Coil Relay Overvoltage

[akaf24]

Hello, I'm a bit new to electronics, so this is probably a simple question. I am looking to to control the heated bed of my 3D printer with a relay. I'll buy the correctly rated relay (I have a 12 V power supply for the coil, and I'll have the contacts connected to a 24 V power supply for the bed). However, I do happen to have a relay laying around my house. The coil is rated for 12VDC (great!). But the contact terminals are rated for 80A/14VDC... My question is: does this voltage matter, and if so, why? The most current my heated bed will ever use is at most 10 A so does it matter if I have 24VDC across the contacts if only 8 A is passing through the contacts when it can handle 80A? Thanks. Here's the data sheet:

https://www.citrelay.com/Catalog Pages/RelayCatalog/A3.pdf

 

[che12345]

Dear MR.akaf24
Q1. "... I do happen to have a relay .... The coil is rated for 12VDC . But the contact terminals are rated for 80A/14VDC... does this voltage matter..."
A1. Based on the data sheet: a) max switching voltage 75 VDC, b) Dielectric strength : coil to contact 500Vrms min. and c) Dielectric strength : contact to contact 500Vrms min.
This relay would be [suitable] for operation on a 24VDC system , switching current 10ADC.
Q2. " ... if so, why? The most current my heated bed will ever use is at most 10 A so does it matter if I have 24VDC across the contacts if only 8 A is passing through the contacts when it can handle 80A?..."
A2. Data sheet:
a) max switching voltage 75 VDC. This indicates that the [contact gap] (during opening) is able to withstand the [arcing voltage] of a 75VDC system,
b) and c) also indicate that the clearance and creepage distances are adequate for a 24VDC system.
A3. The [current rating] is dependent on the (contact size and contact pressure). As for this case the contacts are adequately rated.
Che Kuan Yau (Singapore)

 

[itsmoked]

Keep in mind that every time whatever you have controlling the relay coil interrupts the relay coil it will get a large harmful inductive spike. You should include a "freewheeling diode" or make sure your controller includes one, in or with, whatever is turning on/off the relay coil.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[akaf24]

Thank you so much for the responses. Yep, this relay is just switched on and off by the on board transistor and I understand what you're saying about the fly back diode. Maybe it's hard to say without knowing more of the system, but would you advise a small resistor in series with the coil or with the diode so that the current isn't free-wheeling? My guess is that it really isn't necessary.

 

[waross]

The greater the resistance in the free wheeling circuit, the higher the surge voltage.
If you feel that you must use a resistor, a Zero Ohm resistor may be the best choice.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[che12345]

Dear Mr. akaf24
Q. "... I understand ... about the fly back diode.... would you advise a small resistor in series with the coil or with the diode so that the current isn't free-wheeling? My guess is that it really isn't necessary.
A1. The usual [free-wheeling] diode is a diode of (sufficient voltage and current rating) only (without any resistor in series) is [connected across the relay coil]. BTW: the diode is wired in the reverse direction.
A2. Numerous other surge suppressors viz., transil diode, varistor, RC type methods are also used connected across the relay coil. In many cases, it is "unnecessary". It is like buying insurance.
Che Kuan Yau (Singapore)

 

[akaf24]

Thank you! I'll stick to just the diode.

 

[che12345]

Dear Mr. akaf24
Q. "... Thank you! I'll stick to just the diode."
A. I agreed with your learned decision. The diode say any 1N XXXX would be a very small and very low cost item. It cost nothing, occupies no room and easily connected across the relay coil. A small premium for an insurance coverage. Have a nice day.
Che Kuan Yau (Singapore)

 

[LionelHutz]

The resistor would allow the relay to open faster because it dissipates the magnetic energy in the coil which makes the magnetic field collapse faster. Without the resistor, the magnetic energy gets dissipated in the resistance of the coil. That doesn't sound necessary in this application.

I recently had to use a resistor and diode to get a magnetic clutch to drop out quicker. It is easy to calculate the resistor. Knowing the coil current at turn-off and the maximum reverse voltage allows you to pick the resistor.

 

[waross]

Be aware that Lionel's solution is trading time for voltage;
Less time but a higher induced voltage.
An inductor opposes any change in current.
If a diode is inserted with no additional resistance, the current and voltage will both decay to zero.
If a resistance equal to the coil resistance is inserted in series with the diode, the current will still decay from the steady state value but the current is now passing through twice as much resistance and so the induced voltage will be doubled.
The doubled voltage will discharge the energy in the magnetic field twice as fast and so the drop out-time will be cut in half.
It is a compromise.
The diode eliminates the inductive kick, and the voltage decays from the applied voltage to zero.
By adding the resistor, we are allowing a controlled amount of over-voltage due to inductive kick in order to achieve faster drop out.
The maximum current seem by a free-wheeling diode is the steady state current.
When resistance is added to the free-wheeling circuit, both the coil and the driving circuit should be evaluated for their ability to withstand the increased kick voltage.
By the way, Lionel, do you remember the ratio of added resistance that you used?
Thanks.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[che12345]

Dear All,
I am of the opinion that:
1. For the intended use of the relay asked by Mr. akaf24, a free-wheeling diode alone would be a good choice. It cost "nothing" , very small in size, easy installation and generates no heat. [Very short delay] on the (drop-out time), as the (forward resistance) is very low.
2. A (permanent resistor without any switching) connected across the relay coil would i) delay the drop-out time, ii) consumes energy (i² r), iii) generates far more heat than a diode.
3. A (diode in series with a resistor) connected across the relay coil would delay the drop-out time.
4. The delay in the drop-out time (usually very short) may or may not be acceptable, depending on the application.
BTW: a) in most applications, "fast" (or very short delay in the drop-out time) does not cause any problem or is the preferred.
b) the higher the resistor value the longer the delay in the drop-out time but lower the power consumption and the heat generated.
Che Kuan Yau (Singapore)

 

[3DDave]

It better generate heat. There's a current with a 0.7V drop -> P = IE, right? I expect there isn't enough to obliterate a typical diode, but I bet the diode still gets warmer.

All the energy in the coil is going somewhere and that somewhere is heat.

 

[LionelHutz]

The clutch operating voltage was < 100VDC. I calculated for about 800V across the resistor, but was picking from standard available 250W resistors so got a little less.

che12345 - No, a series reverse biased diode and resistor will decrease the drop out time. And as Dave pointed out, the diode does dissipate some power so there is some heat.

 

[waross]

Thanks for sharing your first hand experience, Lionel.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[che12345]

Dear All,
1. "...There's a current with a 0.7V drop -> P = IE, right? I expect there isn't enough to obliterate a typical diode, but I bet the diode still gets warmer."
A1. On a 24Vdc source the diode [leakage current] is in the region of say ( mA/uA?). P= 0.7V x (0.001?)A = 0.00007 W !.
A1.1 During [free wheeling], the current circulates within the diode and the coil. The "momentary" current may be in the region of say (A?) but for a [very short duration] ( of say (ms?). As the (forward resistance) of the diode is [very low], the power loss/heat generated is say < 1W for a very short duration . When the back emf is =0V , current =0A; therefore Power = 0 W within say (ms?).
A2. Connecting a resistor directly across the coil (without a blocking diode) would have to be rather high value say in the region of (xx Ohm). It shall not be too low that may affect the coil to operate and the [continuous power loss] say (xx W?) is much higher than a diode.
A3. When the [reverse diode is in series with a resistor], the higher the resistor, (the longer will be the discharge time). The longer the discharge time would result to longer drop-out time.
BTW: [Drop-out time] is measured from the time taken for the N.O. contact to reset to N.O. position, when the coil is [switched off].
Che Kuan Yau (Singapore)

 

[waross]

The formula for the time constant of an RL circuit:
T=L/R

A3. When the [reverse diode is in series with a resistor], the higher the resistor, (the longer will be the discharge time). The longer the discharge time would result to longer drop-out time.

Not so.
More resistance gives a higher voltage and a shorter time.

Current: The current of an inductive "kick" will not exceed the steady state current. The "kick" current will decay from the steady state current to effectively zero in five time constants. (0.67%)

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

A3. When the [reverse diode is in series with a resistor], the higher the resistor, (the longer will be the discharge time). The longer the discharge time would result to longer drop-out time.

Still wrong.

The coil in question is 12V @ 80 ohms with a dropout time of 5mS. So, the current is 0.15A and the reverse diode would see a peak of 0.11W . If you put a 80ohm resistor in series with the diode, then the resistor would see a peak of 1.8W. In both cases, the power would decay to zero in milliseconds. You could use a 1/4W resistor and be fine unless you are switching the coil a lot.

 

[che12345]

Dear All,
Thank you for sharing your learned opinion on point A3. See my mail dated 9th Oct.
I try to go through it and see whether it make sense?
1. the relay coil rated 12Vdc Ri=80 Ohm is connected across a 12Vdc source. A reversed diode (only) is connected across the coil. When the source is off, voltage across the coil is 0V, I=0A, N.O. contact is in open position,
1.1 when the 12V source is switched on, the [transient voltage] across the coil [begins to rise] from 0V to 12V. After a short time delay (depending on L/R)* when the voltage reaches 8.4V; the relay picks up. The N.O. contact closes. However, the voltage continues to increase passing 8.4V and finally reached [steady voltage] 12V. At 12V, the steady current(coil) I=12V/80 Ohm = 0.15A , Power (coil)=I² x Ri = 1.8W. Current flow through the reversed diode is +- =0A , Power (diode)=0W. This state continues until the source is [switched off], see 2,
2. When the 12V source is [switched off] with the (coil steady voltage) at 12V, the [generated back emf] is dependent on L di/dt. It is possible to >12V. Therefore, the [transient current] due to the back emf is possible to > 0.15A. As the only resistance is Ri, while the diode forward resistance is +- 0 Ohm,
2.1 When the 12V source is [switched off], the coil voltage from steady 12V starts [to decrease] (depending on R/L)** from 12V to 0V. The only resistance is Ri=80 Ohm. At 1.2V the relay releases (= drop-out) i.e. the N.O. contact resets to the open state. When the [diode is in series] with a resistor Rs, see 3,
3. The coil voltage from 12V would [decrease at a slower rate] (depending on R/L)**. The resistance would be =(Ri + Rs). The voltage would take a longer time to decrease from 12V to 1.2V, i.e. the N.O. contact drop-out at a longer time delay.
Che Kuan Yau (Singapore)

 

[waross]

You have omitted a basic factor, well known to any and all who work, hands on, with inductive circuits.
A review of the simple, basic, formula for time constants may be in order.
T (time) = (L {inductance in Henrys} divided by R {resistance})
It is easily seen that the Time Constant is inversely related to the Resistance, and as the Resistance increases, the time constant decreases.
Explained another way:
The current does not increase.
The current always decays from the steady state current.
The peak kick-back voltage is the product of the resistance and the steady state current.
As the resistance in the circuit is increased, the peak induced voltage is increased.
At any given current, an increased instantaneous voltage will result in an increased instantaneous power or Watts.
More Watts results in a faster discharge of the stored energy.
The back side of this is an increased induced voltage.

This is not as serious as it may seem in the case of actual automotive components.
Transient over-voltages due to "load dump" are common in automotive applications.
When the alternator is supporting a heavy load, and the load is abruptly turned off, there will be a transient over-voltage until the voltage regulator is able to reduce the alternator field excitation.
Imagine a pick-up truck attached to a large trailer with a large number of clearance lights.
The lights, including the high beam headlights, driving lights and marker and clearance lights are turned off with one switch.
The load dump transient may reach 100 Volts or more on a nominal 12 Volt system.
All automotive electrical components are capable of withstanding transient voltages in excess of 100 Volts.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

Your theory is completely wrong starting at point 1.1 and onwards. You need to stop posting wrong information and learn how it works when you don't know the theory.

At point 1.1, when switched on the coil voltage immediately goes to 12V. It is the coil current that rises to the steady state current.

The current in the coil after turn-off will not increase based on the coil voltage at turn off. You have it completely backwards. At turn-off, the current "tries" to continue flowing at 0.15A and that is what will dictate the voltage across the coil terminals.

At turn-off with just a simple diode, the 0.15A current will generate -0.7V across the coil terminals.

At turn-off with a series diode and 80 ohm resistor, the 0.15A current will generate -12.7V across the coil terminals.

The relay mechanically switching off is actually dependent on collapsing the magnetic field in the coil, not on the coil voltage. The relay will mechanically begin to open when the magnetic field is too weak to hold it closed. Same applies to any DC relay or solenoid.

Since the magnetic field in the coil represents a stored energy, the only way to collapse the field is to remove or dissipate the energy from the coil. With a diode, the only place for the energy to be dissipated is in the coil resistance. With a series diode and resistor, the energy can be dissipated in both the coil resistance and the external resistance.

With a diode only, the coil energy dissipation starts at 1.8W and decays from there.

With a diode and 80ohms series resistor, the coil energy dissipation starts at 3.6W and decays from there.

It should be obvious that 3.6W > 1.8W, so the energy will dissipate faster in the second case.