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First of all, you have to recognize that diode can either be forward-biased (FB) or reverse-biased (RB) depending on the circuit and the component values. However, when you are writing your KVL equations you don't know ahead of time which state they will be in.
What one normally does is to "guess a solution" ... what this means is that you assume that the diode is either in FB or RB and based on your assumption complete the circuit analysis then you use your solution see if the constraints for FB or RB is satisfied.
When in FB, the diode anode-cathode voltage is about 0.7V and the anode to cathode current must be positive.
When in RB, the anode-cathode voltage must be negative and the anode-cathode current is zero (open circuit).
For example, if you assume that the diode is FB and do your analysis and find out that the anode-cathode current is *negative*, then you violate the constraint of the FB and therefore the diode must be RB.
hi, all i can say is, you first analyze the circuit, you have to determine which diode will be ON (forward biased) or OFF (reversed biased) then after analyzing.... remember when the diode is OFF, there is no current (theoretical)... and those diodes which ar e ON, they have current, but also remember those diodes in a unbiased condition are also are no current but it is represented as a close switch, then thats the time u can make a KVL or even a KCL equation. in my personal oppinion i do not do the "asumption of direction of currents" i always stick on the conventional flow of current... i hope i help u about this... tnx
i think kvl does not apply to non-linear models. so approximate the diode as an equivalent linear model(resistance and battery) and write the ohm's law or whatever. refer sedra and smith or any other electronic book for furthur insight into this.
KVL and KCL have nothing to do with linearity at all. You can still write a KVL equation around a loop in which there are nonlinear elements. If there are nonlinear elements, their voltage-current relationship will not be as simple as v=iR like in a resistor. So to solve these equations, you usually have to use numerical techniques since closed-form solutions usually are not possible. Circuit simulation programs like spice do exactly this -- they are used to solve nonlinear sets of equations which are formed by electrical circuits.
sivamit,
if you have a specific problem in mind, go ahead and post it and I will
show you how to proceed.
I think diode is the only device that resist the current upto a certain limits like 0.7 drop across it,so KVL is not applied to it nd also it is a non linear device so KVL for it is impossible.
if u r operating DIODE in forward bias then u can consider as cutin voltage drop across the diode and then sum the other drops across linear elements in the loop. if u operated it in reverse bias then take voltage drop in reverse mode then same proceedure for FW bias
I think diode is the only device that resist the current upto a certain limits like 0.7 drop across it,so KVL is not applied to it nd also it is a non linear device so KVL for it is impossible.
i dont understand the relation between KVL and linearity.
most of the books that i have studied to learn the electronics always use KVL and KCL
please can you explain the relation between them.
KVL has nothing to do with linearity -- all it says is that the algebraic sum of all voltages around a closed loop equals zero -- that is it. Simlarly, KCL says that the algebraic sum of all currents flowing into a node equals zero. There is no mention of linearity here at all. Given any point in time, both of these laws must hold no matter what the circuit elements are.
As I posted before, linearity only comes in when you are trying to solve for voltages and currents that is contained in these equations. See the attached circuit.
You can easily write the KCL equation to solve for the unknown node voltage \[v_x\]:
\[ I_1 = i_{D1} + \frac{v_x}{R}\]
Now we note that \[i_{D1}=I_s (e^{v_x/n V_T} - 1)\]. If you substitute with the values given above, you get
\[ 500mA = I_s (e^{v_x/n V_T} - 1) + v_x/1\]
This is a nonlinear equation, but it is nonlinear because of the diode. Just because you have a nonlinear device does not mean you cannot apply KCL.
Here \[I_s\] is the saturation current, \[V_T\] is the thermal voltage (about 26mV at room temperature), \[n\] is the emission coefficient (usually 1 to 2). Now, you can see that you have to solve this numerically -- that is, try to get a value of the voltage \[v_x\] so the right hand and left hand sides are equal. If you use \[n=1, V_T=0.026V, and I_s=6.9\mu A\]. So the equation to solve is
\[ 500mA = 6.9\mu A (e^{v_x/0.026V} - 1) + v_x\]
Solving this I get, \[v_x=271mV\]. Again, KCL has to do with writing down the first equation. Its actual solution (be it linear or nonlinear) is not a limitation of KCL itself at all.
I hope this is helpful.
By the way, this is how spice would solve the circuit shown. Just put the circuit in
and do an operating point (.op) analysis. Just put the diode of your choice in there in case you don't have the same model available.
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