Diode across coil?
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petersonra
Senior Member
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It is to handle the inductive kickback when the voltage is removed from a DC coil.
When energized, the coil has stored energy that has to go somewhere when it is de-energized. The diode provides a path for the stored energy to fizzle out around the coil instead of damaging the external switching mechanism.
you size it based on how much energy the coil stores and the voltage it will release that stored energy at. In most cases it is not a lot and a 1N4005 diode will handle almost any coil.
When energized, the coil has stored energy that has to go somewhere when it is de-energized. The diode provides a path for the stored energy to fizzle out around the coil instead of damaging the external switching mechanism.
you size it based on how much energy the coil stores and the voltage it will release that stored energy at. In most cases it is not a lot and a 1N4005 diode will handle almost any coil.
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I thought it was an MOV not a diode?
A diode for DC coils.
dfmischler
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Early relay logic systems mainly protected contacts from inductive spikes with capacitors (and resistors as needed). Diodes were used later. Much the same type of circuits are now used to protect transistors/MOSFETs/SSRs driving inductive loads from similar damage.
Contact Circuit Protection
Contact Circuit Protection
- Location
- Massachusetts
Thanks. I thought it funtioned more like resistor than a diode but willing to learn.
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- Retired PV System Designer
For a DC coil a diode will work fine but will allow the magnetic field to remain for a longer time, slowing the dropout of whatever is moved by the coil.
A well chosen series resistor will maximize the magnetic field decay by dissipating energy.
An MOV will combine the functions of the diode and resistor since it will not conduct at the normal applied voltage.
For an AC coil a diode will not work but an MOV or a pair of zener diodes will.
You can also use a snubber circuit of a capacitor and resistor in series placed across the coil.
Just a capacitor will also limit the voltage spike on either an AC or DC coil, but can produce a resonant ringing instead and will slow the dissipation of the stored energy.
junkhound
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And , if you want the contacts to open in the absolute minimum time, you size a capacitor for a ringing frequency of the LC circuit of about 70% of the mechanical resonance of the contactor armature and spring. That way the ringing does not have enough drive to turn the relay back on, but drops the magnetic holding force the fastest.
- Location
- Massachusetts
Thank you.
One other question, I would expect a diode to be in series with the load and a resistor to be in parallel with the load while the MOVs I have seen are in parallel with the load.
So if we dumped the MOV for a run of the mill diode would it be in parallel and series.
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- Placerville, CA, USA
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- Retired PV System Designer
Any snubber circuit has to be in parallel with the control contacts or in parallel with the coil.Thank you.
One other question, I would expect a diode to be in series with the load and a resistor to be in parallel with the load while the MOVs I have seen are in parallel with the load.
So if we dumped the MOV for a run of the mill diode would it be in parallel and series.
When I was talking about various snubber options, they would all be configured as a two terminal network attached in one of those locations.
The resistor would be in series with the diode so that the load resistance on the disconnected coil is optimal for energy absorbtion rather than just acting as a short across the inductance of the coil. The short would leave only the internal resistance of the coil to absorb the energy more slowly.
gar
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- EE
160121-2223 EST
iwire:
This is where a basic understanding is important.
Consider a series circuit of a battery, switch, and an inductor. Assume the inductor inductance is constant. In the real world there will be resistance in series in this circuit. Assume the dominate series resistance is the internal resistance of the inductor coil. Let the reistance be R and the source voltage V.
Also note that the inductor current can not instantaneously change. This means that, if just before time T1 the current is I, then just after T1 the current must = I. From any energy stored in the inductor, if an attempt is made to change that current, then the inductor will generate whatever voltage across its inductor that is required to maintain that same current in the same direction. Theoretically this could mean infinity voltage, but in the real world something will breakdown to limit the voltage.
The loop voltage drop for this circuit is V = i*R + L*di/dt where L is the inductance, and di/dt is the rate of change of current with respect to time. If the starting point is where inductor current is zero, then at closure of the circuit all the voltage drop is across the inductance because at that time the current is 0. After a long time the steady state current builds exponentially to a value of I = V/R. One uses calculus to determine the shape of the current curve.
Theoretically voltage sources have zero internal impedance. This means if the voltage source is reduced to 0 volts, then the voltage source looks like a 0 resistance wire.
Next instantaneously change the voltage source from V to 0 volts. Just before this voltage change the current thru the inductor is flowing from the + end of the inductor thru the inductor and out the negative terminal. Just after the source voltage change the inductor current must remain the same and flow in the same direction. This means the inductor becomes a voltage source and thus the inductor terminal polarity reverses to maintain current flow in the same direction. What was positive is now negative. An exponential current discharge curve occurs that is like the charge curve, but reversed in the vertical axis.
Placing a reversed biased diode across the inductor and the inductor's internal series resistance provides a new series path. When battery voltage is applied the diode is reversed biased and negligible current flows thru the diode. If a switch is in series with the battery loop, then when the switch is opened the current from the battery goes to zero, but the current in the inductor must continue. Since the inductor voltage reverses this forward biases the shunt diode, and it becomes a ahort across the inductor.
The time constant of an L-R series circuit is a function of L/R. The lower R is relative to L the longer is the time constant and the time for the current to increase or decay.
To reduce the time constant to obtain a shorter dropout time one makes R greater, but then the terminal voltage at turn off increases. Non-linear devices such as Transorbs and MOVs become useful as mentioned in other posts.
See http://www.electronics-tutorials.ws/inductor/lr-circuits.html don't be confused by my use of L/R vs R/L you see in the equation. What is in the exponent of the equation is really the ratio of two times --- an independent variable t divided by a time constant (L/R).
.
iwire:
This is where a basic understanding is important.
Consider a series circuit of a battery, switch, and an inductor. Assume the inductor inductance is constant. In the real world there will be resistance in series in this circuit. Assume the dominate series resistance is the internal resistance of the inductor coil. Let the reistance be R and the source voltage V.
Also note that the inductor current can not instantaneously change. This means that, if just before time T1 the current is I, then just after T1 the current must = I. From any energy stored in the inductor, if an attempt is made to change that current, then the inductor will generate whatever voltage across its inductor that is required to maintain that same current in the same direction. Theoretically this could mean infinity voltage, but in the real world something will breakdown to limit the voltage.
The loop voltage drop for this circuit is V = i*R + L*di/dt where L is the inductance, and di/dt is the rate of change of current with respect to time. If the starting point is where inductor current is zero, then at closure of the circuit all the voltage drop is across the inductance because at that time the current is 0. After a long time the steady state current builds exponentially to a value of I = V/R. One uses calculus to determine the shape of the current curve.
Theoretically voltage sources have zero internal impedance. This means if the voltage source is reduced to 0 volts, then the voltage source looks like a 0 resistance wire.
Next instantaneously change the voltage source from V to 0 volts. Just before this voltage change the current thru the inductor is flowing from the + end of the inductor thru the inductor and out the negative terminal. Just after the source voltage change the inductor current must remain the same and flow in the same direction. This means the inductor becomes a voltage source and thus the inductor terminal polarity reverses to maintain current flow in the same direction. What was positive is now negative. An exponential current discharge curve occurs that is like the charge curve, but reversed in the vertical axis.
Placing a reversed biased diode across the inductor and the inductor's internal series resistance provides a new series path. When battery voltage is applied the diode is reversed biased and negligible current flows thru the diode. If a switch is in series with the battery loop, then when the switch is opened the current from the battery goes to zero, but the current in the inductor must continue. Since the inductor voltage reverses this forward biases the shunt diode, and it becomes a ahort across the inductor.
The time constant of an L-R series circuit is a function of L/R. The lower R is relative to L the longer is the time constant and the time for the current to increase or decay.
To reduce the time constant to obtain a shorter dropout time one makes R greater, but then the terminal voltage at turn off increases. Non-linear devices such as Transorbs and MOVs become useful as mentioned in other posts.
See http://www.electronics-tutorials.ws/inductor/lr-circuits.html don't be confused by my use of L/R vs R/L you see in the equation. What is in the exponent of the equation is really the ratio of two times --- an independent variable t divided by a time constant (L/R).
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But usually not by a lot. The coil has resistance. It has to otherwise the current would be infinite.
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- Massachusetts
dfmischler
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- Western NY
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- Facilities Manager
AKA: When you stop applying current to the coil, its magnetic field collapses and induces a reverse voltage in the circuit. This reverse voltage may be much higher than the original voltage applied to the coil, and could damage other parts of the circuit. So you want to safely dissipate the energy. A diode in parallel with the coil, and installed so that it won't conduct when the original voltage is applied to the coil, will conduct when the coil induces the reverse voltage spike. The diode will absorb the electrical energy and radiate it away as heat. Other snubber circuits could also be used.
Then perhaps a little more direct.
Coils can and do store energy. For AC, there are natural zeroes. DC does not. When you interrupt the flow of current to the coil that energy gets dissipated. Either in an arc across the contacts which the contacts may not be rated for and get destroyed. I've seen that happen. Or you can provide an alternative means. A (flywheel) diode across the coil allows the current to circulate within the coil until it dissipates in the coil resistance.
Hence the diode.
gar
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- Ann Arbor, Michigan
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- EE
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iwire:
My goal is to try to help you understand how the circuit works.
A relay coil has both inductance and resistance that can be represented by a schematic series circuit of a pure (ideal) inductor and resistor. These two components in series are what you see looking at the relay coil terminals. In the real world these two elements are not separable. Thus, I can not put a scope across the resistive component of the coil, but I can measure the input current to the coil and from that deduce the voltage drop across the internal resistance vs time.
I can determine the value of the coil's internal resistance by a DC resistance measurement by waiting long enough for any transient affect to die out. For a typical small relay that is probably less than one second. Using appropriate instrumentation I can measure the coil's inductance independent of the coil's DC or AC resistance. An LRC bridge is that type of instrument. The LRC bridge with AC excitation can also be used to measure the inductance.
For a DC relay there are pull-in and drop-out voltages and currents. For one sample of a P&B KUP 11D15 24 VDC relay some measured values are:
1. 480 ohms DC coil resistance using a Fluke 27.
2. 476 ohms DC coil resistance using a General Radio 1650-A bridge.
3. 0.950 Henrys Q = 1.6 armature open at 1 kHz inductance using GR 1650.
4. 1.050 Henrys Q = 1.4 armature hand held closed with same instrument.
More later I have to leave.
.
iwire:
My goal is to try to help you understand how the circuit works.
A relay coil has both inductance and resistance that can be represented by a schematic series circuit of a pure (ideal) inductor and resistor. These two components in series are what you see looking at the relay coil terminals. In the real world these two elements are not separable. Thus, I can not put a scope across the resistive component of the coil, but I can measure the input current to the coil and from that deduce the voltage drop across the internal resistance vs time.
I can determine the value of the coil's internal resistance by a DC resistance measurement by waiting long enough for any transient affect to die out. For a typical small relay that is probably less than one second. Using appropriate instrumentation I can measure the coil's inductance independent of the coil's DC or AC resistance. An LRC bridge is that type of instrument. The LRC bridge with AC excitation can also be used to measure the inductance.
For a DC relay there are pull-in and drop-out voltages and currents. For one sample of a P&B KUP 11D15 24 VDC relay some measured values are:
1. 480 ohms DC coil resistance using a Fluke 27.
2. 476 ohms DC coil resistance using a General Radio 1650-A bridge.
3. 0.950 Henrys Q = 1.6 armature open at 1 kHz inductance using GR 1650.
4. 1.050 Henrys Q = 1.4 armature hand held closed with same instrument.
More later I have to leave.
.
Lost_RFTech
Member
- Location
- IL., Ia., Mo.
Hopefully simple and straight-forward:
The diode installed across a DC relay coil in what you'd view as a reverse biased fashion is called a free-wheeling diode (at least they used to be). It's purpose is to protect any solid state switching device (e.g. transistor) from the colapsing field of the coil when the transistor turns off to de-energize the relay.
Without the diode, a voltage is developed at the transistor/coil connection that may exceed the voltage rating of the switching device. With the diode installed, the induced voltage can never exceed the absolute value of the system power supply plus 0.7VDC forward biased diode voltage drop.
The diode installed across a DC relay coil in what you'd view as a reverse biased fashion is called a free-wheeling diode (at least they used to be). It's purpose is to protect any solid state switching device (e.g. transistor) from the colapsing field of the coil when the transistor turns off to de-energize the relay.
Without the diode, a voltage is developed at the transistor/coil connection that may exceed the voltage rating of the switching device. With the diode installed, the induced voltage can never exceed the absolute value of the system power supply plus 0.7VDC forward biased diode voltage drop.
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- Placerville, CA, USA
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- Retired PV System Designer
Yes.
Bingo! Then selecting tty he properly tasted ZNR diode is the next thing. This is a common application.
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