2
\$\begingroup\$

Using this link as reference.

Question 1:

From my understanding, if the applied forward voltage across a diode is much greater than its rated forward voltage the diode will potentially fail.

Based on above, won't the flyback diode fail? Since the inductive kickback voltage is so much greater than the forward voltage of the diode? The only reason I can think of the diode not failing is because the kickback is instantaneous.

Question 2: Referencing the same link above, I do not understand why the voltage is below zero in section 'e' of the waveform of the series diode-resistor circuit.

\$\endgroup\$
1
  • 3
    \$\begingroup\$ If the applied forward voltage is much higher than the rated forward voltage, you will probably be exceeding the forward current rating by several orders of magnitude. Do you mean the reverse voltage rating? \$\endgroup\$
    – vir
    Commented 2 days ago

2 Answers 2

5
\$\begingroup\$

After it turns on, the forward voltage across the diode will be the forward voltage of the diode at the current that was just going through the inductance. This is a characteristic of the diode at that particular current.

For a silicon diode that will typically be in the 1V range. Take a typical diode eg. UF4004:

enter image description here

If the current through the inductance was 1A then just after the switch turns off, there will be 1A or so going through the diode and the forward voltage will be a bit over 900mV at 1A.

If you actually applied a 'stiff' voltage source to the resistor of higher than that to the diode, enormous current would flow and the diode would be destroyed.

However the inductive load does not behave like a voltage source- it behaves more like a current source that decays towards zero at some rate. For something like a relay coil, it will be close to zero in some milliseconds.

In a similar vein, the second part of your question, refer to the diagram with current flow shown:

enter image description here

When the switch turns off (opens) the inductance will cause the current to attempt to continue to flow just as it did before the switch opened. The voltage at the top of the coil will will drop below zero almost instantaneously until it finds a way to get that current to flow. When the diode (and no resistor) is there, the current will flow at about -1V wrt ground (assumed to be at the bottom of the circuit). If there was no diode (and no resistor) then the voltage would spike way below ground until something 'else' happened. Part of that will be arcing at the switch. It could break down the insulation in the coil, but more often it will just 'ring' with a decreasing amplitude with the distributed capacitance in the inductance winding. As such it will behave a bit differently than shown in the tutorial and will go positive again (but not as much), then negative, and so on until the oscillations are lost in noise.

\$\endgroup\$
2
  • \$\begingroup\$ Oh okay thats much more clear. I was not considering the current of the inductor. I was just considering the voltage the moment switch opens. Thank you \$\endgroup\$
    – brye
    Commented 2 days ago
  • 2
    \$\begingroup\$ @brye the voltage across the indcutor only goes up to a ridiculous value when you force current to zero. With the diode there, that doesn't happen. \$\endgroup\$
    – hobbs
    Commented 2 days ago
7
\$\begingroup\$

At the instant of commutation (the voltage reversal), the inductor tries to maintain the exact same current flow. It doesn't try to impose a voltage anywhere, it just forces that current through whatever is connected to it at the time. The voltage that develops is determined by the components connected across the inductor, through which that current must flow.

Take the following as an example. I close a switch to establish a known current through the inductor. I then open the switch to break that current path:

schematic

simulate this circuit – Schematic created using CircuitLab

Initial inductor current (while SW1 is closed) is shown on AM1, and set by R2:

$$ I = \frac{V_1}{R_2} = \frac{12V}{10\Omega} = 1.2A $$

When the switch opens, the inductor will continue to pass 1.2A downwards, through the only path available to it, which is upwards through R1. We can predict what voltage will appear across R1:

$$ V_{AB} = -I \times R_1 = -1.2A \times 100\Omega = -120V $$

This can be seen in a simulation, in which the switch opens after 10μs. This is a plot of voltage \$V_{AB}\$ across the inductor:

enter image description here

Inductors oppose change in current, which is why current remains at 1.2A immediately following the opening of the switch. That can be seen if we plot current:

enter image description here

Resistor R1, therefore, is responsible for whatever voltage will appear between A and B, and inductor L1 is only providing the current, having no influence on voltage at all.

It's the resistor, or diode, or whatever else you connect across the inductor, that determines what voltage will appear there. To obtain a "spike" of 6V across the inductor, choose an appropriate resistance that will develop 6V when 1.2A flows through it:

$$ R_1 = \frac{6V}{1.2A} = 5\Omega $$

schematic

simulate this circuit

enter image description here

Your question is about the case where a diode is connected across the inductance, not a resistor, so we are interested in finding what voltage will appear in that case instead:

schematic

simulate this circuit

We know the current that will flow immediately following switch opening, 1.2A, so we could use the diode equation to reveal the voltage that a diode would develop with 1.2A flowing through it. This is the equation rearranged to make voltage the subject:

$$ V = ηV_T \ln \left(\frac{I}{I_S} + 1\right) $$

Setting parameters roughly according to the 1N4001 diode, thermal voltage \$V_T=25mV\$, saturation current \$I_S=8\times 10^{-11}A\$ and ideality \$\eta=1.5\$, this is what we get:

$$ \begin{aligned} V &= 1.5 \times 25mV \times \ln \left(\frac{1.2A}{8\times 10^{-11}A} + 1\right) \\ \\ &= 0.88V \end{aligned} $$

You already knew that a forward biased diode has somewhere around 0.7V across it, so that should be no surprise. Don't forget that the cathode, node A has the lower potential, since the diode is forward biased, so we expect to see \$V_{AB} = V_A - V_B = -0.88V\$:

enter image description here

That little "wobble" at the end is no doubt due to the diode's junction capacitance resonating with the inductance.

The short answer, then, is that the inductor doesn't choose what voltage it generates, only the current that it attempts to maintain. The voltage is determined by the element through which that current flows, which in the case of a diode is 0.7V or so.

In other words, the inductor does not impose some huge voltage across the diode, it only forces current through it, and the voltage that results will depend on the diode's response to that current, according to its I-V curve.

These same principles apply to your second question. You can use Ohm's law and the diode equation to calculate the total voltage across a series-pair of diode and resistor, to work out what voltage to expect across them both, when the kick occurs.

Combining the two last examples, by putting R1=5Ω and the diode in series, which would then both be momentarily passing 1.2A, we would expect:

$$ V_{AB} = -(0.88V + 6V) = -6.9V $$

schematic

simulate this circuit

enter image description here

\$\endgroup\$
1
  • 1
    \$\begingroup\$ That was a fantastic answer. I appreciate the clear examples. Definitely helped me get a better understanding of how an inductor behaves \$\endgroup\$
    – brye
    Commented yesterday

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.