How To Reverse Polarity On A Generator? The 128 Correct Answer

DC generator voltage polarity

When no external voltage is applied to the DC motor but the rotor of the motor rotates, this motor works as a DC generator. Depending on the direction of rotation of the rotor, the output voltage is positive or negative. To demonstrate the polarity of the output voltage, two light emitting diodes (LEDs), one red and one green, were connected to the motor contacts with opposite polarity. Every LED emits light when the voltage is applied in direct polarity and does not emit when the voltage is in reverse polarity. As a result, the direction of rotation of the rotor is indicated by switched-on red or green LEDs.

The experiment demonstrates this concept by visually observing the change in light color as the car moves forward or backward.

Yes, if you accidentally reverse the polarity on an outlet, the device you are plugging into the outlet is unsafe and could cause a short circuit, electric shock, or fire. Attention – it is highly dangerous!

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Home » Technical Resources » Are DC Motors Reversible?

Are DC motors reversible? Simply put, DC motors can rotate in either direction (clockwise or counterclockwise) and can be controlled simply by reversing the polarity of the applied voltage. In fact, the motors can actually generate a force in either direction. We make this important distinction because some applications, such as B. haptic feedback, use “brakes” to control the motor without actually rotating it in the opposite direction. When a motor is already moving, the applied voltage can be inverted and the motor will quickly decelerate and eventually stop. If the voltage is still present, the motor will start rotating again according to the polarity of the voltage. Fleming’s Left-Hand Rule and DC Motors The direction of force and thus the rotation is explained using Fleming’s Left-Hand Rule for Motors. First we use a (very) simplified model of a motor – imagine two magnets with opposite poles (N and S) separated by a small air gap, with a wire in between carrying an electric current. This is essentially how a motor is constructed, although in this simplified example we are imagining unipolar magnets of infinite length to avoid introducing complications such as the commutator. This concept is perfect for explaining the important part of the theory. Speak to a member of our team. Contact us Motor catalog Looking for our products? Reliable, low-cost, miniature mechanisms and motors to meet your application needs. Check out our motor products

When the wire is free to move and current passes through the magnetic field, a force acts on the wire, causing it to move. In a motor, the coils can be attached to the rotor so that the force acting on the wire causes the shaft to rotate. In our simplified diagram we can say that the wire moving to the left corresponds to the motor rotating counter-clockwise, and moving to the right it is clockwise.

Now we apply Fleming’s left hand rule to determine the direction of the force. The resulting force is perpendicular to both the magnetic field and the direction of the current. Using the hand position in the image at the top of the article, you can position your left hand to replicate the image below. You might want to wait until you’re alone in the office because you’re going to look pretty weird!

Your first finger represents the magnetic field and points directly to the ground.

Your middle finger represents electricity and points to the computer screen.

Your thumb represents the resultant force pointing to the left.

This tells us that the current flowing through the cable “into” the computer screen is causing a left-pushing force. In our model, this corresponds to counterclockwise rotation of the motor.

What we’re most interested in now is how to change the force so the wire moves in the opposite direction, causing our motor to spin “backwards”. We can use Fleming’s left-hand rule again, with the same magnetic field, but this time we use our thumbs to point to the right instead of the left. As a result, your middle finger should now be pointing at itself, showing the current flowing out of the screen.

This shows that in order to make the motor rotate clockwise, we need to reverse the current flow (i.e. changing the current flow changes the direction of the force 180 degrees).

Of course, the direction of the current is controlled by the polarity of the voltage. So, to change the direction of rotation, we can simply reverse the voltage, causing the current to flow in the opposite direction, changing the force 180 degrees and driving the motor “backwards”.

Practical Implications – How to reverse the voltage

If you are unfamiliar with electronics, changing the polarity of the voltage may sound more difficult than it really is. In fact, you’re more concerned with the control logic – that is, deciding and commanding when to reverse polarity. You can easily drive the motor in both directions with a single chip, but that depends on your application.

Let’s take two example applications that drive a motor in both directions, a locking mechanism and a haptic feedback device.

The locking mechanism uses a geared motor driven in both directions to lock or unlock a door. When a motor actually needs to turn both clockwise and counterclockwise, one of the most popular drive chips is called an H-bridge. These are discrete components that house 4 transistors that act as switches. One pair of switches are used to drive the motor in one direction while the other two are used for reverse. The control for motor direction (often simple GPIO signals) is separate from the driver voltage, which controls speed, so you can vary them independently.

Conversely, haptic feedback devices implement “active braking” which is used to stop the motor faster and improve the sharpness of the effects. The motor does not actually turn in the opposite direction at any point, but we use the effect of a counter-voltage to control the motor more precisely. Many haptic chips implement active braking by default, either as an on-chip setting or as part of a pre-programmed waveform — making implementation very easy.

If we inspect homes where an amateur has done some electrical work, chances are we’ll find reverse polarity outlets. This happens when the hot and neutral wires are flipped at an outlet or upstream from an outlet. Reversed polarity presents a potential shock hazard, but is usually an easy fix. Any $5 electrical tester will alert you to this condition, provided you have a properly grounded three-prong outlet. However, if you have a three-prong outlet that isn’t grounded, your electrical tester won’t be able to tell you if the polarity is correct or not.

A brief definition of hot and neutral wires: On a standard outlet, technically called a duplex outlet, there are two wires that carry power. One of these wires is connected to earth or grounded, so that wire is called the grounded conductor. This wire is commonly referred to as the neutral wire and should always be white. That’s what the larger slot on your outlet is for.

The other wire is not connected to ground and is called the ungrounded conductor or hot wire. This wire can be any color except white or green, but is usually black or red. That’s what the smaller slot on your outlet is for. Since the hot wire completes a circuit by making contact with earth, when you touch a hot wire and make contact with earth (which is pretty much always the case), you become part of the circuit. In other words, you will be shocked.

To make electrical devices safer, some plugs are polarized. This means that one blade is larger than the other and the larger blade will only fit in the neutral side of an outlet.

Shock Risk Scenario #1: I’m toasting an English muffin and it gets stuck in my toaster. I look in the toaster and see that the heating elements are off, so I assume it’s safe to stick a knife in the toaster to get my muffin. I should be doing this for sure because the switch that controls the flow of electricity to the heating elements in the toaster shuts off the hot wire. Unfortunately, my toaster is plugged into a reverse polarity outlet, so the switch on my toaster turns off the neutral instead of the hot. As a result, there is always current at the heating elements. As soon as some dope sticks a knife in the toaster, electricity flows up the knife, through my body, and back to earth. ruined breakfast. Your mom was right when she told you never to do this, even if the toaster is off.

Electrocution Hazard Scenario #2: I’m using an old emergency lamp and my finger accidentally touches the outside of the metal socket that holds the bulb. The outlet is always connected to neutral, so no big deal…unless the light is plugged into a reverse polarity outlet. In this case I get a shock. If this happens while I’m lying on the garage floor working on my car, there’s a good chance I’ll get electrocuted. This can also happen with old table lamps that have exposed metal fittings.

Solution: Get an electrician. I’d like to say that the solution is as simple as swapping the wires at the outlet, but that’s not always the case. If the wiring seems correct, the wiring error occurred somewhere before the outlet. Now the electrician has to trace the wires to find out exactly where something went wrong and fix it there. Simply swapping the wires at the outlet would not be an acceptable solution.

The bottom line is that reverse polarity on outlets poses an electric shock hazard, but electronic devices plugged into a reverse polarity outlet will still work. Don’t assume your outlet is wired correctly just because it “works”. You can test your sockets for reverse polarity with an inexpensive socket tester that you can find at any hardware store.

And if you have reverse polarity outlets, have an electrician fix the wiring.

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not a good idea. A short circuit with bad contact becomes a fire hazard.

But if you’re just temporarily plugging in a device with a 2-pin connector, maybe nothing.

In my previous home I found that installers had used neutral to package the ground for a stair switch when I converted it to a 2 way switch. The neutral wire is grounded inside or outside the house at the distribution transformer, and the ground is a safety wire that also carries line filter cap currents of up to 0.5mA each.

Old TVs used to use 2 pin plugs which were polarized and a neutral grounded chassis.

The potential (no pun) for danger is great, but theoretically it might not matter in very limited circumstances.

2-prong plugs for outlets are polarized in America for a reason for some products that rely on the neutral being close to 0V to keep the winding insulation close to the metal shell.

Such as; The metal Edison socket is neutral with the center contact switched. If you swap L-N and turn off the lamp, the metal collar would become an unswitched line.

NEW: Learn electronics? Submit your questions on the new Electronic Questions & Answers website hosted by CircuitLab.

Basic Electronics » Why is reverse polarity bad?

October 10, 2011

by Glendon

Hello, I read that a device that uses a battery often contains a diode that protects the device if the battery is inserted the wrong way round. Can anyone share why such a simple reversal of polarity would damage an electronic component? Many Thanks.

October 10, 2011

by BobaMosfet

Glendon, This is because in most cases the device is designed to handle currents controlled by specific voltage levels in only one direction. If you apply this voltage in the opposite direction, the “reverse” breakdown voltage of one or more components within a circuit will literally overheat, short out, and conduct in the wrong direction until they slag. Such slagging can result, for example, in a cloud of smoke, a small volcano in an IC, or an exploding capacitor. When we say “device” we are actually talking about a circuit that contains many paths – some polarized, some not. The unpolarized paths don’t care, but the others do. For example, anything based on a diode (transistor, IC, Scr, rectifier, LED, etc.) can be damaged by passing too much current through it in the reverse direction. Other components, like unpolarized capacitors or plain old resistors don’t care, they work the same in either case. It’s a really good idea to learn how to put your own diode across your input terminals coming into whatever circuit you make to protect it for this simple reason. Components aren’t nearly as expensive as they used to be – but there’s no point wasting them by accidentally reversing the connection. Especially with the increasingly complex breadboard designs you are making. A second accident can cause hours or days of frustration and rebuild a toasted circuit. Hope this helps BM

October 10, 2011

by Glendon

Hello BobaMosfet, thanks for your answer! Sorry, but I’m still a bit unclear. If we send reverse voltage it means we send -5V instead of 5V. Is there such a thing as negative voltage (actually causing the damage)? If not, would reversing the polarity emit nothing (i.e. 0V) unrelated to the electronic component “receiving” it?

October 10, 2011

by mongo

It seems so right? In fact, many integrated circuits have a built-in diode that essentially shorts out the power rails when wired backwards. Unfortunately, they don’t handle much power and the PSU wins the battle. I blew up ICs on the board which fused a smelly little pile of plastic onto the board. Other circuits conduct a little in the reverse direction, and since it’s more or less a resistive load, it gets hot. That heat, so close to the rest, does the damage. A negative voltage is really subjective. However, since most TTL circuits use a +5 volt supply, reversing the polarity would present a -5 volt source. However, there are some circuits that use both a +5 volt and a -5 volt supply. (usually because there is either a linear circuit there or it is a programming voltage for a memory chip). Then there are others that use -12, +12, -5, and +5 volt supplies. (Also usually memory chips). Some EPROMs use a 25 volt pulse to write to the chip. Voltage levels generally refer to the “ground” side, in most cases it is the common between two voltage levels like a +5 and +12 volt supply. They share the negative lead, so it would be a positive dual voltage supply. For a single-ended power supply, this is usually the negative side for ground. What you usually get out of a circuit when you reverse the power supply might be a little smoke.

October 11, 2011

by bretm

It’s a really good idea to learn how to put your own diode across your input terminals that will come into whatever circuit you make, not across the input terminals but in series with one of them. If you put a diode across the power supply terminals and apply the current backwards, the diode would shunt the current away from the circuit and protect it for the short time before the diode melts. In series, the diode would simply block reverse current flow. Reversing polarity in a semiconductor circuit is a problem simply because many of the junctions are intended to be reverse biased and block current in normal use. If they are reversed, they conduct electricity and there must be no current-limiting resistor in the circuit to prevent overloading. Example: This circuit works perfectly. It doesn’t do much, but it doesn’t fail: o———-o | | — + —- – /\ — / \ Diode – – —- | | o———-o But if you put the battery in the wrong way round: o———-o | | – – —- — /\ – / \ Diode — + —- | | o———-o Now the diode is conducting current, and since current flow in a forward-biased diode increases exponentially with voltage, it will overload and melt.

October 11, 2011

by bretm

Oh, and another issue is electrolytic capacitors. If one of these is on the input side of a DC regulator and you reverse the polarity, you will destroy it. There is a very thin layer of oxidation on the aluminum plate of an electrolytic capacitor, which is created by an electrochemical process. This layer of oxide insulates the positive and negative plates of the capacitor. If you reverse the polarity, you reverse the electrochemical process that created the oxide layer and it begins to erode. As soon as electricity begins to leak somewhere, the increased current flow accelerates the chemical reaction in the vicinity of the leak, destroying the oxide layer even faster, resulting in even more current flow, and a runaway chain reaction ensues. The capacitor will short circuit and may explode.

October 11, 2011

by Vern

Some components just can’t take the battery swapping.

For example an electrolytic capacitor used in power supply filtering. Apply the wrong power (+ and – vice versa) and this bad boy will get really hot or explode. Zener diode circuits tend to be reverse biased. They just don’t work properly when forward biased. Some circuits protect against reverse polarity by placing a forward-biased diode across the power input. This is great for low power circuits, but when we draw significant current it becomes a problem. Put simply. What would happen to your car if the battery terminals were reversed? Pay attention to polarity when powering the electronics.

October 11, 2011

by Glendon

Hello everyone, thanks for all your lovely replies! I think I’m getting the picture now. So, here’s what I have in mind: the battery will still provide the flow of electrons, even in a reverse polarity scenario. The electrons flow to the battery due to the voltage difference as long as the circuit is closed. Since the gates we are using are n-channel and p-channel mosfets, they are virtual switches that turn on and off above a certain threshold current. If reverse polarity occurs, the electron flow (wrong direction) will cause the gates to turn on/off with the opposite intention, which can cause cascade errors in the circuit. To prevent this, diodes are used as fail-safe devices. Usually, however, the “reverse” current is too strong (or on for too long), causing the diodes to heat up. Consequently, the generated heat damages the electronic component. Am I in the stadium for this?

October 11, 2011

by BobaMosfet

bretm- Thanks for the correction, yes the diode should be in series, not parallel. Glendon- When you apply too much voltage in the reverse direction, your current forces it to literally blow (dissolve) through paths in components that cannot withstand the force of the current in that direction. The current sees everything between itself and what it wants as resistance, regardless of direction. Components heat up because they resist the excessive current flow (simply from the chemistry and materials they’re made of and how those things are put together to control the current flow). If that flow is too great, they will melt (you will smell ozone when it happens). It can be slow, it can be fast, it can be explosive. And it has nothing to do with goals. The part is disassembled. It’s not about time, it’s about quantity. The amount of current that flows through any part of a circuit is directly proportional to the voltage between two points in the circuit where that current flows. Most components can handle different maximum amounts of current flowing through them, depending on the direction. Cross it both ways, and the portion frys. Frying it briefly can damage other components as well. Each component has characteristics and power dissipation curves that describe how the device should behave based on the applied voltage and current. For this reason, it’s a good idea to use a diode in series with your input voltage – it will normally allow enough current to flow in one direction to fully power your circuit, but almost completely block it in the other direction (up to a certain maximum) such as 600-PRV – also known as Peak Reverse Voltage). bm

October 12, 2011

by Glendon

Thank you BobaMosfet. I think I’m quite clear now. By the way, I mentioned logic gates because I got the impression that every electronic component consists of different types of gates arranged in a specific order to serve their purpose. This gives me the idea of ​​cascading faults in the circuit when the current is reversed… 🙂

October 12, 2011

by BobaMosfet

Glendon, yes they have goals but no ‘cascading errors’ occur. The gates are literally melted and blown apart. Current doesn’t know or care what a goal is – it seeks all paths and sees everything as an obstacle in its path. bm

October 12, 2011

by bretm

“The amount of current that flows through any part of a circuit is directly proportional to the voltage between any two points in the circuit where that current flows.” The current flowing through a forward-biased PN junction actually increases approximately exponentially with voltage, not proportionally. This is why reverse polarity is so bad for semiconductor circuits. The proportional current-voltage relationship only applies to resistive materials such as simple resistors.

October 12, 2011

by BobaMosfet

bretm- Don’t confuse the PN junction voltage drop with the actual voltage across the diode. And you are describing a diode, not a transistor. A diode is not an amplifier. It is a rectifier with asymmetrical breakdown voltages. Normally, the forward voltage is about 0.6V to 0.7V (silicon) or 0.2V (germanium), and the reverse voltage is smaller than the breakdown voltage, which is much higher (PRV). bm

October 12, 2011

by bretm

i am not confused Transistors have pn junctions and current flow is exponential, not linear. I’m not talking about gain, just the basic IV curve of almost all pn junctions. These are non-resistive materials, so the current/voltage ratio is not proportional.

The forward voltage drop of a silicon junction is only 0.6 to 0.7 volts for a given current range. If you apply a large voltage to the junction and the current is not limited, then this voltage range will easily be exceeded.

October 13, 2011

by BobaMosfet

bretm: What you described was a diode. I replied to that specific description. I agree that the transistor doesn’t represent a linear relationship between E, I and R (due to the transconductance – see Ebers minor), but that wasn’t entirely relevant to the whole concept I was trying to express. Anything between the poles of a power supply can be thought of as resistance or impedance (Thievnin). If you apply enough voltage difference between the poles, it doesn’t matter which direction you go, the electron stream will blow through everything like a shotgun through a target. It does not matter. All I was trying to say was that the OP shouldn’t consider the reverse current as something that finds an orderly path through the circuit and introduces computational errors across the cascade. It’s a lot more brutal. I sometimes find it easier to give someone a generalization at the outset to get their mind moving in a certain direction; then you can later add details, sentences, why and why if the whole thing is not so easy to grasp. It’s a bit like saying, ‘The sky is blue.’ (Most people will agree on that) But then comes bretm “No, sometimes it’s gray and cloudy or black with stars or red in the morning or green at night depending on where you are.” You are absolutely right, but is it being offered to be helpful or just to be right? bm

October 13, 2011

by BobaMosfet

bretm- Because you like accuracy, your pointed example isn’t entirely accurate: Quote: Example: This circuit works fine. It doesn’t do much, but it doesn’t fail: o———-o | | — + —- – /\ — / \ Diode – – —- | | o———-o But if you put the battery in the wrong way round: o———-o | | – – —- — /\ – / \ Diode — + —- | | o———-o Now the diode is conducting current, and since current flow in a forward-biased diode increases exponentially with voltage, it will overload and melt. :endquote The first example works because the diode acts as an open circuit – no current flow (other than leakage). The second example doesn’t work, NOT because the current flow increases exponentially with voltage, but because it’s now a short circuit where nothing opposes the current flow, so the power supply delivers full current at full voltage (minus the small .7 drop due to the diode) over the diode. So if it’s a 9V battery: 9 – 0.7 = 8.3V at about 0.500mA (or 4.15A), which will roast the diode. bm

October 13, 2011

by bretm

I agree that “side comments” can sometimes distract from the original post, but this discussion forum doesn’t support threaded discussions, so I’ll just do it. ==== WARNING – THIS IS A SIDE DISCUSSION – DON’T GET CONFUSED === But other than that I don’t think we’re on the same page at all and I think that’s ok. My comments had nothing to do with transconductance. “Everything between the poles of a power supply can be thought of as some form of resistance or impedance (Thievnin).” No, that is not Thevenin’s theorem. Specifically, Thevenin’s theorem only applies to linear circuits – resistors and constant sources. “If you apply enough voltage difference between the poles, it doesn’t matter which direction you go, the electron stream will blow through everything like a shotgun through a target.” Absolutely correct, but we’re only talking about -9V here. This is not a breakdown voltage situation. -50V, sure. +50V too – polarity doesn’t matter anymore in this situation. For purposes of this thread, it’s not why reverse-inserting a 9V battery causes problems in semiconductor circuits. -1V is sufficient for some semiconductor circuits due to the disproportionality of the current flow of PN junctions. Reversing the battery destroys semiconductors that should not be excessively forward biased, such as B. the overvoltage protection diodes on the input pins of the MCU. The reason you can’t go beyond Vcc +0.5V or Vee-0.5V and not Vcc +1V or Vee-1V is because of the exponential I-V curve. “The second example doesn’t work, NOT because the current flow increases exponentially with voltage, but because it’s now a short with nothing resisting the current flow” I totally disagree. It’s not a short circuit. Put a 0.7V battery instead of a 9V battery and it would not damage the diode. Why? Because it’s not really a short circuit. This is due to the exponential IV curve. It is not true that “there is nothing to prevent the flow of electricity”. There’s a diode in there that resists the flow of current. It does an admirable job of resisting current flow at 0.7V and a poor job at 9V because of the exponential I-V curve. “So the power supply delivers full current at full voltage (minus the small 0.7 drop through the diode) through the diode. So if that’s a 9V battery: 9 – 0.7 = 8.3V” It’s not 8.3V. There is no 8.3V reading anywhere in this circuit. If you find a 0.7V voltage drop somewhere in this circuit please tell me where to measure it because I’d love to see it. (I almost went straight into “Please tell me where to place the probes”. Phew, that was close.) Here’s the proof: Set up this circuit with an adjustable power supply instead of a battery and 0.9V instead of 9 V in. Measure the voltage across the diode. It will be 0.9V, not 0.7V. Why won’t it be 0.7V? Because you put 0.9V there. Measure the current. It will be great value e.g. B. between 500mA and 1A for a 1N4001. Now reduce the power supply to 0.45V and again measure the voltage drop across the diode. It will be 0.45V. Now measure the current. Will it be half the original amount because the voltage is half the original 0.9V and the current is proportional to the voltage? No, it will be a tiny amount because of the exponential I-V curve. Now adjust the supply to 1V if the diode and the supply can handle it. (Come on, it’s only 0.1v measly more volts, what harm can that do?) Measure the voltage drop. Is it 0.7? No, it’s 1V. Measure the current. Is it a ninth more than at 0.9V? No, because the IV curve is not linear or because the diode is dead. Myth: A sufficiently forward-biased silicon diode or silicon base-emitter junction has a voltage drop of 0.6V to 0.7V. Truth: The voltage drop is 0.6V to 0.7V for a wide and typical one Range of current levels found in small signal circuits because the exponential I-V curve is so steep in this range of currents. But whether it is actually in that range or not depends entirely on the current flow through the junction. If you apply 0.5V or 0.8V across it, the voltage drop is 0.5V or 0.8V. Rule of thumb: A sufficiently forward-biased silicon diode or silicon base-emitter junction has a voltage drop of 0, 6V to 0.7V. There is another myth about zener diodes and breakdown voltages that I see a lot, but I think there is a limit to how I should stray from the topic.

October 13, 2011

by bretm

OMG, I just noticed this part: “But then comes bretm…is it offered to be helpful, or just to be right?” Unbelievable. The OP asked why putting the battery in the wrong way round was bad. I’m giving the correct explanation after some people mistakenly gave “breakdown voltage” as the explanation, but somehow I’m the problem.

October 13, 2011

by bretm

According to an admittedly unreliable source called Wikipedia: The breakdown voltage is a parameter of a diode that defines the largest reverse voltage that can be applied without causing an exponential increase in current in the diode. As long as the current is limited, exceeding a diode’s breakdown voltage will not harm the diode. But I’m done with this thread.

October 13, 2011

by BobaMosfet

bretm- Instead of reading Wikipedia, try reading something like ‘Art of Electronics’ or ‘Understanding Circuits & Op-Amps’ or MIT dissertations like I do (as well as many other sources). There is a little more to the breakdown voltage than what Wikipedia claims. Since the OP was discussing a battery (which we all assumed was 9V), discussing subvolt reverse voltages was not relevant. But let’s go back to the beginning. Forget everything said so far and stick a 9V battery with reversed polarity on any diode, such as an LED. Hmmmm…. blew it apart (I just did). I think the breakdown voltage has been exceeded. bm

October 13, 2011

by bretm

That’s great. I hope you had fun. I just did it with a silicon diode and it didn’t blow apart. From my understanding the Atmega168 uses silicon and does not contain LEDs. But guess what, you’re absolutely right. I was talking about the nerdkit all along when the OP wasn’t talking about the nerdkit at all. Still, the insult was inappropriate. I actually love being wrong. It’s the only time I learn anything.

October 13, 2011

by BobaMosfet

bretm: No insult was pronounced. Just an observation specific to you. It wasn’t mean or awkward. It was very true. You are often the most productive poster here. However, sometimes someone needs to get used to the forest before anyone else can point out the trees, the bushes, the creatures, and so on. The ATMEGA168’s gates don’t have the same level of protection as a single silicon diode, so I used an SMT LED- it’s a substrate device. Your silicon diode probably has a PRV (aka breakdown) rating of probably 40, 100, 400, or 600 volts. I don’t know because you didn’t give the number. On a positive note, however, most devices like the ATMEGA today have protection diodes which (within limits) emit some overvoltage (in either polarity) to the rails. In fact, that’s one of the tricks I use to power an ATMEGA in special circumstances when I can’t connect VCC or GND. No hard feelings, BM

October 14, 2011

by Ralphxyz

Well I’d like to thank Glendon for asking this question, what an interesting discussion. Also, of course, thank you bretm and BobaMosfet, you guys just blow my mind. Too bad this forum doesn’t support thread discussions or even [quote] for that matter. In fact, Markdown syntax supports [Blockquotes] Blockquotes can contain other Markdown elements, including headers, lists, and blocks of code: > ## This is a header. >> 1. This is the first item in the list. > 2. This is the second list item. > > Here is an example code: > > return shell_exec(“echo $input | $markdown_script”); But if you want a different background color to make the quote stand out, someone has to set that up. Speaking of digress geesch!! I really wanted to thank Bretm and BM. Ralph

October 24, 2011

by Hexorg

I see the question has already been answered, I just wanted to have a little input. Glendon, Matter of electric current through a wire as of liquid current through a tube. If you direct a stream of water through a nozzle, it will work no matter what direction the water is flowing. But now imagine a heart valve opening/closing at a certain rate. When the blood flows through the valve in the desired direction, it is working properly. However, if you force the blood in the opposite direction, not only will it affect the function of the valve, but it may also rupture. PS A differential potential (voltage) is like a pressure in the liquid. A negative voltage means that the electrons will simply flow in the opposite direction.

October 24, 2011

by Hexorg

*think

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Polarizing your 12 volt alternator is necessary for your alternator to work in conjunction with the regulator. In contrast to an alternator, a generator can be operated with both positive and negative current. A regulator always runs with positive current. Therefore, if the generator is not polarized positively and positive current is fed into the terminals, the generator can be damaged, as well as the other motors, and turn on the same current. Fortunately, polarizing a generator is very easy.

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In a way, torque is produced that is proportional to the applied load resisting rotation, as clearly must be the case to cause mechanical power to flow from the prime mover to the machine rotor, but in a grid-connected machine, the speed set by the grid frequency, not the load on the individual machine (ok, there can be some funny dynamics, but assume a steady state).

The dynamics are compounded by the fact that rotor speed (and hence frequency) is the integral time of the net torque.

Imagine an ideal three-phase machine connected to a large network. If we start with everything in sync and with no net power flow, the angle between the rotor field and the grid-tied induced field of the stator is zero and no torque is produced. Now if we open the throttles, what happens? The prime mover tries to accelerate, which causes the rotor to start leading the fields generated by the grid-tied stator, and when this angle starts to increase, torque is eventually produced as the stator’s magnetic field pulls on the rotor. This torque equals the torque, generated by the prime mover, and we end up in a (hopefully) stable state where the rotor leads the grid just enough to match the torque of the prime mover.

The interesting thing is that when the grid is massive compared to the generator, the RPM hasn’t changed (as determined by the grid frequency), only the rotor angle has changed, but we are now generating power.

If the overall power consumption is too low, the grid frequency will drop (and all generators will slow down), if the total power consumption is too high, the grid frequency will increase, but in all cases the speed of all machines corresponds to the grid frequency.

Note that no mention was made of voltage, which is normally controlled to set reactive power, not real power.