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Plitch
04-23-2007, 08:57 PM
All:

As part of my project, I am using a Basic StampPLC equipped with a Stamp2 and a MAX1270 A/D converter.

One of the tasks that I must perform is the control of battery power to a couple of devices. I have chosen to do this via relays attached to two digital I/O ports on the StampPLC. The relays will be DS2E-S-DC24V items with a minimum operating power of approximately 100 mW and a nominal operating power of 200 Mw. I will use only one of the switch sets in each relay. Both will be have the coil side operated by 24 VDC. It may be possible to stagger their operation so that only one is drawing power at any moment. Each one of these devices will only operate for about 20 seconds per hour on a regular cycle of 20 seconds on, 1 hour off.

I note that on page 6 of the Stamp PLC documentation there is a diagram of exactly the setup I intend to use. The +24 VDC (from a battery bank) is supplied to the "Dout V+" terminal on the PLC, and the coil side of the relay is operated by switching the Dout 1 digital output pin on the PLC. The other side of the relay coil is connected to ground and the "Dout GND" pin of the PLC. Of course, with 2 relays I would also employ an additional Dout port.

I understand that the PLC employs "high side drivers", optical and electrical isolation, and thermal protection of these digital output pins using a IPs512G device. The last page of the documentation is a spec sheet for this and related devices.

My concern is that I may be exceeding the current carrying capacities of the StampPLC design with my circuit. Not being an electrical engineer, I looked at those specs for the IPs512G and I guess it can carry .8 amps "per leg." I have no idea what that means, or even if that is the appropriate spec to measure my problem.

I know that if I were just connecting a relay to a "naked" Basic Stamp2 I would have to employ a transistor to switch the relay while keeping the current across the Stamp digital out pins in-bounds. However, it appears from a read of this data sheet that my use of (even two simultaneously) a direct connect from the PLC to the relay drawing 200 Mw at 24 VDC would not exceed the capacity of the PLC output pins and therefor the circuit shown in the diagram on page 6 of the PLC documentation is OK for me to use.

So rather than hook it all up to and determine the limits by damaging something, I am asking for some feedback that the PLC circuit as drawn on page 6 will probably work for my project, or perhaps an alternative can be suggested. I thank you for taking the time to read this long winded question!

Pieter
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jeffjohnvol
04-24-2007, 04:07 AM
Pieter, I don't know the answer to this, but I was wondering the exact same thing. I was about to post the question until I came across your topic in the search.

I believe the high side drivers means that it can't absorb current like some of the other inverters can do, it can only push current.

I'm looking forward to hearing the answer to your question myself (can each leg support 800 ma, or enough to drive a relay).

Jeff

Paul Baker
04-24-2007, 05:05 AM
Pieter, it can carry 0.8 Amps. Your relay requires 200mW power on the coil, at 24V thats 8.33 mA current (I=P/V) which is easily handled by the IPS512G. I couldn't see what resistance the coil has so you should measure this and using I=V/R calculate what current will flow, you may need to amount some resistance to the line to reduce the current.

Jeff, a high side driver means that it acts like a single pole single throw switch with one side tied to the + supply (high side), you use the other side to provide power to your load.

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Paul Baker (mailto:pbaker@parallax.com)
Propeller Applications Engineer
[/url][url=http://www.parallax.com] (http://www.parallax.com)
Parallax, Inc. (http://www.parallax.com)

jeffjohnvol
04-24-2007, 05:10 AM
Paul Baker (Parallax) said...
Pieter, it can carry 0.8 Amps. Your relay requires 200mW power on the coil, at 24V thats 8.33 mA current (I=P/V) which is easily handled by the IPS512G. I couldn't see what resistance the coil has so you should measure this and using I=V/R calculate what current will flow, you may need to amount some resistance to the line to reduce the current.

Jeff, a high side driver means that it acts like a single pole single throw switch with one side tied to the + supply (high side), you use the other side to provide power to your load.


Thanks for the correction on the high side explanation, Paul. To clarify is that 0.8 amps per output, or total for all outputs?

Pieter, if using a relay, a small signal diode should be added across the coil to suppress back current..

Thanks.
Jeff

Chris Savage
04-24-2007, 05:20 AM
Jeff,

That rating is per leg…As for a diode, you could use a 1N4001 across each relay coil with the cathode (banded side) toward the positive supply rail.

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Chris Savage
Parallax Tech Support

Plitch
04-24-2007, 07:11 AM
Thanks for all your replies. Most helpful!!

Paul: The relay is a 24 volt and has a coil resistance of 2880 Ohms according to the charts in the spec sheet. I believe that this ( I=V/R) comes out to 8.333 milliamps, does it not? Do I need a resistor?w

I appreciate the tip about the diode and will add one to each relay circuit.

Pieter

Paul Baker
04-24-2007, 07:29 AM
You are correct, no resistor is needed.

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Paul Baker (mailto:pbaker@parallax.com)
Propeller Applications Engineer
[/url][url=http://www.parallax.com] (http://www.parallax.com)
Parallax, Inc. (http://www.parallax.com)

jeffjohnvol
04-24-2007, 09:00 AM
Paul Baker (Parallax) said...


Jeff, a high side driver means that it acts like a single pole single throw switch with one side tied to the + supply (high side), you use the other side to provide power to your load.


Upon second thought... Aren't we sort of saying the same thing?

Thanks.

Paul Baker
04-24-2007, 09:33 AM
After re-reading what you wrote, yes it is the same.

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Paul Baker (mailto:pbaker@parallax.com)
Propeller Applications Engineer
[/url][url=http://www.parallax.com] (http://www.parallax.com)
Parallax, Inc. (http://www.parallax.com)