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I am not shure we understand this correctly. What is current through diode if we apply forward voltage on diode which is exactly 0.6V. Is it 0 or ∞ or undefined value in between.
Above circuit can be analyzed this way:
D2 conducts. Voltage drop on R1 will be 2.4V because diode subtracts .6V from 3V battery. Current through D1 can not flow because of R2. Equivalent characteristics of series connection of resistor and diode is that it has zero current up to .6V and it has 1kΩ ramp after. At defined point .6V current through such connection is 0. So D1 is off and D2 is on, voltage on R1 remains 2.4V.
My idea with simulation is exactly what trigger74 said, do make experiments and simulation is nothing but that.
Isn't this what both j2005 and I already said before? I thought we were discussing the approach - weather to jump straight into simulation or to do some anayisis before that.
In respect to resulting voltage on R1 and the condition of diode D1, the results are equal. But explanations are different. I think, problems like this are solvable exactlu only by using mathematical model of diode. This diode model is discrete fonction which has 0 value up to 0.6V and 0 to ∞ at 0.6V. This function can be represented by series connection of 0.6V battery and ideal diode. Connecting additional resistor in series with such diode model gives function (characteristics) which has 0 value up to 0.6V and ramp with slope equal to R to infinite voltage.
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