How does the diode protect the transistor in this circuit?

since the forward voltage drop is lower the current decreases slower

No not right. Think about what happens at each limit, with your thinking if the voltage drop on the reverse diode was zero would the discharge rate be slower? If the voltage drop were 100V would it be faster? See it doesn’t make sense.

Grumpy_Mike: No not right. Think about what happens at each limit, with your thinking if the voltage drop on the reverse diode was zero would the discharge rate be slower? If the voltage drop were 100V would it be faster? See it doesn’t make sense.

Sure, if the voltage drop were zero and resistance of the coil were zero (superconductor) the current would flow "forever". If you use an air gap instead of the diode a very short spark will be produced - the field collapse "immediately" all energy used to charring contacts of the air gap.

If you use an air gap instead of the diode a very short spark will be produced -

Yes that is because a plasma is not a linear resistance. Once a plasma is established it's impedance goes from being very high to very very low. In fact almost like a short circuit.

Lets face it you are often wrong on things you say here, this is no exception.

I'll tell you what, you have a good google and find some references to uphold your proposition.

Grumpy_Mike: Yes that is because a plasma is not a linear resistance. Once a plasma is established it's impedance goes from being very high to very very low. In fact almost like a short circuit.

YOU are missing basics of basics of electronics.

Maybe you know P=VI. If resistance of an electric arc is "like a short circuit" it means V ~ 0 and then P ~ 0. Where do you get the power for welding and cutting? Or more humble task: where do you get the power to keep the air hot and ionized? There is surely a lot of power needed to keep the arc conducting so there must be considerable voltage drop somewhere in the arc. (Maybe electrodes/air junction?)

Also you may know for ideal inductor dI/dt=V holds. It means when voltage drop is small change of the current is slow.

And how you explain loop of superconductor can keep current "forever"?

Sorry, it is correct that the lower the forward drop on a flyback diode, the longer it takes for current to decay. Power is used up as VI, so if V is lower, power is lower, it takes longer to dissipate the energy of the collapsing magnetic field.

It can be a real problem with relays dropping out too slowly and burning the contacts. So a resistor may be added in series with the protection diode. The transistor has to withstand a higher peak voltage, but the field collapses faster.

http://sound.whsites.net/articles/relays.htm#s4

The second trace shows the release time and voltage spike when a diode and 270 ohm resistor are used to get a higher release speed.

If you search hard enough, you will find it mentioned in a few places, and it's been pointed out [ 8 ] that simply using a diode can cause the relay to release too slowly to break 'tack welding' that can occur if the contacts have to make with high inrush currents. This can happen because the armature's physical movement is slowed down, and it doesn't develop enough sudden force to break a weld. It's far more complex than just an additional delay when a diode is placed in parallel with the coil.

The zener diode scheme shown above may be a bit more expensive than a resistor, but it allows the relay to deactivate much faster. The most common arrangement will be to use a zener rated for the same voltage as the relay's coil and supply. In the example, the release time was 2.6ms, and that's significantly faster than obtained using a resistor and diode (4ms). A higher voltage zener will be faster again, with a 24V zener giving a drop-out time of 1.84ms.

Higher voltage, faster current decay.

Steve, you are spot on with those circuits, yes that is what they do, however your analysis is missing something.

A capacitor and an inductor both store energy. So let's suppose you had a charged capacitor for a moment. What would be the quickest way to discharge that capacitor, with a 100K resistor or a 1K resistor? I think you would agree that a 1K resistor would discharge it quickest would you not?

The voltage on the capacitor would discharge in an exponential curve who's characteristics are that the most rapid rate of change occurs at the start and the rate of change in voltage slows down as the capacitor becomes increasingly discharged. Lets say the capacitor's initial charge was 40V but you are interested in making the transition from 5V to 1V as quick as possible and you were not worried about the time it took to reach 5V. So one strategy would be to discharge the capacitor until it reached 5V with a resistor and then short out that resistor. Of course in the real world there is no such thing as a short only varying degrees of small resistor. So let's modify that by saying discharge with a 10K resistor and when 5V has been reached add an extra 1 ohm resistor. The newly introduced resistor would introduce a new exponential curve in the rate of change in voltage resulting in a more rapid transition between the two voltages you are interested in.

Now a capacitor and inductor are very similar in that they both store energy and in both cases the way to remove that energy is to put as bigger load on it. However with a relay you have a pull in current and a smaller hold current. This corresponds to two strengths of magnetic field. What you want to do to make the contacts break quickly is to minimise the time it takes to go from the hold magnetic field to zero. So one way to do this, just like the capacitor, is to drain the energy ( magnetic field ) from the inductor slowly so that it stays above the holding field for as long as possible. Then when the field drops below the holding field their is very little energy left in it and will quickly dissipate.

So by introducing a resistor in series with the diode you are reducing the discharge rate of the inductor's field and making it hold up longer so that when it drops their is little energy left to keep the field close to the holding field thus making the contacts open more quickly. You care little about the overall discharge time because a relay is slow anyway. So you delay or make longer the discharge time of the inductor's energy so that the drop off at the end is quicker.

Grumpy Mike you missed one important fact: inductor is NOT capacitor. In fact inductor is mostly presented as "inverse" of capacitor because their behaviour is "opposite". Capacitor tries to keep voltage constant and sources (or sinks) "any" current needed to do so until it runs out of energy. So you discharge it via a small resistor to force it to provide large current. OTOH an inductor tries to keep current constant and generates "any" voltage to do so. So when you short circuit the inductor it does nothing - the current flows in loop unchanged, inductor "is happy" and generates no voltage so wasting no power. But when you connect it via a large resistor it needs to generate large voltage to force the current to flow - loosing energy quickly. (Ofc since real inductors have non-neligible resistance the current does not keep flowing for long - unless you make everything from a superconducting material. It IS used "commonly" to make super strong magnets - such as MRI - or power storage.) TLDR: your previous post is pointless because an inductor is NOT a capacitor

@polymorph: interesting link, thanks

EDIT: to make it crystal clear I will do some quick calculation for you. Let assume the inductor and diodes are ideal, meaning for the inductor holds dI/dt = V/L and for diode its forward voltage drop Vf is constant for any forward current, zero when no current flows and infinite reverse breakdown voltage. Assume the inductor has inductance 1H and some current I(0) flows through it (it is charged). Suddenly we turn off the path that was charging the inductor and the current have to flow through the flyback diode. How long it will take the current reaches zero? We know V and L is constant so it is easy to "solve" the differential equation above: dI/dt = -Vf I(t) - I(0) = - Vf.t I(t) = I(0) - Vf.t Since we are interested in time needed for the current to stop we need to find such t that I(t)=0. It is easy: 0=I(0) - Vf.t t=I(0)/Vf Maybe even Grumpy_Mike now sees t is longer when Vf (forward drop of the diode) is smaller ;-)

BTW Grumpy_Mike you say you are (used to be?) a professional. If you don't understand this topic you may try to simulate it with Spice and/or make a circuit and look at it with an oscilloscope. I am just a hobbyist so I don't know how to use Spice and don't have access to an oscilloscope so I am unable to do it for you.

Smajdalf: YOU are missing basics of basics of electronics.

Maybe you know P=VI. If resistance of an electric arc is "like a short circuit" it means V ~ 0 and then P ~ 0. Where do you get the power for welding and cutting? Or more humble task: where do you get the power to keep the air hot and ionized? There is surely a lot of power needed to keep the arc conducting so there must be considerable voltage drop somewhere in the arc. (Maybe electrodes/air junction?)

Also you may know for ideal inductor dI/dt=V holds. It means when voltage drop is small change of the current is slow.

And how you explain loop of superconductor can keep current "forever"?

That is the most ignorant stupid statement ever made on this forum from someone with over 500 posts.

Where do you get the power for welding and cutting?

From the current you idiot. Or are you saying that a plasma has a high impedance? Think florescent lights. Unbelievable that you have things so wrong. If a plasma has a high impedance then how the hell do you think a neon relaxation oscillator works?

Grumpy Mike you missed one important fact: inductor is NOT capacitor.

In the respect that it stores energy then yes it is the same. An inductor is an "upside down" capacitor.

I am just a hobbyist so I don't know how to use Spice and don't have access to an oscilloscope so I am unable to do it for you.

You are unable to do it precisely because you do not know what you are talking about.

to make it crystal clear I will do some quick calculation for you.

Yes you have made it crystal clear that you can't do maths either.

Only insults? Where is my math wrong? I KNOW it is not 100% correct because it would be much longer and for readability it would need more than a plain text but I tried to make the idea understandable.

Grumpy_Mike: From the current you idiot.

You claim plasma = short circuit = resistance 0 Ohm. V=I.R=I.0=0 P=I.V=I.0=0 Your "short circuit" plasma generates no power. Or you must admit there is some voltage drop over the plasma. What is this drop? It may be only a few volts but it is a lot compared to a diode forward drop.

Please, make a Spice simulation to prove I am wrong. It must be a few seconds work for someone as skilled and experienced as you are.

Smajdalf: Please, make a Spice simulation to prove I am wrong. It must be a few seconds work for someone as skilled and experienced as you are.

http://www.falstad.com/circuit/ may be enough for you.

but I tried to make the idea understandable.

It was the idea that was wrong, so no amount of maths manipulation is going to correct that.

You claim plasma = short circuit = resistance 0 Ohm.

No, can't you even read either. I said:-

Once a plasma is established it's impedance goes from being very high to very very low. In fact almost like a short circuit.

See that word "almost" in there? Ever tried googling "almost"? It says:-

almost adverb: almost

not quite; very nearly. "he almost knocked Georgina over" synonyms: nearly, just about, about, more or less, practically, virtually, all but, as good as, next to, close to, near, nigh on, not far from, not far off, to all intents and purposes, approaching, bordering on, verging on, nearing;

From This date booklet from a major manufacturer of gas discharge tubes.

During the glow region the increase of current flow will create an avalanche effect in gas ionization that will transition the GDT into a virtually short circuit mode and current (dependent on the impedance of the voltage source) will pass between the two conductors. The voltage developed across the GDT with a short circuit condition is called the “Arc Voltage” VARC.

See how they use the word virtually which is a synonym of my word almost?

Your "short circuit" plasma generates no power.

Plasmas do not generate power, they carry the power from the current across the tube. Voltage is not an issue, you can not use the W = IV relationship in any meaningful way and you have NO control over the V figure.

inductor is NOT capacitor. In fact inductor is mostly presented as "inverse" of capacitor because their behaviour is "opposite". Capacitor tries to keep voltage constant and sources (or sinks) "any" current needed to do so until it runs out of energy. So when you short circuit the inductor it does nothing - the current flows in loop unchanged, inductor "is happy" and generates no voltage so wasting no power.

So look what you said here "when you short circuit the inductor it does nothing - the current flows in loop unchanged" That is total and absolute rubbish. Both a capacitor and inductor store energy. A capacitor stores it as electrons gathering on the plates and an inductor stores energy as a magnetic field. When you short out an inductor you remove the current that keeps up the magnetic field and it collapses. I think you have a very shaky idea about superconductors. In the real world at real temperatures the wire forming a coil has a resistance. When current goes round this, the current causes heat to be generated as it flows through the coil and this takes energy out of the magnetic field which is busy collapsing anyway. The collapsing field causes a voltage to be generated in the reverse polarity of the voltage that caused the field. This is basic physics.

Only insults?

You started it.

I am sorry I got dragged into discussion about "impedance of plasma". I know little about it and a quick Google search did not help me. You may be right on this topic but in either case it is only distracting us from the main topic of this (hijaked - sorry OP) discussion. I really want to have this settled. PLEASE let's stand back, forget insults and start again from basics. I claim when (ideal) inductor with inductance L is being discharged via "ideal" diode (meaning it has constant forward drop Vf at any forward current) the following holds: I(t) = I(0) - t.Vf/L for any t greater than 0 and lesser than T where T = I(0).L/Vf meaning I(T) = 0. Ofc I(t) = 0 for each t > T.

You have said those equations are rubbish. Please show your equations and I will either show you where yours defy fundamental laws of physics or we find we are speaking about different scenario.

You may be right on this topic

What do you mean "may be"? Are you implying their is any doubt that Bourns do not know what they are talking about?

I claim when (ideal) inductor

Stop right their, their is no ideal inductor here so that makes a nonsense of everything that follows.

the following holds: I(t) = I(0) - t.Vf/L

Can you say in words what you mean by that, it sounds wrong. The dimensions of that relationship do not add up. Time multiplied by voltage divided by inductance does not have the units of current.

The dimensions add up perfectly (A=V/H*s), of course it makes sense to look at an ideal inductor to discuss a principle and I would like to know your version, too.

Grumpy_Mike: What do you mean "may be"? Are you implying their is any doubt that Bourns do not know what they are talking about?

No. I am implying maybe you don't know what they are talking about. But first things first - let's settle the inductor issue and then we can talk about this.

Grumpy_Mike: Stop right their, their is no ideal inductor here so that makes a nonsense of everything that follows.

Let's follow standard scientific procedure - start with idealized components which are easy to understand. When we both agree on "ideal" behavior we may step further and try to determine how parasitic properties of real parts influence this behavior.

Grumpy_Mike: The dimensions of that relationship do not add up. Time multiplied by voltage divided by inductance does not have the units of current.

Inductance is measured in henry (H). Wiki claims H=V.s/A; time is measured in seconds (s), voltage in volts (V) and current in ampere (A). s.V/H = s.V/(V.s/A) = A which contradicts you. What units "time multiplied by voltage divided by inductance" have if not ampere?

Grumpy_Mike: Can you say in words what you mean by that, it sounds wrong. The dimensions of that relationship do not add up. Time multiplied by voltage divided by inductance does not have the units of current.

Since the definition of inductance is given by the differential equation. Vdt = LdI, clearly you're wrong!

I think you are thinking of inductive reactance, not inductance.

ba58smith: One could argue that no one ever chooses anything - that everything is pre-determined, and we're just playing out a script. And yet, we still use the word "choose", because it simplifies the discussion. That's what I chose to do in this case - simplify the discussion. But then, I think you knew that.

sorry to diverge from the inductance debate :P

to bring the water analogy back into it; this is how i understand the "diode protection" happening; imagine two pipes connected to a water source (5m up from the ground) - one runs off into a basin 4m up from the ground, while another runs off straight onto the ground. a trickle would probably reach the basin, but most of the water would flow through the pipe down to the ground.

am i correct in understanding this analogy of electricity/water flow ?

BabyGeezer: sorry to diverge from the inductance debate :P

to bring the water analogy back into it; this is how i understand the "diode protection" happening; imagine two pipes connected to a water source (5m up from the ground) - one runs off into a basin 4m up from the ground, while another runs off straight onto the ground. a trickle would probably reach the basin, but most of the water would flow through the pipe down to the ground.

am i correct in understanding this analogy of electricity/water flow ?

1) Water source 5m up from ground is a good analogy for charged capacitor or chemical battery. Not for an inductor. Analogy for an inductor is possibly some rotaty pump or peristaltic pump with small friction (= resistance of the inductor) and large mass (= inductance). If the pump has no external power to do the pumping it is analogy for an inductor. If the pump is powered it may be considered as a current source. 2) Magnitude of the water flow depends not only on potential difference but on width of the tubes (= resistance). If you have thick tube to the 4m basin you will get higher current flow than very narrow tube leading into the 0m basin.

Wow... Grumpy Mike, when you are wrong, you really double down.

These examples use zener diodes because the effect of current decay is more pronounced than with a resistor.

Switching Inductive Loads with Safe Demagnetization - Application Note - Maxim

The diode must be able to handle the initial current at turnoff, which equals the steady-state current flowing through the inductor when the switch is closed. In addition, the voltage rating for the diode needs to handle the swing between positive- and negative-voltage levels. A rule of thumb is to select a diode rated for at least the amount of current the inductor coil draws and at least twice the voltage rating of the operating voltage of the load. For many applications, especially those found in industrial applications that have many output channels per IO card, this diode is often physically quite large and adds significant extra cost to the BOM.

The other major disadvantage of the simple freewheel diode approach is that it lengthens the decay of current through the inductor. As explained in "Coil Suppression Can Reduce Relay Life," this slow decay of current can create problems such as "sticking" between relay contacts. For applications where the current must decay quicker, an alternative solution is to use a Zener diode as shown in Figure 4, which gives a faster current ramp rather than an exponential decay. When the switch opens, the current is shunted through the general-purpose diode and Zener diode path, maintaining a voltage equal to the Zener voltage (plus forward diode drop) until the inductor energy is dissipated.

Coil Suppression Can Reduce Relay Life

This diode shunt provides maximum protection to the solid state switch, but may have very adverse effects on the switching capability of the relay. It is important to realize that the net force available to cause the armature to open is the difference between the magnetic restraining forces and the spring opening forces, that each of these is varying in a manner to cause the net force to vary both with time and armature position. It is this net force which gives rise to the armature system velocity and energy of momentum as it attempts to effect armature and contact spring transfer.

A slowly decaying magnetic flux (the slowest is experienced with a simple diode shunt across the coil) means the least net force integral available to accelerate the armature open. In fact, rapid loss of the opening forces supplied by stiff NO contact springs, coupled with slowly decaying magnetic forces, can actually cause a period of net force reversal where the armature velocity is slowed, stopped, or even momentarily reversed until the flux further decays, finally permitting available spring “return” forces to cause transfer to continue.

...

The more rapidly the coil current decays, the less the magnetic hold back, and thus the greater the armature momentum and contact stick “break-ability.”

Obviously, this is optimized when no suppression is used. However, near optimum decay rate can be obtained by using a Zener diode in series with a general purpose diode. When the coil source is interrupted, the coil current is shunted through this series arrangement, maintaining a voltage equal to the Zener voltage (plus forward diode drop) until the coil energy is dissipated. This is illustrated in Fig. 3.

The Zener voltage value is chosen to limit the coil switch voltage to a level acceptable to the switch rating. This affords the best compromise both to coil switch protection and relay switching performance, and should be employed to assure maximum relay performance and reliability while providing protection to the control circuit from coil induced voltages.

Inductive Loads and Diode Protection

The best solution is to put a diode across the inductor, as shown at left. The diode must be able to handle the initial diode current, which equals the steady current that had been flowing through the inductor; something like a 1N4004 is fine for many cases.

When the switch is on, the diode is back-biased (from the dc drop across the inductor's winding resistance). At turn-off the diode goes into conduction, putting the switch terminal a diode drop above the positive supply voltage. The only disadvantage of this protection circuit is that it lengthens the decay of current through the inductor, since the rate of change of inductor current is proportional to the voltage across it.

For applications where the current must decay quickly (high-speed impact printers, high-speed relays, etc.), it may be better to put a resistor across the inductor, choosing its value so that Vsupply + IR is less than the maximum allowed voltage across the switch. For fastest decay with a given maximum voltage, a zener could be used instead, giving a ramp-down of current rather than an exponential decay.

Source: "The Art of Electronics" by Horowitz and Hill, Cambridge University Press, copyright 1980, ISBN 0-521-23151-5

Grumpy_Mike, you said I am an idiot and what I am telling is rubbish. Yet you didn't manage to bring anything to support your claims so far. I and other forum members brought evidence you are wrong. I expect you either present something to explain yourself or you apologize. Or are you so coward you will ignore this thread and pretend it never happened?