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### Topic: How does the diode protect the transistor in this circuit? (Read 2712 times)previous topic - next topic

#### BorislavLukanov

#15
##### Jan 09, 2018, 09:07 am
is not water to put a nozzle on it or to divert the flow in another pipe : D

#### Grumpy_Mike

#16
##### Jan 09, 2018, 11:59 amLast Edit: Jan 09, 2018, 12:16 pm by Grumpy_Mike
is not water to put a nozzle on it or to divert the flow in another pipe : D
Well in this case it actually is because you do not have a linear circuit but a circuit with a threshold where no current will flow until that threshold is reached. If component 1 has a threshold of 1V and component 2 has a threshold of 2V and they are wired in parallel, no current will ever flow through component 2 because component 1 clamps the applied voltage to its own threshold.

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are you sure that an inductor generates voltage while is applied voltage to it?
No one in this thread ever suggested such a thing. However when a voltage is removed an other voltage of opposite polarity is generated.

#### ba58smith

#17
##### Jan 09, 2018, 01:30 pm
the electricity doesn't CHOOSE where to pass ... ! It follows precise lows of physics!
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.
Brian Smith
Arduino Newbie 2015

#### BorislavLukanov

#18
##### Jan 10, 2018, 07:37 am
You may get distracted by shit : D and also is a little bit stupid to comment, but I'm not an expert or sure so I would ask if the positive side of an inductor with accumulated magnetic field on it and no external voltage applied could interact with the ground to make possible the circuit to be closed even if the switches are?

#### Smajdalf

#19
##### Jan 10, 2018, 09:50 am
These days a Schottky diode can have an even lower voltage of between 0.15 to 0.45V depending on the current and construction. https://en.wikipedia.org/wiki/Schottky_diode So these are sometimes used for a quicker quenching of the back EMF.
If I understand this right you claim Schottky diodes are used because they cause the field in the coil to collapse faster. But I believe the converse is true - since the forward voltage drop is lower the current decreases slower and so it takes longer for the field collapses (the current stops flowing). This is an advantage when you are PWMing the coil. Another advantage of Schottky vs. "normal" diode is shorter reverse recovery time - but I think it applies only when you want to turn on the inductive load again.

#### Grumpy_Mike

#20
##### Jan 10, 2018, 09:55 amLast Edit: Jan 10, 2018, 09:56 am by Grumpy_Mike
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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.

#### Smajdalf

#21
##### Jan 10, 2018, 10:23 am
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.

#### Grumpy_Mike

#22
##### Jan 10, 2018, 07:19 pmLast Edit: Jan 10, 2018, 08:09 pm by Grumpy_Mike
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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.

#### Smajdalf

#23
##### Jan 10, 2018, 09:19 pm
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"?

#### polymorph

#24
##### Jan 10, 2018, 11:24 pm
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

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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.

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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.
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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 Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8
gammon.com.au/blink - gammon.com.au/serial - gammon.com.au/interrupts

#### Grumpy_Mike

#25
##### Jan 11, 2018, 10:41 amLast Edit: Jan 11, 2018, 10:43 am by Grumpy_Mike
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.

#### Smajdalf

#26
##### Jan 11, 2018, 12:15 pmLast Edit: Jan 11, 2018, 12:54 pm by Smajdalf
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.

#### Grumpy_Mike

#27
##### Jan 11, 2018, 01:25 pmLast Edit: Jan 11, 2018, 01:27 pm by Grumpy_Mike
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.

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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?

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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.

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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.

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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.

#### Smajdalf

#28
##### Jan 11, 2018, 03:07 pm
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.
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.

#### ElCaron

#29
##### Jan 11, 2018, 04:20 pm
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.

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