Diode characteristics - Page 1
Forum section: electronics / beginners

author: nForce
author_rank: Frequent Contributor
timestamp: January 25, 2016, 04:06:45 pm
content: This is the circuit I am trying to analyze.I have to find voltage drop on a diode, where on the right I have diode characteristics. If I apply Ohm's law, I get that through the diode flows 3 mA of current. If I look at the characteristics 3mA of current means, 1.5 V drop.But it's wrong, I need to plot a line through the point 3mA and 3 V, and where it crosses with the characteristics there is the voltage, which really drops.How?, Can someone explain this. Thank you.

author: Paul Price
author_rank: Super Contributor
timestamp: January 25, 2016, 04:15:58 pm
content: What's all this nonsense using a complicated circuit to measure voltage drop of the diode. All you need is a resistor and a voltmeter and a variable voltage source and you make all the measurements you need. Or else use LTSpice.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 25, 2016, 04:30:58 pm
content: Yes, that is the graphical method of load lines. It's the solution for a system of two equations: the diode drop (which is shown as a straight line above 0.5V, conveniently enough*) and the resistor V(I). Kirchoff's law requires the slopes to be complementary, guaranteeing an intersection, and a unique solution. At least for the conditions shown.*Real diodes, of course, are exponential. This leads to a transcendental equation (an expression of the form x = e^x), which has no analytical solution, though a quite reasonable approximation can be calculated pretty easily by iteration. A graphical solution doesn't care about the curve's function, of course, and finds the solution just as easily.Tim

author: orolo
author_rank: Frequent Contributor
timestamp: January 25, 2016, 06:19:47 pm
content: You can't use Ohm's law because the diode is nonlinear. The idea is that the diode and the resistor share the same current, I, while their added voltage drops amount to 3V. This means, IR + V(I) = 3, where V(I) is the diode drop, a function of the intensity through it. Taking the Ohm's term to the other side of the equation, V(I) = 3 - IR. The right-side term is a straight line, with negative slope -R (note the scale in milliamps in the graphic, in that scale the slope is just -1, for R = 1KOhm.) That line intersects the y axis at I=3, V=0, and the x axis at V=3, I=0. When that straight line intersects the diode I-V characteristic, the equation V(I) = 3 - IR is satisfied, and you get the diode current and voltage at equilibrium. It's easy to see that V(I) = 1V, I = 2mA.The circuit is quite like an inverting voltage adder, save for the diode. Using the load line analysis, we get a diode drop of 1V for a current of 2mA (I'd expect about 0.6V for a typical diode, though). So the current through the feedback resistor is 3 + 2 = 5mA, and the output voltage -5V.If you try a spice simulation of the circuit, with a 1N4148 diode, you get an output voltage of -5.375V. The discrepance is due to a voltage drop of 0.625 V through the 1N4148, more typical than a 1V drop.

author: Audioguru
author_rank: Super Contributor
timestamp: January 26, 2016, 02:18:16 am
content: The diode has a voltage drop of about 0.7V so the voltage across the 1k resistor feeding it is 3V - 0.7V= 2.3V so the current in the diode is 2.3mA, not 3mA.Your circuit was completely wrong so I fixed it, made the opamp one with Jfet inputs so its input resistance is high and used high value 1M resistors for the opamp so the diode voltage is not loaded down.You did not say which diode. I found that a 1N4001 matches your graph.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 26, 2016, 02:47:14 am
content: Correct me if I'm wrong, but 2/4 posters missed that this is a homework question, not a measuring circuit.Tim