Need Relay help...

[CarveJunky]

Hello All,

I have a question on relay ratings... I am building a power supply using Gecko controllers. I have ~71 VDC@10.5 A. out from the Filter Cap. I want to be able to turn the power on and off for each controller. I also have a 12 VDC transformer in the box for my fans, controllers and box. I was looking at using relays to turn the voltage on and off to the controllers.

This is were I am kinda stuck. I cand find some relay's rated 110 VDC for the contacts, but they are pretty pricey. I also found some auto relays, that seem good but they rate the contacts at 20A, 30A, and 40A. They don't give a voltage rating for the contacts so I was wondering if they would be able to handle the 71 VDC I am putting out?

TIA

Chuck

[Al_The_Man]

The thing to observe with relays is the type of load you are switching, relays are usually quoted for AC resistive load and de-rated if the load is AC inductive, with a DC inductive load, these relays cannot be used at all with anything else but very small loads due to arc over, if you really want to switch high current DC then specialized relays are necessary with arc blow-out magnets etc.
It is preferable to switch the AC side wherever possible.
The automotive relays you mention are most likely rated for that current at not much higher than 12v due to the previous mentioned arc conditions that can occur.
Al.

[CarveJunky]

Al,

Thanks for the quick reply... I kinda figured I wouldn't be able to use the auto relay's, seemed to simple. Looks like I'll have to use the higher rated relays or not at all.

Chuck

[gar]

050718-2015 EST USA

CarveJunky:

You can use a relay to switch your load as I will describe later.

First, why do you want to turn off power to an individual motor? If we assume this is to allow you to adjust the lead screw position by hand, then you simply need to inhibit the driver with a logic signal. If you want to disconnect a motor electrically, then do as Al suggested.

To switch a dc inductive load with a switch or relay you need to dissipate the energy stored in the inductor some where other than between the switch contacts. Two broad methods --- a snubbing RC network, or a reverse biased diode.

We have for years (about 35) used an RC snubber with a P&B KUP relay with silver cadimum oxide contacts to switch a large inductive load. The load is a 1000 #-ft electric clutch. The excitation is full wave rectified 120 v ac or about 108 v dc average. That ratio comes from 0.636/0.707 = 0.900 . The approximate current is 1 ampere. This is used to build axle pinion preload. On average the relay may cycle every 5 seconds, 24 hrs/day, most of the year. Typical life is about 6 months. The failure mode is metal transfer in one direction creating a cone and a conical cavity. This relay was used for mechanical response time, a few milliseconds. Special relays showed no net advantage. The RC snubber was optimized for clutch drop out time rather than relay life.

The better way if response time is not an issue, this is your case, is the reverse biased diode. Draw the following circuit --- a battery, battery positive terminal to a switch, the switch output to the cathode of a diode, the anode of the diode to the negative terminal of the battery, and across the diode put an inductor in series with a resistor.

Start with the switch open. Close the switch. The diode is reverse biased so no current flows thru the diode while the switch is closed. The current thru the inductor will be zero initially and thereafter will exponentially rise to a steady state value of Vbat / R.

Once current is flowing in the inductor there is energy stored in the magnetic field.

A characteristic of an inductor is that you can not instantaneously change the current nor current direction thru the inductor.

Suppose the diode is not in the circuit, then we have only a series circuit of battery, switch, resistor, and inductor. Open the switch, it is now essentially an infinite resistor. At this point the inductor will produce whatever voltage is necessary to keep the same current flowing in the same direction. This causes an instantneous change in the voltage polarity across the inductor and enough voltage to breakdown the air gap in the switch causing an arc.

Put the diode back in the circuit. Now there is a path thru the diode, forward direction because of the polarity reversal across the inductor, and over a period of time, defined by the L/R time constant, the stored energy is dissipated in the resistor R. Thus no arc at the switch.

Parallel the contacts on a KUP11D15 (DPDT relay) and you will be fine with the diode.

.

[ViperTX]

CarveJunky....1st question...."Why do you want to turn power off to each controller?" 2nd question..."What do you mean by 'I'm building a power supply using Gecko controllers..'?"

[gar]

050720-0847 EST USA

We have not heard back from you.

As ViperTX ask --- why do you want to turn power off to one or more individual motors?

That question is basic to any other discussion.

.