Need help with Relay Diagram and use

[Iawia]

Hi,

I would like to use reed relay SILO5-1A85-76L4k.

I am not much of an electronics wiz, but surely enjoy learning about them.
The diagram shows a diode connected. Does that mean I need place a diode there myself, or it is already there inside the relay?

Why are diodes placed in this fashion? Is it protecting something???

 

[Boncuk]

Hi Iawia,

diodes across an inductive load (relays, solenoids, etc.) prevent damage to the switching transistor.

The collapsing magnetic field around a coil generates a voltage of opposite polarity and a much higher level (called back EMF) than the switching voltage used for the coil, e.g. coil voltage=5V, back EMF might be as high as 30V, destroying the switching transistor.

Observe the direction of the diode. Connection must be with the cathode as shown in the datasheet you provided.

Does that answer your question?

Boncuk

 

[Iawia]

Thanks Boncuk,

hmm. ok good to know about back emf. just from a theoretical perspective, why would a solednoid or relay become damaged? is it because of reversed current? or is it because of the fact that one simply cannot apply 30 volts to this component without destorying it? The sheet says 40-50 volts for breakdownn voltage.

Also, does this mean that the diode is already there? Or must i place one there myself? Also, if the diode was indeed already internal, would placing another diode there affect funtionality?

 

[Boncuk]

Thanks Boncuk,

hmm. ok good to know about back emf. just from a theoretical perspective, why would a solednoid or relay become damaged? is it because of reversed current? or is it because of the fact that one simply cannot apply 30 volts to this component without destorying it? The sheet says 40-50 volts for breakdownn voltage.

Solenoids and relays normally don't become damaged by back EMF! Moreover they generate it when being de-energized. The switching element (transistor or similar) might get too high opposite voltage (higher than rated) than required to activate the relay and become unusable.

Also, does this mean that the diode is already there? Or must i place one there myself? Also, if the diode was indeed already internal, would placing another diode there affect funtionality?

There are reed relays (like the one shown) containing a free wheeling diode already in the encapsulation. Placing another (external) diode you must observe the polarity of the relay (positive terminal connected to diode cathode)

Boncuk

 

[tvtech]

Hi Iawia,

diodes across an inductive load (relays, solenoids, etc.) prevent damage to the switching transistor.

The collapsing magnetic field around a coil generates a voltage of opposite polarity and a much higher level (called back EMF) than the switching voltage used for the coil, e.g. coil voltage=5V, back EMF might be as high as 30V, destroying the switching transistor.

Observe the direction of the diode. Connection must be with the cathode as shown in the datasheet you provided.

Does that answer your question?

Boncuk

Hi Hans

You have a way of explaining things that I am incapable of. You have "the knack"

Make something understandable by all here.

God given gift. This is not something that can be acquired. Or taught. Ever.

How are you friend???

PM me

tvtech

 

[Iawia]

Hi Boncuk,

I have hooked up the relay. However, although I send it a 5 v signal from pic to close the switch, nothing happens. I have attached a simple drawing showing how I have it set up.
I tested the pic signal and LED to make sure they are working. What am I doing wrong?

(The manufacturing .pdf is in the initial post)

 

[cowboybob]

My bad. Completely mis-read the schematic on my first post.

Second try:

Try reducing the 1K resistor: it's limiting the current to the point that the relay won't close.

 

[Boncuk]

Hi Iwia,

you are not supposed to connect a resistor into the relay solenoid circuit.

Given the relay draws about 8mA (without resistor) adding 1KΩ resistor into the connection will reduce the relay coil current to an amount which won't suffice the relay to be sufficiently energized.

(With a 1KΩ resistor in the coil circuit the relay activation current decreases to about 3mA).

Moreover I do not recommend to connect a relay directly to an MCU ouput pin.

PICs can stand about 20mA sink/source current per I/O, but connecting more than two loads (relays, etc.) will cause the chip to run hot.

(I like cool electronic components.)

A better (and safer) way is using a switching transistor to activate the relay.

I made up a little schematic how I would hook up a relay to an MCU.

R1 (100Ω) is meant to limit transistor gate current upon switching, R2 (10KΩ) takes care of fast discharge of the transistor gate when the MCU output pin becomes low.

Transistor source pin is connected directly to ground while the drain is connected to one relay pin and +5V is connected to the other.

The sample circuit shows pin connections of the MEDER M5-7175 which should be similar to yours.

Boncuk

 

[Iawia]

Thanks Boncuk,

I like cool components as well! I do however have some questions about the schematic you have written as I am incredibly curious about what each component's purpose i. Now let me unveil how clueless I am about electronics. You say 'fast discharge of the transistor gate', what exactly does this mean? Will the circuit still work without the 10k, but this 10k is simply making it more efficient??? How did you determine this value?

From what I understand, the capacitor from the source to ground, is the purpose of this is to reduce transient voltages??? How did you determine this value or would most caps work?

The 100 ohm resistor, how did you get this value? did you have to compute this based on being parallel with the 10k? If you can't already tell I am not from EE, thus I am a little fuzzy about R values.

What program did you use to write the schematic? I really like it! It is super clear what is going on. Thanks for all your help, Boncuk.

Also, cowboybob, I also was thinking the same thing as the coil 'typ power' from the datasheet says 167 mW, so I reduced the resistance to 100-200 ohm, but it still did not work. I will try your suggestions Boncuk and update.

 

[Joe G]

About the diode, check this out. https://www.electro-tech-online.com/blogs/hayato/80-relay-snubbers.html, AMAZING

 

[Boncuk]

Thanks Boncuk,

You say 'fast discharge of the transistor gate', what exactly does this mean? Will the circuit still work without the 10k, but this 10k is simply making it more efficient??? How did you determine this value?

If the switching cycle is long enough between cycles you might omit the gate discharge resistor. MosFet transistor's gates can be considered small capacitors which have to be discharged, e.g. if the controlling device does not go low, but enters high-Z state (high impedance state) when turning off. In that case the gate will discharge only caused by leakage inside the transistor.

From what I understand, the capacitor from the source to ground, is the purpose of this is to reduce transient voltages??? How did you determine this value or would most caps work?

Sorry, there is no capacitor from source to ground. All there is a capacitor from VDD of the MCU to ground which is a decoupling capacitor. Its purpose is to compensate power surges, e.g. when the relay or a relatively high current load is turned on in the +5V rail, thus preventing the MCU to crash. The value is used in all known circuits and therefor I have no reason to calculate it.

The 100 ohm resistor, how did you get this value? did you have to compute this based on being parallel with the 10k? If you can't already tell I am not from EE, thus I am a little fuzzy about R values.

The 100Ω resistor is there to limit gate current to a value the MCU will survive in case the transistor becomes faulty and causes the MCU to ground (dead short). I don't see it connected parallel with the 10KΩ resistor.

What program did you use to write the schematic? I really like it! It is super clear what is going on. Thanks for all your help, Boncuk.

I used EAGLE to create the schematic. You might want to download EAGLE from http://www.cadsoft.de

Also, cowboybob, I also was thinking the same thing as the coil 'typ power' from the datasheet says 167 mW, so I reduced the resistance to 100-200 ohm, but it still did not work. I will try your suggestions Boncuk and update.

I will also answer that question on behalf of cowboybob.

At 167mW power of the relay coil the coil resistance is 149.7Ω. A resistor (100-200Ω) connected in series with the relay coil will reduce the relay activation voltage to approximately 50% of the necessary voltage. At that voltage the relay can't work at all to pull in. It won't even suffice to keep the relay pulled in.

Do you expect a fart from a dead pork?

Boncuk

 

[Iawia]

Hi Boncuk,

Your setup has worked perfectly. The relay now pulls in great. All of your comments are well noted. thanks!

Joe G's link was also very interesting.

 

[Boncuk]

Hi Iawia,

glad to help you. You're always welcome.

Boncuk