[SOLVED] Will I break this relay if...

I have a DS2E-S-DC1.5V and am wondering if ill break it if i send 6vdc through the coil and/or if it will switch, It says in the data sheet that the max allowable voltage at 50°C is 200% of the Nominal coil voltage (so 3V), but on the next page it says under Rating/Max. switching voltage is 220VDC.

So will it still work with 6V even though its supposed to be used with 1.5V, or will that break it?

I'm not sure if i can link to the datasheet but if you just google "DS2E-S-DC1.5V datasheet" its right there.

Hi,

The nominal coil voltage is 1.5v, which according to the datasheet has a max of 3V for a DS2E-S-DC1.5V; the contact or switched voltage has a max rating of 220V DC or 250V AC. These are two entirely different entities.

OOOOOOOhhhhhhhhhhhhhhhhh.............. Ok that clears that up, so wuold you advise against sending 6v through it? i am getting 3 of them so i could rig something up where it only sends 3V or less through one at any given time.

I would absolutely advise against it! Depending on what is driving the coil, not only could you eventually damage it, you could damage the device that is driving it. When the voltage is removed from the coil the magnetic field collapses and produces a back EMF in the opposite direction of the initial voltage.

What if it was a capacitor so it quickly became within normal limits? I understand it migh still be damaging but would it break within the first time or could i do this several times since its only too much for a little while?

btw Thank you for your help

I'm not sure what you mean by driving it with a capacitor. If you're testing the relay with a power supply or battery, why not use a simple voltage divider to reduce the voltage to an acceptable level and drop a diode across the coil to pass through the Back EMF?

Here are a couple of links which might help:




Notice the diode across the coil to pass the Back EMF, also you may have to recalculate values for the 1.5V Coil Voltage in the following example:

Interfacing Relay to Microcontroller

If you only have a 6V supply, then you could use a resistor in series with the coil to drop some of that voltage, and leave ~1.5V on the coil. There are a few different versions, so the coil resistance is different on them, but here's how to calculate the resistor value.

So, here is a lot of use of Ohms Law, V = I*R.

Vsupply - Vcoil = Vresistor, so 6V - 1.5V = 4.5 V
If the coil resistance is 5.63 ohms, then the current it needs will be:
Vcoil / Rcoil = Icoil, so 1.5V / 5.63 ohms = 0.266 amps

Now, we calculate the series resistor value...
We know it'll drop 4.5V at 0.266 amps, so R is....
Rseries = Vresistor / Iresistor, so 4.5V / 0.266A = 16.7 ohms

It will also need to handle the dissipated power...
Power = V*I, or V^2/R, or I^2*R.
So, Presistor = 4.5V * 0.266A = 1.2 watts. You'll need a big resistor, or several of them in parallel.

For the voltage kick-back when the relay is turned off, use a diode (any will work) in reverse across the coils of the relay. Like this:

The line on the diode package corresponds to the line on the schematic (cathode of the device).

I had thought about a voltage divider or just decreasing the voltage to 3v but what i want is to keep the relay on for the longest time i can with a 6000uF cap, what i mean is "how long can i keep the relay on with a 6000uF cap, 6VDC, assorted resistors, and some switches?" To be honest this is for a Science project and one of the limitations is no diodes ( ). This is my circuit so far...
Another question is how can i calculate what size resistor to use for maximum time.
EDIT:: at some point the relay needs to turn off, thats why im using capacitors

Fworg64 said:

I had thought about a voltage divider or just decreasing the voltage to 3v but what i want is to keep the relay on for the longest time i can with a 6000uF cap, what i mean is "how long can i keep the relay on with a 6000uF cap, 6VDC, assorted resistors, and some switches?" To be honest this is for a Science project and one of the limitations is no diodes ( ). This is my circuit so far...
Another question is how can i calculate what size resistor to use for maximum time.

Click to expand...

To get the maximum on-time, you need to overcome the turn-on voltage, which on the spec sheet is about 1.05V for that part. Additionally, you'll want to discharge as little energy as possible (so your cap voltage stays up, longer). The drop-out voltage on that part is pretty low, too... I think I read around 0.15V??? So the relay will stay closed until that threshold is hit for the coil voltage.

Having a higher battery voltage will mean a bigger resistor to keep from blowing up the relay coil, which means it will be burning off power too. It sounds like you'll have to strike a balance of max voltage, extra energy lost to the resistor, and current draw rate. I'd start with a simulation or write down the fundamental equations and try several implementations to find the best balance.

You could set up a simple transient analysis in SPICE to run the numbers and try some values. I know LTSPICE is freeware, maybe you could start there, rather solving the RC charge/discharge equations by hand (or use Excel)... lots of options to experiment with values.

Since you aren't concerned about the turn-off transient (going until the supply voltage dies), the diode is a moot point.

Can you provide the project mission statement, guidelines, etc? Just want to be sure we're heading in the right direction.

You too have been more then helpful, Thank you.
So would you for sure recomend just lowering the voltage to 3v (just 2 AA batteries) so i dont have to worry about another resistor?

Then my only numbers are
1. The voltage it takes to turn on (1.05V)
2. The voltage it must drop to to turn off (.15v)
3. The current it takes to stay on (133.3mA)
4. The resistance of the coil itself (11.3Ω)
5. Any additional resistance to fine tune it (XΩ)
6. Starting Voltage (Vs = 3v)
7. Total Capacitence (6000uF) -semiadjustable

In order for it to run it must
1. Start above 1.05v - Starts at 3 when t=0
2a. Stay above .15v - (0 = (Vs*e^(-t/RC))-Vc) Vc = .15 How long it takes the capacitor to discharge to .15v will be determined by the total resistance (R = 11.3Ω + XΩ)
2b. Stay above 133.3mA - (0 = (Vs/R)e^(-t/RC))-I) I = .1333

In order to find how many seconds it stays on i will graph 2a and 2b and when either one hits 0 thats how many time constants it will run, number of timeconstants * timeconstant = time in seconds

Only 2 vars are R and t.

What happens if i solve the system? -Ill try it now.

Thank you once agian for your helps(plural).

The RC circuit has a exponential decay:

RC Circuit Exponential Decay

I would leave no stone unturned in your quest for maximum on-time. A slightly higher voltage may be a trade-off you could play against energy lost in your series voltage-limiting resistor.

For a cap, energy stored (in Joules)... U = 0.5*C*V^2. So, if you double the voltage, your stored energy goes up by the square (2^2 = 4x). Quick physics info on caps

In the end, you are trying to make the best use of your stored energy. You have a store of X, and a bleed off of Y per second. X and Y are dependent upon your initial voltage, size of the cap, and how quickly you remove that energy from the storage device.

If the equations get too ugly, you could plug them into Excel and use that to do some of the calculations, then look at some specific measurements, like you identified before... voltage cut-out limits, and when they are reached.

You are on the right path, now that you know the "knobs" you can adjust on this system (voltage, resistance). Now you start making permutations and see if you can find a trend... then track it to the peak value. A very cool problem... if I had the time, I'd be working it right along with you!

After many hours of work i have found that the equation:
Y=-1.255488X^2 + 34.624096X - 143.07392
Y=Seconds on
X=Resistance
accuratly represents the data.

Well at least it matches all the math i did, the relay doesnt actualy get here till tommorow.