Led Reverse Voltage

[Leto Corrales]

Irealize this question is answered elsewhere on the interwebs such as wikipedia, but I am looking for a short summary that is less of a technical discussion and more a useful tip to someone using diodes in a hobby circuit.

This is usually much higher than the forward voltage. As with forward voltage, a current will flow if the connected voltage exceeds this value. This is called a "breakdown". Common diodes are usually destroyed but with Z and Zener diodes this effect is used deliberately.

Reverse voltage is sort of the same thing - it's the voltage where current starts to flow when the diode is in the normally non-conducting region - this is also the point where the diode is likely to turn into a charred mess as all the internal semiconductor stuff turns to mush (choose a value somewhat larger than the largest PEAK [not RMS] AC voltage the diode will see)

Forward-bias is when the anode (the pointy part of the symbol) is positive and the cathode (the bar) is negative. Reverse-bias is when the anode is negative and the cathode is positive. A lot of current flows when the diode is forward-biased, provided that the voltage is higher than 0.6V or so for a silicon diode or 0.3V or so for a germanium device. A very small amount of current flows if a diode is reverse-biased.

If you have a DVM and some diodes, you can check it for yourself. Diode cathode leads are usually identified with a band, so if you switch the DVM to a low resistance setting, and connect the leads across the diode in both directions, you should see a low resistance in one direction and a high resistance in the other direction, provided that the DVM is supplying a high enough voltage. Some DVMs have a special diode test setting that is easier to use.

Just so that the information is condensed here and I like to know where to find my offspring, I'd add typical forward voltages for common LED as a quick reference for everyone. (And also because I like to dig an old thread on the 18th December.)

Although you mention "voltage," I believe you mean bias. If this is correct, then "forward bias" is the application of a voltage in such a way that the diode "shows" a low resistance. "Reverse bias" causes the diode to show a high resistance.

For perfect junction diodes there is a relation between the current and the voltage given by Shockley's diode equation I = Is(exp(Vd q/nkT). You can solve this to get Vd, the diode voltage, as a function of I. But when you have a resistor in series with the diode, you can NOT solve the circuit; you must use successive approximations. (Unless you have a WP-34 calculator with the built in Lambert W function, of course.) Ordinarily, you can replace nkT/q with about 0.6 volts, and saturation current Is with about 1 mA, and your voltage calculation will be in the ball park.

What I'm not understanding is that, when a reverse voltage spike occurs, if the flyback diode is meant to protect the transistor, what is stopping the current that is being pushed in the opposite direction to travel to the 9V battery and damage it, or to travel to the transistor through the capacitor?

If you look at the circuit and think about current flow, the current flows out of the positive battery terminal and from positive-to-negative through the load (1) but inside the battery and from negative-to-positive.

Inductors (coils) "resist changes in current flow" (2). When power is removed, the magnetic field collapses. The changing magnetic field creates a voltage, turning the motor or solenoid (temporarily) into a generator.

The current through the coil is flowing the same direction as before but suddenly it's now a generator so like a battery the current is flowing negative-to-positive through the coil and the voltage is reversed.

Without the diode, when the transistor, there is (almost) no path for the current (nearly infinite resistance). You can get a very-high voltage spike as the coil tries to keep current flowing through an infinite resistance (Ohm's Law). That voltage can easily be high enough to fry the transistor or something else.

(1) This is "conventional current", which is just a "concept". The electrons are flowing from negative to positive (through the load). If you take an electronics class they teach conventional current flow but if you take a chemistry or physics class they teach current flow.

Hi,

summing up,

every inductive load, and motor is an inductive load, generates an inverse voltage spike when removing the voltage across it.

In your schematic while the motor receives the current coming from the 9V through the transistor, it turns and there is a voltage on it.

By stopping this current by switching the transistor, the voltage across the motor goes to zero, and this causes a reverse voltage between the motor pins.

As this current stop can be very fast, the reverse voltage can be a spike of values much higher than 9V.

As the diode is also connected inverted between the poles of the motor, it "kills" this high voltage spike protecting the transistor.

The Capacitor is also in this circuit to dampen "electrical noise" generated by the motor in motion.

@jremington

Thank you all so much for the replies. I really appreciate as my professor was not much in understanding my question. So would it be safe to say that the circuit more or less works like this? The thing I really want to understand is the current path when to the collector pin of the transistor when the base pin is receiving the 5V from the Arduino vs. not:

Reverse voltage protection circuits prevent damage to power supplies and electronic circuits in the event of a reverse voltage applied at the input or output terminals. Reverse voltage protection is implemented at the input of the power supply or onboard of the custom, multiple output redundant power supplies. This is important in most electronic applications such as laptops, computers, CMOS circuits, etc.

The protection ensures that the components are not damaged by accidental swap of the power supply connections. There are various methods that differ in operation, efficiency and complexity. While some like a diode or circuit breaker provides only the reversal voltage protection, others such as the protection ICs provide the reverse voltage, over current, and overvoltage protections.

To block negative voltages, designers usually place a power diode or a P-channel MOSFET in series with the power supply. One drawback of the series diode is that it takes up board space and has high power dissipation at high load currents.

On the other hand, the MOSFET dissipates less power even though it requires an extra drive circuitry which increases the cost. Both solutions affect low power operations and especial the series diode. In addition, the solutions may not be suitable at very high load currents.

The diode is connected in series with the load and only allows power to reach the load only when in forward bias. If the voltage is reversed, it blocks the voltage and the reversed power does not reach the load. Using the diode is the simplest method and has the advantage of low cost.

The disadvantages of using the diode are; the forward voltage drop which can be significant in low voltage applications, the high power dissipation in high load currents and low efficiency. A Schottky diode is sometimes used due to its fast response and low forward bias voltage drop.

A better protection uses a MOSFETS which have an advantage of very low on resistances. The method involves using a high-side P type MOSFET on the power path, or a low-side NMOS FET in the ground path.

When the supply voltage is reversed, the PMOSFET gate voltage is high and this prevents it from turning on hence preventing the reverse voltage from reaching the load. For an NMOSFET, the gate voltage is low.

The breakers are used in high power applications of 500 W to several Kilowatts. At these high currents, it is not practical to use the diodes, or even schottky diodes due to the high power dissipation and inefficiency. The electronic circuit breakers are used together with a power shunt diode.

The circuit is expensive, bulky and requires manual resetting of the circuit breaker, hence not suitable for remote installations. In addition, the accuracy of the circuit breaker may be inadequate in applications that require precise current limiting.

Protection ICs such as the LTC 4365 are designed to protect sensitive circuits from reverse polarity, over current and over voltages. The ICs blocks the undesired current or voltage and only allows the safe voltages to pass through.

You will need to protect VBUS and VIN+ and VIN- as they cannot go below -0.3V as well and I assume they could go the voltage if connection is reversed. You can add protection diodes but be aware that if they have leakage during normal operation to account for that in the error. If you add resistance be aware going to large to could cause additional error.

Protecting the differential inputs is more difficult. A bit unsure if the diodes and the high input impedance of the pins are going to work together well. And the voltage drop across the two diodes may not match well.

That circuit would not work as the diodes would not prevent the voltage at the inputs to go below GND. The diodes in this configuration only prevent the direction of the current. A diode from IN+ which would be your source to GND reverse bias may do the whole job but the diode could get damaged if it cannot provide the current. Normally there is a switch also involved for this. Below are a few resources with different approaches.

The peak inverse voltage is either the specified maximum voltage that a diode rectifier can block, or, alternatively, the maximum voltage that a rectifier needs to block in a given circuit. The peak inverse voltage increases with an increase in temperature and decreases with a decrease in temperature. [1]

In semiconductor diodes, peak reverse voltage or peak inverse voltage is the maximum voltage that a diode can withstand in the reverse direction without breaking down or avalanching.[2][3] If this voltage is exceeded the diode may be destroyed. Diodes must have a peak inverse voltage rating that is higher than the maximum voltage that will be applied to them in a given application.

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