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Basic Electronics » Why is reversing polarity a bad thing?

October 10, 2011
by glendon
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Hi,

I read that a device that uses a battery often contains a diode that protects the device if the battery is inserted backwards. Can someone share why would such a simple act of reversing the polarity damages an electronic component?

Thanks.

October 10, 2011
by BobaMosfet
BobaMosfet's Avatar

glendon,

It's because, in most cases, the device is designed to handle currents controlled by specific voltage levels in only one direction. If you apply that voltage in the opposite direction, the 'reverse' breakdown voltage of one or more components within a circuit will literally overheat, short, and conduct in the wrong direction until they slag.

Such slagging can result in a puff of smoke, a small volcano in an IC, or an exploding capacitor, for example.

When we say 'device' we are actually talking about a circuit that has many paths in it- some are polarized, some are not. The unpolarized paths don't care, but the others do.

For example, anything that is based on a diode (transistor, ic, scr, rectifier, led, etc), can be damaged by forcing 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 either way.

It is a really good idea to learn to put your own diode across your input terminals coming into any circuit you make, to protect it for this simple reason. Components are not nearly as expensive as they used to be- but there is no sense wasting them by accidentally reversing polarity on the connection. Particularly with the increasingly complex breadboard designs you make. One uSec of accident can cause hours or days of frustration, rebuilding a toasted circuit.

Hope that helps,

BM

October 10, 2011
by glendon
glendon's Avatar

Hi BobaMosfet, thanks for your response!

Sorry, but I'm still a little unclear. If we send a reverse voltage, does it mean that we are sending a -5V instead of a 5V. Is there such thing as a negative voltage (which actually causes the harm)? If not, wouldn't reversing the polarity send out nothing (i.e. 0V), which does nothing to the electronic component that is "receiving" it?

October 10, 2011
by mongo
mongo's Avatar

It seems that way huh? Actually, many integrated circuits have a diode built into them that essentially shorts out the power rails if they are wired backwards. Unfortunately, they don't handle much current and the power supply wins the battle. I have had IC's blow up on the board, leaving a stinky little pile of plastic fused to the PC board. Other circuits do conduct a little bit in the reverse direction and because it tends to be a resistive load, more or less, it gets hot. That heat, so close to the rest does the damage.

A negative voltage is really subjective. But since most TTL circuits use a +5 volt supply, swapping the polarity would constitute a -5 volt source. However, there are some circuits that use both a +5 volt and -5 volt supply. (usually because there is either a linear circuit there or it's 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. The voltage levels are generally referenced 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 dual voltage positive supply. On a single-ended power supply, it is typically the negative side for the ground.

Usually what you get out of a circuit when you reverse the power supply is maybe a little smoke.

October 11, 2011
by bretm
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It is a really good idea to learn to put your own diode across your input terminals coming into any 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 power backwards, the diode would shunt the current away from the circuit, protecting it for the short amount of time before the diode melts. In series, the diode would simply block the backwards current flow.

Reversing the polarity in a semiconductor circuit is a problem simply because many of the junctions are intended to be reverse-biased in normal usage, and block current. If the polarity is reversed, they conduct current and there may not be any current-limiting resistance in the circuit to prevent them from overload.

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 backwards:

 o----------o
 |          |
 -  -      ----
---         /\
 -         /  \  diode
--- +      ----
 |          |
 o----------o

Now the diode conducts current, and because the current flow in a forward-biased diode increases exponentially with the voltage, it overloads and melts.

October 11, 2011
by bretm
bretm's Avatar

Oh, and another issue is electrolytic capacitors. If there is one of these on the input side of a DC regulator and you reverse the polarity, you'll destroy it.

The aluminum plate in an electrolytic capacitor has a very thin layer of oxidation on it created by an electrochemical process. This oxide layer 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 will begin to erode. Once it starts to leak current anywhere, the chemical reaction near the leak will accelerate from the increased current flow, destroying the oxide layer even more quickly, causing even more current flow, and a run-away chain reaction will occur. The capacitor will short-circuit and may explode.

October 11, 2011
by Vern
Vern's Avatar

Some components just can't take the battery being reversed.

For example, an electrolytic capacitor used in power supply filtering. Apply the wrong power (+ and - reversed) and that bad boy will get real hot or explode.

Zener diode circuits like to be reversed biased. They just don't work correctly if they are forward biased.

Some circuits protect against reverse polarity by putting a forward biased diode in the power input. This is great for low power circuits, but when we draw considerable current this starts to become a problem.

Simply put. What would happen to your car if the battery terminals were reversed? Watch the polarity when you apply power to electronics.

October 11, 2011
by glendon
glendon's Avatar

Hi everyone, thanks for all your kind responses!

I think I'm getting the picture now. So, here's what I have in my mind:

  1. The battery is still supplying the flow of electrons even in a reverse polarity scenario. To the battery, as long as the circuit is closed, the electrons will flow due to the voltage difference.

  2. As the gates we are using are n-channel and p-channel mosfets, they are virtual switches that turned on/off beyond a certain threshold current.

  3. When reverse polarity occurs, the flow of electrons (wrong direction) are causing the gates to turn on/off with the opposite intent, which could cause cascading errors in the circuit.

  4. To prevent that, diodes are used to act as a fail-safe. But usually the "reversed" current is too strong (or turn on for too long) which causes the diodes to heat up. The generated heat consequently damages the electronic component.

Am I in the ballpark on this?

October 11, 2011
by BobaMosfet
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bretm-

Thank you for the correction, yes, diode should be in series, not parallel.

glendon-

If you apply too much voltage in reverse, your force current to literally blow through (disintegrate) pathways in components that cannot withstand the force of the current in that direction. Current sees everything between it and where it wants to go as a resistance, irregardless of direction.

Components heat up because they are resisting the excess flow of current (simply by the chemistry and materials of which they are composed, and how those things are put together to control the flow of current). If that flow is too great, they melt (you will smell ozone when it happens). This can be slow, it can be fast, it can be explosive. And it has nothing to do with gates. The part is disintegrated.

It is not about time, it is about quantity. The quantity of current flowing through any part of a circuit is directly proportional to the voltage between two points of the circuit where that current is flowing.

Most components can handle different maximum quantities of current flowing through them based on direction. Exceed it in either direction, and the part fries. If it fries as a short, then other components can get damaged as well. Ever component has characteristic curves and power-dissipation curves which describe how the device is designed to behave based on voltage and current applied.

This is why using a diode in series with your input voltage is a good idea- it will usually allow enough current to flow in one direction to fully power your circuit, but will block it almost entirely in the other direction (up to some maximum, like 600- PRV - aka Peak Reverse Voltage).

BM

October 12, 2011
by glendon
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Thanks, BobaMosfet. I'm think I'm quite clear now.

By the way, the reason why I mentioned logic gates is because I had the impression that each electronic component is made up of different types of gates, arranged in a certain order to serve its purpose. That's what gives me the idea about cascading errors in the circuit when the current is reversed... :)

October 12, 2011
by BobaMosfet
BobaMosfet's Avatar

glendon,

Yes, they have gates, but it isn't 'cascading errors' that occur. The gates are literally melted and blown apart.

Current does not know or care what a gate is- it seeks all paths and views everything as an obstacle in its way.

BM

October 12, 2011
by bretm
bretm's Avatar

"The quantity of current flowing through any part of a circuit is directly proportional to the voltage between two points of the circuit where that current is flowing."

The current flowing through a forward-biased P-N junction actually increases approximately exponentially with voltage, not proportionally. That's why reversed polarity is so bad for semiconductor circuits. Proportional current-voltage relationship only applies to ohmic materials, like plain resistors.

October 12, 2011
by BobaMosfet
BobaMosfet's Avatar

bretm-

Don't confuse the voltage drop of the PN junction with the voltage actually 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. Usually the forward voltage is around 0.6V to 0.7V (silicon), or .2V (germanium), and the reverse voltage is smaller than breakdown voltage, which is much higher (PRV).

BM

October 12, 2011
by bretm
bretm's Avatar

I'm not confused. Transistors have p-n junctions, and the current flow is exponential, not linear. I,m not talking about amplification, just the basic I-V curve of almost any p-n junction. These are non-ohmic materials, so the current/voltage relationship is not a proportional one.

The forward voltage drop of a silicon junction is only 0.6 to 0.7 volts for a certain current range. If you apply a large voltage across the junction and the current is not limited, it will easily exceed that voltage range.

October 13, 2011
by BobaMosfet
BobaMosfet's Avatar

bretm-

What you described was a diode. I was responding to that specific description.

I agree that the transistor does not represent a linear proportion between E, I, and R (due to transconductance - see Ebers-Moll), but that was not entirely germane to the whole concept I was trying to express. Everything between the poles of a power-supply may be considered a form of resistance or impedance (Thievnin). If you apply enough difference in voltage between the poles, it doesn't matter what direction you go, the electron current will blow through everything like a shotgun through a target. It doesn't care.

All I was trying to express was that the OP should not think of reverse current as something finding an orderly way through the circuit, causing computational errors via cascade. It's a lot more brutal.

I find it easier sometimes to give someone a generalization to start with, to get their mind moving in a specific direction; then you can hang details, clauses, wherefores and whyfores on it later, when the whole thing isn't so much to grasp.

It's a little like me saying 'The sky is blue.' (which most people will agree upon) 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're absolutely right, but is it offered to be helpful, or just to be right?

BM

October 13, 2011
by BobaMosfet
BobaMosfet's Avatar

bretm-

Because you like accuracy, your pointed example is not 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 backwards:

 o----------o
 |          |
 -  -      ----
---         /\
 -         /  \  diode
--- +      ----
 |          |
 o----------o

Now the diode conducts current, and because the current flow in a forward-biased diode increases exponentially with the voltage, it overloads and melts.

:endquote

The first example works because the diode acts like an open circuit- no current flow (except for leakage). The second example does not work, NOT because the current flow increases exponentially with voltage, but because it's now a short circuit with nothing to resist the flow of current, so the power-supply will supply full current at full voltage (minus the small .7 drop through the diode) across the diode.

So, if that's a 9V battery: 9 - .7 = 8.3V at about .500mA (or 4.15A), which will toast the diode.

BM

October 13, 2011
by bretm
bretm's Avatar

I agree that "side comments" can sometimes distract from the original post, but this discussion board doesn't support threaded discussions, so I just go for it.

==== WARNING - THIS IS A SIDE DISCUSSION - DON'T GET CONFUSED ===

But otherwise I just don't think we're on the same page at all, and I guess that's OK.

My comments had nothing to do with transconductance.

"Everything between the poles of a power-supply may be considered a form of resistance or impedance (Thievnin)."

No, that's not Thevenin's theorem. Thevenin's theorem specifically only applies to linear circuits--resistors, and constant sources.

"If you apply enough difference in voltage between the poles, it doesn't matter what direction you go, the electron current will blow through everything like a shotgun through a target."

Absolutely true, but we're only talking about -9V here. That's not a breakdown voltage situation. -50V, sure. +50V also--polarity doesn't matter any more in that situation. For purposes of this thread, it's not why putting a 9V battery in backwards causes problems in semiconductor circuits.

-1V is enough for some semiconductor circuits because of the non-proportionality of the current flow of P-N junctions. Reversing the battery destroys semiconductors that weren't intended to be excessively forward-biased, such as the over-under-voltage protection diodes on the MCU's input pins. The reason you can't go past 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 does not work, NOT because the current flow increases exponentially with voltage, but because it's now a short circuit with nothing to resist the flow of current"

I completely disagree. It's not a short circuit. Put a 0.7V battery instead of a 9V battery and it wouldn't harm the diode. Why? Because it's not really a short-circuit. It's because of the exponential I-V curve. It's not true that "there's nothing to resist the flow of current". There's a diode in there to resist the flow of current. It does an admirable job of resisting the flow of current at 0.7V, and a poor job at 9V because of the exponential I-V curve.

"so the power-supply will supply full current at full voltage (minus the small .7 drop through the diode) across the diode. So, if that's a 9V battery: 9 - .7 = 8.3V "

It's not 8.3V. There's no 8.3V to measure anywhere in that circuit. If you find a 0.7V voltage drop in that circuit somewhere, please tell me where to measure it because I'd like to see it. (I almost walked right into "please tell me where to put the probes". Whew, that was close.)

Here's proof: Set up that circuit with an adjustable power supply instead of a battery, and 0.9V instead of 9V. Measure the voltage across the diode. It'll 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 some large value, like between 500mA and 1A for a 1N4001.

Now reduce the power supply to 0.45V and measure the voltage drop across the diode again. It will be 0.45V. Now measure the current. Will it be half the original amount, because the voltage is half of the original 0.9V and current is proportional to voltage? No, it will be some miniscule amount, because of the exponential I-V curve.

Now set the supply to 1V, if the diode and supply can take it. (Come on, it's just 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 one-ninth more than it was at 0.9V? No, because the I-V curve is non-linear, or because the diode is dead.

Myth: A sufficiently foward-biased silicon diode or a silicon base-emitter junction has a voltage drop of 0.6V to 0.7V.

Truth: The voltage drop will be 0.6V to 0.7V for a wide and typical range of current levels found in small-signal circuits because the exponential I-V curve is so steep at this range of currents. But whether or not it is actually within this range depends entirely on the current flow through the junction. If you apply 0.5V or 0.8V across it, the voltage drop will be 0.5V or 0.8V.

Rule of thumb: A sufficiently foward-biased silicon diode or a silicon base-emitter junction has a voltage drop of 0.6V to 0.7V.

There's another myth about zener diodes and breakdown voltages that I see a lot, but I guess there is a limit to how off-topic I should go.

October 13, 2011
by bretm
bretm's Avatar

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 it's bad to put the battery in backwards. I provide the correct explanation after some people erroneously bring up "breakdown voltage" as the explanation, but somehow I'm the problem.

October 13, 2011
by bretm
bretm's Avatar

According to an admittedly unreliable source called Wikipedia:

Breakdown voltage is a parameter of a diode that defines the largest reverse voltage that can be applied without causing an exponential increase in the current in the diode. As long as the current is limited, exceeding the breakdown voltage of a diode does no harm to the diode.

But I'm done with this thread.

October 13, 2011
by BobaMosfet
BobaMosfet's Avatar

bretm-

Instead of reading Wikipedia, try reading something like the 'Art of Electronics', or 'Understanding Circuits & Op-Amps', or MIT dissertations, as I do (as well as many other sources). There is a little more to breakdown voltage than what Wikipedia purports.

As the OP was discussing a battery (9V we all assumed), it was not germane to discuss sub-volt reverse voltages.

But, let's go back to the beginning. Forget everything up until now, and let's take and put a 9V battery in reverse polarity on an arbitrary diode, like an LED for example. Hmmmm.... blew it apart (I just did it).

I think it's breakdown voltage was exceeded.

BM

October 13, 2011
by bretm
bretm's Avatar

That's awesome. I hope you had fun. I just did it with a silicon diode and it didn't blow apart. It's my understanding that the Atmega168 uses silicon, and contains no LEDs.

But you know what, you're absolutely right. I was talking about the Nerdkit this whole time when the OP wasn't actually talking about the Nerdkit at all. Still, the insult was out of line. I actually love being wrong. It's the only time I learn anything.

October 13, 2011
by BobaMosfet
BobaMosfet's Avatar

bretm-

There was no insult given. Only an observation specific to you. It was not mean, or unfavorable. It was very true. You are frequently the most prolific poster here. Sometimes however, someone needs to get used to the forest first, before another points out the trees, the bushes, the creatures, etc.

The ATMEGA168's gates do not have the same protection level as an individual silicon diode, that's why I used an SMT LED- it's a substrate device. Your silicon diode likely has a PRV (aka breakdown) rating probably at 40, 100, 400, or 600 Volts- I don't know since you didn't provide the number.

On the bright side, however, most devices like the ATMEGA today have protection diodes, which (within limits) dump some excess voltage (in either polarity) to the rails. In fact, that's one of the tricks I use to power an ATMEGA in special circumstances where I can't connect VCC or GND.

No hard feelings,

BM

October 14, 2011
by Ralphxyz
Ralphxyz's Avatar

Well I want to thank glendon for asking this question, what a interesting discussion.

Also of course thanks bretm and BobaMosfet you guys just blow me away.

It's to bad this forum does not support threaded discussions or even [quote] for that matter.

Actually the Markdown Syntax does support [blockquotes]

Blockquotes can contain other Markdown elements, including headers, lists, and code blocks:

> ## This is a header.
> 
> 1.   This is the first list item.
> 2.   This is the second list item.
> 
> Here's some example code:
> 
>     return shell_exec("echo $input | $markdown_script");

But if you want a different background color to highlight the quote someone has to set that up.

Speaking of getting off subject geesch!!

I really wanted to thank bretm and BM.

Ralph

October 24, 2011
by Hexorg
Hexorg's Avatar

I see the question was already answered, I just wanted to pitch in a little.

glendon, thing of the electrical current through a wire as of liquid current through a pipe. If you put a water current through a nozzle, no matter which way the water flows, it'll work. But now imagine a heart valve that opens/closes at a certain frequency. If blood flows in the direction required bu the valve, it'll work properly. But if you force the blood in the opposite direction, if will not only prevent the valve from working, but may tear it apart.

P.S. A differential potential (Voltage) is just like a pressure in the liquid. A negative voltage means the electrons simply flow in the opposite way.

October 24, 2011
by Hexorg
Hexorg's Avatar

*think

October 24, 2011
by glendon
glendon's Avatar

Thanks, Hexorg!

Also big thanks to bretm and BM who brought a spirited discussion to the topic! :)

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