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schematic

simulate this circuit – Schematic created using CircuitLab

I trying to solve a problem which says: A particular design of a voltage regulator is shown. Diodes D_1 and D_2 are 10 mA units, that is each has a voltage drop of 0.7 volts at 10 mA. Each has n=1. (n is emission coefficient). What is regulator voltage with load 150 Ohm connected?

My way of solving this would be the following: I would find first the current \$I_D\$ through the diodes and the actual voltage with load disconnected by iteration method. Later I would use this equation to find the change in voltage with the load connected. $$\Delta V = - I_L (2r_d || 150)$$ where \$r_d=V_T/I_D\$

My question is: is this this a right way to proceed, or should I solve it some other way?

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  • \$\begingroup\$ Back of the envelope: Vout = 2 diode drops = 1.2 to 1.4 V. To get the answer based on the Shockley equation, plug the circuit in to SPICE (and be sure the diode model is simplified down to just the Shockley equation). \$\endgroup\$ – The Photon Mar 1 '16 at 17:15
  • \$\begingroup\$ I am interested in how to solve it on paper, no SPICE. \$\endgroup\$ – Mykolas Mar 1 '16 at 17:17
  • \$\begingroup\$ Is this homework? \$\endgroup\$ – Brendan Simpson Mar 1 '16 at 17:21
  • \$\begingroup\$ no, self interest \$\endgroup\$ – Mykolas Mar 1 '16 at 17:21
  • \$\begingroup\$ Then use the shockley equation for a diode or look for a forward voltage characteristic of a diode in a data sheet. \$\endgroup\$ – Andy aka Mar 1 '16 at 17:28
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Your approach is right, but just need a slight modification in the equation. You have to consider the 180 ohm resistance in the current equation. You rd is given as 0.7/(10*10^(-3)) = 70ohm. (Given that each diode has voltage drop of 0.7 at 10mA). So, your total resistance of the circuit would be R = (180 + 140||150) ohm. Having found the current as 45/R, you can find the voltage drop across the resistors. Would leave the rest of exercise to you. :)

For problems like this, always consider the full system from the voltage source.

Regards, Priyank

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  • \$\begingroup\$ @Priyank you may want to become familiar with the $$ to format equations. \$\endgroup\$ – Voltage Spike Mar 1 '16 at 19:24
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If you want to solve this on paper, you can use load line analysis.

  1. Calculate the diode parameter \$I_s\$ from the known operating point of 10 mA at 0.7 V.

  2. Plot the characteristic I-V curve of the combination of the two diodes \$I = I_s\left(\exp(\frac{V_{out}}{2 V_t})-1\right)\$. (The 2 in the denominator comes from having two diodes in series)

  3. First find the Thevenin equivalent of the 5 V source and two resistors (2.27 V with 81.8 ohms).

  4. Plot the inverted I-V curve for the Thevenin source circuit (\$V_{out} = 2.27\ \mathrm{V} - I(81.8\ \mathrm{\Omega}\$).

The intersection of the two curves you just drew gives the operating point Vout.

However, if you build the actual circuit, you'll find that variation between components, thermal effects, etc., will result in a slightly different operating point. Likely so much so that you would have been just as accurate to use the back-of-the-envelope result that Vout is given by two diode drops at 1.4 V.

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  • \$\begingroup\$ Thankyou, your answer was helpfull \$\endgroup\$ – Mykolas Mar 1 '16 at 21:14
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1.4 Volts is the answer. 0.7 + 0.7 = 1.4V.

It is then a parallel circuit with the resistor, and Voltage is the same when parallel.

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  • \$\begingroup\$ But this would be to assume that the diode current is 10mA, which is not actually true. \$\endgroup\$ – Mykolas Mar 1 '16 at 17:38
  • \$\begingroup\$ @Mykolas, If the voltage is 1.4 V, then I calculate the diode current to be 10.7 mA, which is darned closed to the 10 mA where the forward voltage is specified. Real world performance of this "regulator" design is not likely to be more accurate than ~50 mV, so trying to get a more accurate theoretical answer is pointless. \$\endgroup\$ – The Photon Mar 1 '16 at 18:13

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