current's rush.

[rainman1]

I have a coil of relay (with 25ohm coil resistance) that consumes 5V in order to changes the Common leg's state.
Most of the time the circuit is open and 5V doesnt fall on coil but on a switch (i have a protection diode in paralel to relay).
So the circut is simple, 5V, coil, diode, and a switch that separates between coil and ground.

I was adviced by an engineer to connect a capacitor in paralel with to the 5V, so when i close the cicruit (by pushing the switch), the coil will receive a rush of current.

How come the capacitor gives a rush of current? what does it mean a rush of current, as opposed to the current that would flow if there was no capacitor?
And why does coil need a rush of current?

Thanks in advance.

 

[crutschow]

If the 5V supply can readily supply the current to the relay coil then I see no reason for the capacitor. It would not give a rush of current. A relay coil is an inductance so, by it's nature, it does not generate a rapid rise (or rush) of current, but rather the current rate of increase is proportional to the inductance.

I believe that engineer needs to review his notes on inductive behavior.

 

[colin55]

What the engineer is saying is this:

If the 5v is being generated by a source that does not allow a high current to flow, it would be a good idea to put an electrolytic across the rails so the impedance of the supply is reduced.

This is also a good idea to prevent spikes produced by relay from entering the 5v line. By spikes I mean dips in voltage due to the low resistance of the coil of the relay. After all, the relay is taking 200mA and this may have an impact on the 5v supply.

 

[rainman1]

To crutschow, so maybe i interepted it wrong, but there must be a reason for that.

To colin, you have mentioned a very good point, could you give some more details?
What do you mean the impedance of the supply?
I understand perfect that it decreases total impedance, but that should be a bad direction to follow because it increases current consumption from Vcc.

Why should a relay create spikes? you mean to the EMF? because the EMF comes from the coil's side that arent connected to Vcc, and it goes through diode protection.

 

[colin55]

The relay causes spikes in the negative direction. It reduces the 5v and devices like microcontrollers don't like any change in voltage.

The impedance of the supply is reduced when an electrolytic is placed across the rails and this does not necessarily mean more current will be drawn. It just means the 5v will remain at 5v when devices such as relays are activated and any extra current required by these devices will be provided by the electrolytic that is close to the device.
This is an important point when the relay is some distance from the 5v supply as the thin tracks on a PC board can reduce the voltage considerably.


This is the main point the enginner was talking about.

(We are not talking about the back emf that is generated when the relay is de-energised. This is snubbed by the diode across the coil.)

 

[rainman1]

I assume you're talking about spikes (BOUNCING?) that happen when energizing the coil, right?
Why do they happen in the opposite direction (negative voltage)?
I remember that when i tested the activiating time of relay, the bouncing were towards the 5V and not towards some negative voltage.

You really made up a good point with this tracks impedance, cause the distance between the relay and the 5V PSU is a few cm, which i assume means large distance?

 

[colin55]

It's not really a spike. It's a dip in the voltage when the relay is energised. This dip is of an unknown duration and unknown value. It's just a precaution to have an electrolytic near a device that is likely to draw a considerable current.
I'm just confirming what the engineer said. I don't now if it is necessary in your case.

 

[rainman1]

Alright thank you.
Dont you think that the capacitor decreases the time that the current reaches it steady-state?
If you compare RC circuit, to R circuit, that in RC circuit the voltage will increase faster (exponential rising versus linear rising).
So lets keep out the fact that reaching to steady-state might not help us here, but in general, isnt it true?

 

[rainman1]

Addition:
Now that i think about that, could you explain please how did you reach the conclusion that the capacitor decreases the supply's impedance?

 

[colin55]

What happens is this:

The electrolytic near the relay acts as a miniature 5v supply with very low impedance. This means it is capable of delivering a high current for a short period of time.

When the relay is activated, most of the current comes from your power supply but if the voltage dips slightly during this interval of time, the electrolytic delivers a small amount of energy as soon as the voltage dips, the overall effect is the supply voltage does not dip as much.

 

[colin55]

The impedance of a supply is a measure of how much current it can deliver.

If it can deliver a high current, it is classified as a low impedance supply.

Take for instance a battery in a car. It can deliver 400 amps. It is classified as low impedance supply.

Take a new battery - such as a D cell. A dry cell can deliver 4 amps. An alkaline D-cell can deliver 10 amps. This is because its internal resistance is lower. In other words its (internal) impedance is lower. We use the word impedance because we are not really measuring the resistance but working out the resistance by knowing the voltage and current capability.

 

[rainman1]

Man, i learn a lot from you, i really thank you.
I see, than not just it helps the relay get provided with 5V, than it helps the 5V regulator to be more stable?

What do you think about my comment of current reaching faster to steady state when capacitor is added?

 

[colin55]

Yes, it helps the regulator produce a steady output voltage.

""Dont you think that the capacitor decreases the time that the current reaches it steady-state?""

Not really. We cannot control the current. The current is determined by the resistance of the relay. We can only deal with the fact that the current decreases the voltage. We deal with this by adding more current via the electrolytic, because the power supply cannot deliver the required current.

 

[rainman1]

The resistance sets the current's steady state, but what about the time it takes to reach the steady state?

Why is it that the voltage of PSU decreases in high loads?
I noticed it many times.

 

[colin55]

If the output voltage of a power supply decreases when a load is applied, it means the power supply is not capable of delivering the current.

This can be solved by either decreasing the impedance of the supply - by increasing the size of the wire in the transformer or increasing the size of the transformer (in the case of switch-mode power-supply), or increasing the size of the electrolytic on the output of the supply.

 

[Chaerl]

Hi all,

If the inductive load (such as relays, motors, etc.) produces negative transient if it is driven by a high side switch, meaning the the switch is at the supply side. If I say switch, it could be mechanical or semicon (BJT, FET, etc.). But if it is drive at low side, it produces positive transients. If you're familiar with automotive EMC, this is part of the electrical disturbances that is produced by inductive load such as alternators or power locks and windows.

With the capacitor in parallel, it is more on a bulk capacitance. When you switch your relay, it draws suddenly more current which in result a dropout voltage in your supply. without the capacitor, it would imediately shutdown. The capacitor actually produces the voltage during this dropout. Some sort of secondary battery when your main power supply can not tolerate the sudden change of load.

The spikes are sometimes called back EMF, inductive kick-off voltage, load dump, transients, etc. This is actually due to the nature of inductance.

V=L*di/dt

So, the voltage at the inductor become so high during the rise of current from 0A to its actual current draw of the inductor (di/dt). Actually, the voltage has no effect to the other device but it is actually the energy that it desipates. Meaning does the the device that is driving the inductive load can withstand this high voltage (power) in such a duration. But nevertheless, you already have the freewheel diode (parallel with the coil) which is actually the one absorbing part of the energy the coil dissipates. I just hope the diode can withstand.

 

[colin55]

Chaerl has some correct assertions and some incorrect details.

Firstly I don't know what he is talking about when he says switching the HIGH side or LOW side of a device will create a different effect or different result.
An inductor only produces a back emf when the magnetic field collapses. In our case this energy is absorbed by the reverse-biased diode or the resistor across the coil.
The relay has a resistance of 25 ohms so it does not draw any more than 200mA. It is not as though it draws a lot more current because it is an inductor.
It does not draw "suddenly more current." It draws current according to its 25 ohms and in fact it "suddenly" draws less current due to its inductance. So the initial inrush of current is less than a 25 ohm resistor.
Most of the last paragraph cannot be understood because it is not presented in any technical way. However the diode will absorb any spikes produced by the coil, when the current-flow ceases.

 

[Chaerl]

colin55, here is an example of what we called in the R&D across the world High/Low side drivers.

https://www.electro-tech-online.com/custompdfs/2009/02/p395.pdf

True, Derived from the law of conservation of energy that upon switching OFF, the energy of the coil will distribute to the diode, the internal coil resistance of the coil and the energy lost due to the magnetic properties of the the relay coil (includes energy lost through magnetic field).

About the sudden drop, if your design is drawing current about 50mA when the switch is off, and then, eventually the switch is turned which has a draw of 200mA. Therefore, there is a change of current from 50mA to 250mA at certain instance as seen in your power supply side. At your switch side for the relay (if your using a semicon switch), it has a leakage current of 1uA (probably, depends on specs) when it is off, and draws 200mA during on.

Let say the transition happened for 1us. So, we can use the V=L*di/dt.

di = change of current = 200mA - 1uA
dt = change of time = 1us - 0us
L = inductance of the coil

so, you can know the voltage at 1us. But if you're interested what is the voltage at <1us, we need to know what is the current at <1us. So we have V as a function of t, V(t). If you plot this into a graph, you will get the actual spike.

The last paragraph is presented in a technical way. It may be clear to others who has the experience in dealing with transient suppression or EMC stuffs. This is more on the automotive electronics engineering.

One thing, if you already know about high side and low side switch, the negative and positive transients is produced due to the flow of current and the position of the load with respect to the supply.

 

[colin55]

I agree, the spike produced by the inductor will depend on the it being switched on the high side or low side but this is not an issue in this case and that's why I did not understand why it was brought up.

Secondly, the writer is trying to say the inclusion of an inductor will produce a larger inrush of current than if a 25 ohms resistor was included. This is not so. The inrush of current will be less for an inductor, due to the characteristics of an inductor creating its own back emf to "slow down" or reduce the inrush of current.

 

[Chaerl]

I brought up this in the first place because the writer ask about negative transients.

I also said earlier (which agrees in your previous statement) that the capacitor serves as a bulk capacitance (temporary engergy storage). I don't know where's the off topic part.

Also, you said there are incorrect details, which part? So that I could explain why and can clear up the misunderstandings. thanks.