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Posted: **Fri Mar 14, 2008 1:54 pm**

by **zmwworm**

dyarker wrote:For that relay, in a vehicle, a 1N4001 (1A, 50V PIV) should work fine. As will 1N4002 (1A, 100V PIV) or 1N4003(1A, 200V PIV).

If using solid state relay driver, a faster diode would be better. Like 1N4148 for coil currents 150mA or less. A 1N4148 would work electrically for this project too, but is not as physically rugged as 1N400x diodes.

I bought a [url=httphttp://

www.radioshack.com/product/index.jsp?pr ... age=search]50 pack of 1N4148s[/url] a while ago, and like I said the relay draws 133 ma, so I'll probably just use them. Although I would definitely go with the 1N400x if I had them on hand, the 1N4148 seems so fragile (looks like glass). What if I used two in parallel for a little more fail-safe design?

Posted: **Fri Mar 14, 2008 2:13 pm**

by **Bigglez**

Greetings Zach,

zmwworm wrote: I bought a 50 pack of 1N4148s a while ago, and like I said the relay draws 133 ma, so I'll probably just use them.

The current in the back-emf diode is unrelated to the coil

holding current (which is limited by the applied voltage

and DC resistance of the coil).

When the coil is de-energized all of the energy stored in

the coil must go somewhere... The diode conducts all of it,

which might be a much higher current (but for a shorter time)

until the energy is removed (as heat in the diode).

zmwworm wrote:The 1N4148 seems so fragile (looks like glass). What if I used two in parallel for a little more fail-safe design?

They are glass, which is a poor conductor of heat. Rectifier

diodes have thicker wires to remove the heat from the junction.

Placing diodes in parallel is hit and miss unless the diodes

are matched. Random selection would result in the lower

forward voltage drop diode conducting the greater share

of current, and the other diode(s) not conducting much or

any current.

If multiple diodes are to be used small ballast resistors

are required in each one to equalize the total current.

Comments Welcome!

Posted: **Fri Mar 14, 2008 2:20 pm**

by **Robert Reed**

Zmwworm

All most all diodes can carry a minimum of ten times their rated current for a short period - several milliseconds - and some over a hundred times rated current. For the extremely low duty cycles you will encounter almost any diode will work. just beware of the peak inverse voltage that the diode will see. Dyarkers suggestion will be fine, allthogh i would tend to go with a minimum of 400 volts on the rating. 1N4004 0r 6.

Posted: **Fri Mar 14, 2008 2:27 pm**

by **Bigglez**

Greetings Robert,

Robert Reed wrote:just beware of the peak inverse voltage that the diode will see. Dyarkers suggestion will be fine, allthogh i would tend to go with a minimum of 400 volts on the rating. 1N4004 0r 6.

The PIV (peak Inverse Voltage) for a back-emf diode

application is equal to the pull-in voltage of the relay coil.

Why waste a high voltage (and more costly) diode when

only 10 - 16V PIV will be encountered? A 1N4001 (50V

PIV) will suffice.

Comments Welcome!

Posted: **Fri Mar 14, 2008 3:43 pm**

by **Robert Reed**

Fairchild 1N4001 - $0.05

Fairchild 1N4006 - $0.05

Typical for all manufacturers

Posted: **Fri Mar 14, 2008 3:46 pm**

by **dyarker**

Bigglez,

The current in the back-emf diode is unrelated to the coil

holding current (which is limited by the applied voltage

and DC resistance of the coil).

The peak current through the diode will be exactly the holding current of the coil at the moment the supply is interupted, and decreases from that instant on. The amount of energy stored in the coil determines how long it takes for the current to decrease to zero, not the peak current. In fact, to make the current decrease faster, put a resistor in series with diode. The resistor increases the power the coil must provide while trying to maintain the current.

zmwworm,

the 1N4148 seems so fragile (looks like glass). What if I used two in parallel for a little more fail-safe design?

That is why I recommended 1N400x for use in a vehicle. Especially because they probably won't be mounted on a PCB, but point to point on the relay terminals. Right?

Cheers,

Posted: **Fri Mar 14, 2008 3:48 pm**

by **MrAl**

Hi again,

Here are some waveforms for a coil that is energized for a long time

and then suddenly deenergized using a switch.

Without a catch diode the voltage jumps to -13kv with an ideal coil

but only lasts for a short time period.

With a catch diode the voltage is clamped to about -0.7 volts and stays

near that level for about 30us for every mH of coil inductance as the

current through the coil decreases exponentially. During the discharge,

the current is always equal to or less than the original coil current while

energized. Also during the discharge the voltage never goes below

-0.8 volts or so. While the coil is energized the reverse voltage across

the diode is 14v (or a little higher for a 14.4 voltage source or course).

I can't say i would trust a 1N4148 diode for this application, but i

would certainly use a 1N4001 diode and these kind are used all

the time for this purpose.

Posted: **Fri Mar 14, 2008 7:20 pm**

by **zmwworm**

dyarker wrote:That is why I recommended 1N400x for use in a vehicle. Especially because they probably won't be mounted on a PCB, but point to point on the relay terminals. Right?

Actually, because it's already been constructed and crammed under the dash, they'll probably be point to point between wires going to the relays after the resistors (which are near the switches in front). I knew that the tiny diode leads would not be ideal, but I figured with enough electrical tape and shrink tubing I could keep them from breaking.

Bigglez wrote:They are glass, which is a poor conductor of heat. Rectifier

diodes have thicker wires to remove the heat from the junction.

MrAl wrote:I can't say i would trust a 1N4148 diode for this application, but i

would certainly use a 1N4001 diode and these kind are used all

the time for this purpose.

Okay, for all of the above reasons I'll just run to Radioshack and buy

these.

dyarker wrote:In fact, to make the current decrease faster, put a resistor in series with diode. The resistor increases the power the coil must provide while trying to maintain the current.

Would this be something I might want to do (considering I have some pretty weak switches (rated 3 amps at 125 AC))? Or will I be safe with just the diodes?

Posted: **Fri Mar 14, 2008 8:41 pm**

by **dyarker**

Or will I be safe with just the diodes?

Yes, the diode protects the switches from arcing.

Posted: **Fri Mar 14, 2008 9:19 pm**

by **Bigglez**

Greetings Dale,

dyarker wrote:The peak current through the diode will be exactly the holding current of the coil at the moment the supply is interupted, and decreases from that instant on.

Agreed, I stand corrected.

dyarker wrote: In fact, to make the current decrease faster, put a resistor in series with diode.

This is counter-intuitive. The lower the loop impedance

during discharge, the

*faster* the energy will be drained

from the coil. VL(t) = V e -t(L/R) seconds.

dyarker wrote:The resistor increases the power the coil must provide while trying to maintain the current.

The amount of energy (power) stored in the coil inductance

is fixed.

If the loop impedance is increased (by adding a resistor

in series with the back-emf diode) the voltage across

the coil during discharge will increase proportionally

to maintain the coil current.

Comments Welcome!

Posted: **Fri Mar 14, 2008 10:19 pm**

by **dyarker**

Yes, it is counter-intuitive.

Yes, the stored energy if fixed.

1N400x diodes have a max forward drop of 1.1V. 133mA is a light load for them, so let's call it 0.8V.

The internal resistance of the coil is the same whether there is a resistor in series with the diode or not. So ignore it for this explanation (though it must be included to calculate actual drop-out time.

Using the relay being discussed.

With only diode, the power dissipated outside the coil is:

0.8V * 0.133A = 0.1064W initially

With diode and 100 Ohm (for example) resistor:

100 Ohm * 0.133A = 13.3V

13.3V * 0.133A = 1.7689W

Power dissipated is: 1.7689W + 0.1064W = 1.8753W initially

Peak reverse voltage: 0.8V + 13.3V = 14.1V

Fixed energy available and fixed initial current, the resistor causes higher voltage. Higher voltage at a current is higher power dissipated. The higher the power, the faster energy is used up. (energy = power times time. ie KiloWattHours or WattSeconds which can be converted joules or calories or whatever units you prefer)

Still counter-intuitive to me too. That's just the way it works out.

Cheers,

Posted: **Sat Mar 15, 2008 2:23 am**

by **MrAl**

Hi again,

Yes, the dissipation of energy in an inductor is a little counter intuitive

because we are probably more familiar with voltages and dissipating

energy from a voltage source than dissipating energy that comes from

a current source. The cap is discharged faster with a low resistance

because it acts like a temporary voltage source, while the inductor

is discharged faster using a high resistance because it acts like

a temporary current source, and also note that a short circuit to

a pure current source results in zero power dissipation, much unlike

what happens with a pure voltage source, which needs an infinite

resistance in order to prevent power dissipation. What would help

us is to take an in depth look at how current sources work as opposed

to the more familiar voltage sources like batteries and wall warts.

Anyway, there are some interesting equations that come from looking

at the discharge of an inductor with various voltage drops.

The energy in an inductance can be dissipated faster by allowing

the voltage across the coil to go higher than only one diode drop.

This happens automatically anyway, because of the windings series

resistance.

The familiar equation for the energy in an inductor is:

W=0.5*i^2*L

but another form can be developed by simply looking at a circuit

with a voltage source and inductor in the frequency domain,

solving for the current, taking the inverse Laplace Transform,

squaring the result, and finally solving for time t. The result

of this effort provides the following form:

t=L*sqrt(2*W)/E

where

t is time

L is inductance

W is energy

E is the voltage across the coil

Since we know the energy W by knowing the energized current I,

we can calculate W from that and stick it into this new equation

and solve for t, or we could simply note that E is in the denominator

of the equation for the time t, so that increasing E always decreases

the time t.

Another way of visualizing this is to think of the energy in the

inductor in terms of "volt seconds" rather than the formula W=0.5*I^2*L.

A higher voltage means more energy, either stored or dissipated.

If the energized current is 140ma and the winding

has 100 ohms resistance, then the starting discharge voltage would

be 100 times 0.14 plus the diode drop. The ending voltage

would be approximately one diode drop. Thus, the average voltage

while the coil is discharging is roughly Vcc/2.

This also gives us a starting point to calculate a rough time period of

the discharge. If we approximate the discharge voltage by E=Vcc/2

we end up with the following formula:

t=2*L*sqrt(2*W)/Vcc

where

W=0.5*I^2*L

where

I is the energized current.

Adding an extra resistance in series with the diode will definitely cause

the coil to discharge faster, but it also gives rise to a higher reverse

voltage which could destroy a device like a transistor if one is used as

the driver for the relay. Note that the coils internal resistance does

not cause this problem because the diode is across the whole coil

which includes the series resistance.

For a graphical example, please refer back to my previous post

where the discharge voltage was as high as 13kv (13,000 volts)

and as low as the diode drop. With the voltage up to 13kv the

discharge time is extremely rapid, while with the diode it takes

much longer to discharge.

Posted: **Mon Mar 17, 2008 12:58 am**

by **Bigglez**

Greetings Mrai,

MrAl wrote: Yes, the dissipation of energy in an inductor is a little counter intuitive because we are probably more familiar with voltages and dissipating energy from a voltage source than dissipating energy that comes from a current source.

Great! Thank you (and Dale in the earlier post) for

the explanation.

MrAl wrote: For a graphical example, please refer back to my previous post where the discharge voltage was as high as 13kv (13,000 volts)

and as low as the diode drop.

I did SPICE a similar circuit to see what would

happen, but you beat me to posting results.

For the switch I used a PMOSFet and a Schottky

power diode, while experimenting with the

series resistor value. As we don't know the

inductance of the actual relay I used 1mH.

We've probably thrashed this topic to death,

and the outcome is more involved than I

would have thought (not sure about the OP,

who just wanted guidance on a vehicle mod).

Comments Welcome!

Posted: **Mon Mar 17, 2008 1:14 am**

by **Bigglez**

Greetings Mrai,

MrAl wrote:Here are some waveforms for a coil that is energized for a long time and then suddenly deenergized using a switch.

Okay, had to share this SPICE result!

1mH coil with only 1milli-ohm DC resistance. Switched

from 12V without a snubber or back-EMF diode.

Comments Welcome!

Posted: **Mon Mar 17, 2008 4:54 am**

by **MrAl**

Hi Peter,

Oh yes, very nice.

It looks like you are using Switchcad? Nice simulator too.

You can also do a sim with the diode connected to see how much

longer the pulse lasts with the catch diode, but you also need to add

some series resistance (say 100 ohms) to that coil to simulate a

real world relay coil (0.001 ohms gives rise to 12000amps), and

either set the initial conditions for the coil current to Vcc/100 or allow

enough 'on' time for the inductor to charge up to full current.

When i did my simulation i also used other clamp voltages (higher than

one diode drop) and realized there would be a formula that would

work out to some degree, that's when i came up with that formula,

although i have since found that in that formula Vcc/2 is a very rough

approximation of how the voltage across the coil (alone) changes when

it's discharging, and that Vcc/3 is a better approximation. Of course

the voltage is exponential so this is just an approximation too.

I also found that when the coil is discharging its series resistance is

in the discharge current path, and that allows the voltage across the

coil to go much higher than the single diode drop, and that allows the

coil to discharge faster. This can be easily viewed by using the simulator

with the coil and series resistance, and looking at the voltage across the

coil alone (not across the coil and the series resistance).

After that it's also a little interesting to connect a diode across the

coil itself, bypassing the series resistance (something that can not be

done in the real world) and see how much longer it takes to discharge

the coil. The time gets very long with only one diode drop.

I tried several coil inductances, like 10mH and 100mH. I figured the

inductance would be rather high because the maker would want to

generate enough force to pull the relay in and since F=k*A*N roughly,

the more turns (N) the more force. I dont think i ever measured the

actual inductance of a relay coil before however, or if i did long time

ago i forgot what it was. I have to wonder now if some manu's might

publish the data for their coils somewhere.

In any case, if you do more simulations it would be nice if you could

share them here with us. The waveform snapshots tell a lot about what's

happening in these circuits.