# Novice question about Relays, Capacitors, and Diodes

• shushi
In summary, the relay is a 1.5VDC 2A DPDT NON-LATCHING SMD Relay that is suitable for single side stable circuits that are protected by a IN4001 diode. The coil is protected by a 50VDC blocking capacity and the contacts are safely insulated to each other up to a voltage of 1.5V.

#### shushi

TL;DR Summary
How to determine specific capacitors and diodes in order to minimize back emfs and arcing of opening contacts.
Hello,

In regards to this relay, TQ2SA-1.5V Panasonic 2 Form C AS Single side stable, 1.5VDC 2A DPDT NON-LATCHING SMD Relay
https://www.mouser.com/ProductDetail/769-TQ2SA-1.5V
PDF 1 of Relay
PDF 2 of Relay (specifically pages 5-6) is attached at the bottom as a pdf file.

I was previously told that I would need to wire a IN4001 diode which have a 50VDC blocking capacity, in parallel to the coil pins of this relay which could produce a back emf surge of up to 50VDC, which was more than enough to protect this specific circuit.
https://www.engineersgarage.com/electronic-components/1n4001-diode
PDF of 1N4001-Diode

I know basic electronics and ohms law theory, but I'm still a bit of a novice to when it comes with electronics, but is this true? Is a IN4001 diode more than enough to protect this circuit?
In the PDF, it says this about the coil and contact electrical characteristics,
>Electrical characteristics

>>Breakdown voltage
>>(Initial)
>>>Between open contacts 1,000 Vrms for 1 min. (Detection current: 10 mA)
>>>Between contact and coil 1,500 Vrms for 1 min. (Detection current: 10 mA)
>>>Between contact sets 1,500 Vrms for 1 min. (Detection current: 10 mA)

>>Surge breakdown
>>voltage (Initial)
>>>Between open contacts 1,500 V (10×160µs) (FCC Part 68)
>>>Between contacts and coil 2,500 V (2×10µs) (Bellcore)

but I'm not sure how to translate this as useful data when looking at the capacities of the IN4001 diode.I was also told that I needed to protect the contact ends of the relay since when they open up, they tend to create an arc that can wear out and damage the relay's performance. I was reading on another forum that in order to minimize this damage that I needed to wire a capacitor and a resistor in parallel to those contact ends, which they provided a "No Math" approach to figuring out the capacity and type of capacitors and resistors that I needed, which was the following method,
for the farad value of the capacitor,
0.5 to 1 µF per switching current (or per 1 Amp)
and for the voltage of the capacitor,
Use a capacitor with a dielectric strength between 200 and 300 V

and for the resistance value of the Resistor,
0.5 to 1 O per switching voltage (or per 1 Volt)

Based on my schematics, are 0.04µF 300VDC capacitor + 5 Ohm 5 Watt resistor approprate for a circuit that has total of 22mA and 5VDC, and a 0.02µF 300VDC capacitor + 5 Ohm 5 Watt resistor for a circuit that has total of 40mA and 5VDC?

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• relay schematics.png
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• relay.pdf
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Serious case of overdesign. Those 5W resistors will be bigger than the relay, you know

For the coil, that 1N4001 is more than enough. The back EMF is generated by the coil of the relay, its size/energy depends on the coil size and its polarity is in reverse than the original polarity of the current. So during a back-emf situation that diode will face it with the forward voltage, which is ~ 0.6V for most basic diodes anyway. If there would be significant energy then bigger diode would be needed due the current, not the voltage... The reverse voltage capacity of 50V matters only against the battery there. 50V vs. 1.5V - more than enough.

About the breakdown voltage: that means that the coil and the contacts are safely insulated to each other up to that voltage. Directly not related to sparks or arcs.

Regarding the arcs on the contacts: actually, how long lifetime are you aiming for? That design approach would be adequate to make those contacts last till the Coming of the Horsemen

For this low load those contacts will be fine without any protection at all, unless you have significant inductance in the switched circuit.
If you want to stick with the cap&resistor, then the most simple SMD would be more than enough. At least that won't cost more than the relay... Also, the voltage of the cap is good at 25-50V: any common ceramic SMD will do.

Ps.: just noticed that you plan to use a LED to provide the necessary voltage drop between the relay and the battery? That's not a good idea, battery voltage does not count as constant. You might get that LED burned if the battery is full, while it just won't work when the battery is low.

shushi
Rive said:

Serious case of overdesign. Those 5W resistors will be bigger than the relay, you know

For the coil, that 1N4001...

Although I'm someone that likes to play things safe, I had anticipated in the back of my mind that I had overthought this all, but I do appreciate you pointing that out, and of course in the charitable and well meaning manner that you did :)

but in regards to the capacitor, and resistor, I will definitely look for smaller sizes since the previous explanations that I found in other forums don't necessarily apply to my low signal/load application xD

But I do appreciate your explanation, that it has to do more with amperage (which I'm learning is pretty important, as well as the ampacity capacity of the medium through which said current passes through), and I shouldn't worry about breakdown voltage/insulation for this small load application (although I am curious about your statement on significant inductance in the switched circuit, specifically what would be specific cases where this would likely be a factor to be concerned about (is it relative to a specific current/voltage level, or a relationship between power and something else)? Are there any articles that you could point me towards to read up more on that? I apologize for asking more questions!)

I just had one more question about figuring out a specific way to calculate the size/energy of the back emf using the coil size. The closest formula that I could relating to this calculation was (what I believe is called) the Capacitive Current formula
I = C dv/dt

Is this correct or what other formulas could I use to more accurately estimate back emf values?

Thank you again for all of your help, it helped clear up so many things, it is greatly appreciated! And thank you for the greetings!

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Rive said:
Ps.: just noticed that you plan to use a LED to provide the necessary voltage drop between the relay and the battery? That's not a good idea, battery voltage does not count as constant. You might get that LED burned if the battery is full, while it just won't work when the battery is low.

Ahahaha the entire circuit of the led button that controls the relay isn't directly connected to the battery (which is a portable charger; although there's another led button switch that is, that is connected in parallel with the switch that controls almost the entire device I'm building by acting as an on and off switch), that specific led button that controls the relay gets its power from a usb hub port that provides at most 2.4amps (which through my estimations, it seems only one and a half amp will be available to that hub) and I believe that hub has smart current limiting technology, but I'm not really positive on that, I have to double check since I'm still touching up on some areas in this total project.

in the schematics, the usb port that has a star is where the led switch and relay get their power from. But about current draws, although a battery or power source may provide more amperage than is required for a specific circuit, doesn't the resistance from the devices and the voltage of the circuit limits current draw on their own? Or in other words, that circuits and the devices will only draw on the current that they need despite what the battery provides? (and if this was the case, would that damage batteries since they are designed to drain at an appropriate rate?) Or is the amperage that the battery or power source provides forced onto the entire circuit which will wear out the entire circuit at a faster rate?

Oh! and here's the pdf specifications for the led button switches,

PDF for the Galco AL8 Illuminated Pushbuttons

Thank you for raising an important question, I'm going to look into this further!

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shushi said:
I am curious about your statement on significant inductance in the switched circuit, specifically what would be specific cases where this would likely be a factor to be concerned about?
Capacitive loads are relatively easy, since any capacitor would be able to store the energy deposited after the circuit disconnect still at the same voltage: but inductive loads are different since they 'try' to maintain current without any concern about the voltage needed, and while doing this they have the backing of the energy stored in the magnetic field. Without suppression this might get pretty violent. A simple light tube choke coil can produce fat, 10kV arcs after disconnecting just a 5V battery from it... Of course real loads are usually not this thorny, but for example switching a small motor or an electromagnet needs some care.

shushi said:
Are there any articles that you could point me towards to read up more on that?
For simple cases any search for 'flyback diode' will do. For advanced studies, it is 'snubber'. What I know about this is limited to the first part, with knowing when to call for an expert

shushi said:
although a battery or power source may provide more amperage, doesn't the resistance from the devices and the voltage of the circuit limits current draw on their own?
With a battery you have to design for the maximal and the minimal voltage too, not just for the nominal (for a 12V lead-acid battery, it is 14V, 12V, 9V... Quite a range). And in case of LEDs it is important, since those are not linear with V and A. Slight change in voltage might cook the thing.
Just one more small remark on the USB charger. These are chargers: things are not necessarily expected to actually work when connected. You better check the output voltage and noise.

shushi
To use the "USB" label, devices must meet the USB specifications. here is an excerpt fromUSB Serial Bus Specification Revision 2.0, April 27, 2000, page 175 of 650:

"7.2.2 Voltage Drop Budget
The voltage drop budget is determined from the following:
• The voltage supplied by high-powered hub ports is 4.75 V to 5.25 V.
• The voltage supplied by low-powered hub ports is 4.4 V to 5.25 V."
So that is the voltage range you must accept.

Included in the same zip file is Universal Serial Bus Power Delivery Specification (394 pgs)

The full standard is available at:
http://www.usb.org/developers/docs/usb_20_070113.zip

It is a 37MB zip file of about 2000 pages.

Have Fun!
Tom

shushi

## 1. What is a relay and how does it work?

A relay is an electrical switch that is controlled by an electromagnet. When a current flows through the coil of the electromagnet, it creates a magnetic field that pulls a metal armature towards it. This movement of the armature closes or opens the switch, allowing or stopping the flow of current through the circuit.

## 2. What is the purpose of a capacitor in an electrical circuit?

A capacitor is an electronic component that stores electrical energy in the form of an electric charge. It is used to filter out noise and stabilize voltage in a circuit. Capacitors are also commonly used to store energy and release it in short bursts, which is useful in many electronic devices.

## 3. How do diodes work and why are they important?

A diode is an electronic component that allows current to flow in one direction only. It works by using a semiconductor material to create a depletion zone, which prevents current from flowing in the opposite direction. Diodes are important because they are used to convert AC to DC, which is necessary for many electronic devices to function properly.

## 4. What is the difference between a relay and a switch?

A switch is a mechanical device that is manually operated to open or close a circuit, while a relay is an electronic switch that is controlled by an external signal. Relays are typically used to control high voltage or current circuits, while switches are commonly used for low voltage or current circuits.

## 5. Can capacitors and diodes be used together in a circuit?

Yes, capacitors and diodes are often used together in electronic circuits. Capacitors can be connected in parallel with diodes to smooth out voltage fluctuations and reduce noise. They can also be used in series with diodes to create circuits that release stored energy in a controlled manner.