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 failsafe design?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.
Question about relay voltage
Greetings Zach,
holding current (which is limited by the applied voltage
and DC resistance of the coil).
When the coil is deenergized 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).
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!
The current in the backemf diode is unrelated to the coilzmwworm 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.
holding current (which is limited by the applied voltage
and DC resistance of the coil).
When the coil is deenergized 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).
They are glass, which is a poor conductor of heat. Rectifierzmwworm wrote:The 1N4148 seems so fragile (looks like glass). What if I used two in parallel for a little more failsafe design?
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!

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 Joined: Wed Nov 24, 2004 1:01 am
 Location: ASHTABULA,OHIO
 Contact:
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.
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.
Greetings Robert,
application is equal to the pullin 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!
The PIV (peak Inverse Voltage) for a backemf diodeRobert 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.
application is equal to the pullin 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!

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 Joined: Wed Nov 24, 2004 1:01 am
 Location: ASHTABULA,OHIO
 Contact:

 Posts: 1752
 Joined: Fri Aug 22, 2003 1:01 am
 Location: Izmir, Turkiye; from Rochester, NY
 Contact:
Bigglez,
zmwworm,
Cheers,
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.The current in the backemf diode is unrelated to the coil
holding current (which is limited by the applied voltage
and DC resistance of the coil).
zmwworm,
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?the 1N4148 seems so fragile (looks like glass). What if I used two in parallel for a little more failsafe design?
Cheers,
Dale Y
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.
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.
LEDs vs Bulbs, LEDs are winning.
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.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?
Bigglez wrote:They are glass, which is a poor conductor of heat. Rectifier
diodes have thicker wires to remove the heat from the junction.
Okay, for all of the above reasons I'll just run to Radioshack and buy these.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.
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?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.
Greetings Dale,
during discharge, the faster the energy will be drained
from the coil. VL(t) = V e t(L/R) seconds.
is fixed.
If the loop impedance is increased (by adding a resistor
in series with the backemf diode) the voltage across
the coil during discharge will increase proportionally
to maintain the coil current.
Comments Welcome!
Agreed, I stand corrected.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.
This is counterintuitive. The lower the loop impedancedyarker wrote: In fact, to make the current decrease faster, put a resistor in series with diode.
during discharge, the faster the energy will be drained
from the coil. VL(t) = V e t(L/R) seconds.
The amount of energy (power) stored in the coil inductancedyarker wrote:The resistor increases the power the coil must provide while trying to maintain the current.
is fixed.
If the loop impedance is increased (by adding a resistor
in series with the backemf diode) the voltage across
the coil during discharge will increase proportionally
to maintain the coil current.
Comments Welcome!

 Posts: 1752
 Joined: Fri Aug 22, 2003 1:01 am
 Location: Izmir, Turkiye; from Rochester, NY
 Contact:
Yes, it is counterintuitive.
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 dropout 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 counterintuitive to me too. That's just the way it works out.
Cheers,
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 dropout 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 counterintuitive to me too. That's just the way it works out.
Cheers,
Dale Y
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.
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.
LEDs vs Bulbs, LEDs are winning.
Greetings Mrai,
the explanation.
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!
Great! Thank you (and Dale in the earlier post) forMrAl 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.
the explanation.
I did SPICE a similar circuit to see what wouldMrAl 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.
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!
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.
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.
LEDs vs Bulbs, LEDs are winning.
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