that the voltage at the node between the two diodes will be equal to (0.5) * (the over-all voltage drop across both diodes). Now lets observe what happens when we reintroduce the original resistor.

      The added resistor's resistance is a couple of orders of magnitude larger than the effective-resistance of the diodes when the diodes are under high current. Current division of a high-resistance branch(RL) and very-low-resistance branch(D3) will show that the majority of the current will flow into the very-low-resistance branch(D3) (think of a short circuit). From the perspective of a diode(D1 and D3) with a VERY small intrinsic resistance, this new large "RL" resistor resembles an open circuit. The diodes thus pass an (approximately) equivalent current (\$I_{D1}\$~=\$I_{D3}\$). If they pass an equivalent current, they have an equivalent voltage drop and equivalent effective-resistance. Thus, the voltage drop across one diode = (0.5)*(voltage drop across both diodes)

  3. Result

    • What you will be left with is voltage drop across D3 of 0.5*(3V - 1V) = 1V. The voltage at the Vo node is the Voltage before D3(3V) minus the voltage drop across D3(1V). 3V-1V=2V at Vo. The solution is: Vo = 2.0V, with respect to ground.
  • 1
    \$\begingroup\$ If you allow the voltage drop across D3 to be greater than 0.7 V, how can you assume that the drop across D1 must be exactly 0.7 V? Your method is not consistent. \$\endgroup\$ – Joe Hass Apr 19 '14 at 11:00
  • \$\begingroup\$ I see what you are saying and I agree with you that my original reasoning was off. I edited the explanation to account for the inaccuracy. Thanks for pointing that out @Joe! \$\endgroup\$ – Miron V Apr 20 '14 at 4:51

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