0
\$\begingroup\$

In a circuit having two diodes with different conduction voltages,how to know which diode is in ON condition. Let the circuit be

schematic

simulate this circuit – Schematic created using CircuitLab

If D1 diode has cut in voltage of 0.6 and D2 having cut in voltage of 0.2 then what are the conditions of D1 and D2 whether they are ON or OFF and is the current flowing through R3. Please provide the explanation. Thanks in advance.

\$\endgroup\$
  • 1
    \$\begingroup\$ Subtract the diode voltage from the source voltage in the same leg, then treat the circuit as if the diodes are 'ideal'. Also (if it aids in understanding) feel free to rearrange the order of the source, resistor & diode in each leg because each leg in isolation is a simple series circuit. \$\endgroup\$ – brhans Jun 29 '17 at 17:01
  • \$\begingroup\$ whether D1 and D2 both are in ON condition or Only one of them? \$\endgroup\$ – Eswara Srisai Jun 29 '17 at 17:04
  • 1
    \$\begingroup\$ I don't know - I haven't analyzed the circuit - that's your job. I gave you some hints on how to do that. \$\endgroup\$ – brhans Jun 29 '17 at 17:36
4
\$\begingroup\$

Let me redraw your schematics. (In fact, you should get into the regular practice of redrawing any schematic you don't feel you understand well using well-understood rules.)

schematic

simulate this circuit – Schematic created using CircuitLab

Current flow should be arranged so that the top of a schematic is the most positive and the bottom of the schematic is the most negative. Signal, if applicable, should flow from left to right, with inputs on the left and outputs on the right.

Don't bus voltage rails (or ground) around. You don't need to see all the connections as it doesn't matter (mostly) for understanding a circuit. I've eliminated the useless wiring you added. Pointless. Just label the nodes where you know the voltage.

This will save you a lot of grief.

Finally, it's just fine to swap series-arranged, two-terminal devices. I swapped your diode/resistor series chains to put the diodes closest to the positive node voltages to make it a little easier to understand. I think you should also be able to see that they are both on -- or, at least, that you cannot find an a priori reason why either one of them should be off.

At this point you should be able to work out the voltage for \$V_X\$ and, from that, all the currents involved as well. Then you can go back and verify in your own mind the on/off status of each diode, if you like.

|improve this answer|||||
\$\endgroup\$
0
\$\begingroup\$

Start by analysing the D1 path. D1 has to drop at least 0.6 volts to start conducting hence you might as well say that it conducts at 0 volts and the supply voltage is 9.4 volts instead of ten volts. Next, because R1 and R3 form a 2:1 potential divider you can say that the D1 path can produce no more than 4.7 volts across R3 (in the absense of D2's circuit).

This should then inform you that the D2 path is also conducting current and will lift the 4.7 volts upwards a little bit due to that current. Can you now see that both diodes must be conducting?

After that it's just a case of analysing two voltage sources being joined together with two resistors with a third resistor to ground from that joined junction. Can you take it from here?

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ I think the low conduction diode will ON. And high conduction diode will be OFF.is that true? \$\endgroup\$ – Eswara Srisai Jun 30 '17 at 2:24
  • \$\begingroup\$ You do the same analysis with D1 removed and ask the question what will be the voltage across R3. You should conclude that the voltage across R3 is 2.9 volts and clearly, when D1 is brought into things there has to be current flowing through D1 - both are conducting. \$\endgroup\$ – Andy aka Jun 30 '17 at 8:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.