What happens if you connect a diode with an inductor in series? - Page 1
Forum section: electronics / beginners

author: engineheat
author_rank: Frequent Contributor
timestamp: July 19, 2017, 04:32:17 pm
content: Suppose a 5V battery powers the circuit. Assume the inductor is ideal, will current just rise with a rate given by: L (di/dt) = 5 - diode voltage drop?I'm a bit confused on whether the diode will be forward biased in this case. What if the voltage drop across the inductor is 5V? Is that possible?

author: kalel
author_rank: Frequent Contributor
timestamp: July 19, 2017, 04:37:41 pm
content: Do you mean something like this?http://tinyurl.com/y779udtwHere everything is ideal however and the power source is not a battery but an ideal voltage sourceThat said, I have no idea if the simulation can answer your questions, just wondering if that's the right configuration.

author: Chalcogenide
author_rank: Regular Contributor
timestamp: July 19, 2017, 04:45:24 pm
content: Depends on the amount of ideality of the components.If we were to use the "ideal" diode, with 0 voltage drop, the voltage across the inductor would be 5V and the current would rise at a rate (5 V)/L A/s forever.If we were to use the "turns on at 0.7V" diode, then the current would rise indefinitely forever at a rate (4.3 V)/L A/s.If we were to use the exponential equation of the diode I = Is*exp(V/mKT) the current would ultimately be limited by the condition I=Is*exp((5 V)/mKT) because as the current rises, also the voltage across the diode increases, reducing the potential difference across the inductor, thus slowing the rate of increase of the current, and eventually an equilibrium will be reached with a constant current flowing through the circuit.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 19, 2017, 04:49:04 pm
content: Current ramps up, and yes, the diode will be forward biased. At t < 0, the diode's junction voltage is undefined (you didn't give initial conditions); if we assume zero initial voltage and current, then: the junction voltage will start at zero, and charge up as the inductor begins to draw current. Later, the diode voltage saturates to Vf (the DC value, given as a function of current), as it goes into conduction proper.If the inductor and diode are ideal, and Vf << (5V supply), then the diode voltage will be a pretty nice logarithm, and the current will be a linear ramp. (Well, okay, for an ideal diode -- with no junction capacitance -- the diode forward voltage will appear immediately, without taking time to charge the capacitance.)In a real circuit, with resistance, the current will saturate, according to the L/R time constant, and the diode voltage as well. As \$t \rightarrow \infty\$, the system reaches steady state.The voltage drop across the inductor, at t = 0, will be exactly 5V, because Vf(If = 0) = 0. The derivative of Vf and If depends on capacitance: if capacitance is zero, the derivatives are undefined at t = 0. If nonzero, all derivatives exist, and the first derivative is zero. I think?Clear as mud?Tim

author: engineheat
author_rank: Frequent Contributor
timestamp: July 19, 2017, 07:16:22 pm
content: Quote from: T3sl4co1l on July 19, 2017, 04:49:04 pmThe "Vf" is the 0.7V voltage drop common in a lot of real diodes in forward biased mode right? I really like how you analyzed t<0 and t=0, it really makes sense. The "the junction voltage will start at zero, and charge up as the inductor begins to draw current" really clarifies how things work. thanks

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 19, 2017, 07:20:13 pm
content: Yes, Vf is the forward voltage. It's not actually "0.7V" of course, but a function of current; it's a shallow function, though, so that at most currents (~mA to ~A), for silicon PN diodes, it's in the 0.5-1.0V range.Cheers,Tim