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I am trying to calculate the current of the circuit but I am unable to find the correct answer. I was able to solve the same circuit without the diode witch gave me 2mA. I am unsure of what to do with the diode added to the circuit. Here is my circuit.
simulate this circuit – Schematic created using CircuitLab
The diode polarization is direct with a voltage of 0,7V. How do I solve the current of this circuit? Any help would be greatly appreciated.
For problems like this, the first step is to assume the state of the diode. Is it conducting or not?
Once you assume a state, then you redraw your circuit to match the assumption. For example, if you assume the diode is not conducting, what can you replace it in the schematic with? How about if you assume it is conducting (hint: remember you said it has a voltage of 0.7V)?
After making these assumptions, re-drawing and solving, one of the solved circuits will result in a contradiction with the original assumption. Perhaps you calculate the voltage drop across the diode as something smaller than 0.7V by using KVL around the loop formed by R2 and R3. This contradicts the diode being at 0.7V when conducting, which means that assumption was wrong.
The circuit that doesn't result in any contradictions is the correct assumption, and the values from that circuit are the "right" answer.
You might consider converting your voltage source and the \$R_1\$ and \$R_2\$ resistor divider pair into an equivalent Thevenin voltage and Thevenin source resistance. Once replaced that way, the remaining circuit is trivial to understand and quite easy to solve. (It also instantly avoids all this, "assume one way, then assume the other way" testing process.)
simulate this circuit – Schematic created using CircuitLab
All you need to do is work out \$V_\text{TH}\$ and \$R_\text{TH}\$ and decide if \$V_\text{TH}\$ is large enough. I'm sure you can work out the details from there.
Nodal equations are your FRIEND.
Let V_a be the voltage at the diode anode. You can get away with treating the diode as a pure voltage source and ignore its resistance (which will be MUCH less than 10K ohms).
Then:
(30 - V_a)/10K = V_a/10K + (V_a - 0.7V)/ 10K
Solve for V_a, then solve for the branch currents.
Not sure how you got 2mA - Without the diode, using simple circuit analysis and Ohm's Law, the total current would be 1.2mA.
Since the diode is forward-biased, R3 will still be in the DC circuit and the total current will still be 1.2mA with the voltage drop across R3 will just be reduced by 0.7 Volts.
If you turn the diode around, that is a different story where no current would flow through R3 and you would have a simple series circuit with a total resistance of 20kohms and therefore a total current of 1.5mA.
