john61ct said:
At the 3-4Vpc level, takes very little power in watts to get up to very serious amps.
I am unclear as to what you are saying.
3-4Vpc, is that 3 to 4 cells in series (so 3S to 4S) or are you saying 3-4Vpeak, as in HVC (high voltage cutoff)?
Answering both,
Maximizing 3S to 4S equates to 16.8V
Watts = Volts * Amps
Very Little Watts = 16.8V * Very little Amps, but... you have not defined "Very Little" or "Very Serious"
Very Little would typically be mW, say 50mW.
Anything close to 1W would create "serious heat" at the board or component level
Even half a watt (500mW) will easily smoke parts and burn your fingers
So we will call "Very Little Watts 50mW
Plug that in...
50mW = 16.8V * AmpsUnknown
0.050W/16.8V = 2.3mA
2.3mA is a TINY amount of current by the standards we are talking about (Relay Breaking), so, by the proof above, factoring in the assumptions made, I am stating that
your statement is likely incorrect.
....
Generally speaking, most "relays' are "Automotive"
They utilize a 12V coil and it likely takes a watt or two to hold them closed
They are often rated for 30A
They regularly switch loads of 10A, 20A... hundreds of watts on a 12V system
For Ebike Use we consider those unacceptable.
This is not for the current flowing (tho it can be). . . but for the potential on the lines. 36V is much harder to break than 12V.
To understand this, maximize the values.
We assume 1 Amp
We assume two cases: 1V and 1,000,000V
When the relay opens, in case 1, there is very little voltage behind the current. Definitely not enough, even with flyback, to jump the gap and ark.
In case 2, at any current, the potential is so huge that it can just ark right across the relay contacts...
So that tells you (along with the relays being STRICTLY rated for max voltage) that the distance the relay opens drives how how of a voltage potential it can break.
Voltage is POTENTIAL
Current is FLOW
A contactor is breaking a flow of current. Now you can think of this as the potential getting angry... so... Potential + Flyback (additional potential created by breaking the flow of current thru an inductive load), which is outside of the scope of this question.
... relays are also rated (or bound by) their CURRENT CARRYING CAPACITY, and this is a matter of internal resistance at the contact points, size of the contact paddles, size of the wires or studs, thermal capacity of the heat sinking material it is made of. If you exceed the current rating of a relay for short periods of time, or at a low duty cycle, this is often OK. If you do it continuously the I2R (Current Squared Times Resistance) losses accumulate in the form of power, which is heat, and the relay melts.
john61ct said:
A contactor breaking 500W at 1S needs to be just as robust as 12kW at 24S
Amirite?
I dont know, here is the math
W = 500W
1S = 4.2V worst case
500W / 4.2V = 120A
W = 12,000W
24S = 100.8V
12KW/100.9V ~= 120A
"Robust" is far too general of a term to use for the "Rating" of a relay.
Relays are rated by the current they can carry and the voltage they can stand off.
So yes,
The 1S and 24S relays in your example would require the same current carrying capacity.
No, the 1S relay would not have to be nearly as robust as the 24S
Proof:
You could run 4pcs of Automotive relay on the 1S application and they would ... possibly survive.
You could run the same 4pcs with 100V and they would fail, very quickly, if blown open under load.
If switched only at 0V, they may work.
Generally speaking costs go up for Voltage Handling. You will find this is especially true once you jump from something like 12V to 2000V.
There are guidelines for minimum spacing, wire insulation values, material insulation values, plating materials on the contacts, ... many factors.
-methods