Under-Rated Load Resistor Worry 2

[Oaklandishh]

I am working on using USB serial to communicate with a high speed spindle driver to automatically run a spindle.

I am using an ATMEGA to flip transistors and manually set each of the required serial pins to either high or low to get the outcome I want on the spindle driver.

The schematic and text in the operation manual require that each pin get 24VDC and at least 100mA.

The schematic shows some resistors in line with the solid state relays. while the device was turned off I measured their resistance to be 4.3k ohms each, but I was planning on putting my own 250ohm 10W resistors in to be certain about there always being a load on the power supply I sourced.

The device isn't that big, and I am worried that the 4.3k ohms of resistance is not rated for the 50W it would need to be for the specified power they called for.

Am I going to completely mess this driver up? Does my plan even make sense and am I reading the schematic correctly?

Please see attached picture or the manual:

http://www.nsk-nakanishi.co.jp/industrial-eng/download/pdf/ispeed3_en.pdf

Pg.20

Thanks,
-Eric

 

[MacGyverS2000]

I'm not sure what you're getting at... if you're measuring the resistance of the SS driver, you're measuring the LED in the driver. The resistors inline with the drivers are likely voltage drops (and an extreme waste, IMO). The power @50mA, however, is only 1.2W... I have no idea where you got 50W.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[Oaklandishh]

If I am running 100mA through the solid state relays, and I measured the resistance to be 4.3k ohms wouldnt the power dissapation be P=I^2R so .1A*.1A*4300ohm = 43 watts?

http://www.wolframalpha.com/input/?i=100mA*100mA+*+4300ohm+

I measured with a digital multi-meter between points 1 and 2 and I assumed the item in the box was a resistor, was this not a fair assumption to make?

http://i.imgur.com/7tLD9ti.png

I am just worried their resistors in line with the SS-relay, which you called voltage drops, are not rated for what I thought was the amount of power dissipation they needed. I really don't want to blow anything out on this part.

Thanks for the quick reply!

 

[IRstuff]

BUT, and a BIG BUT, 100mA * 4.3kohm = 430V, which your circuit couldn't possibly supply. The correct thing you can calculate is that it's probably in parallel with the load, and will only siphon off 5.6mA from the load.

your 250ohm resistor, however, will siphon off 96mA, or almost all of the current capacity of the supply.



TTFN
faq731-376

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Of course I can. I can do anything. I can do absolutely anything. I'm an expert!

 

[Oaklandishh]

That is a good point, but because these resistors(?) are in the internals of the control unit, I wouldn't be able to put it in parallel I only have access to put them in series and even if I could put them in parallel I need the 100 mA to go through the resistors in order to get the solid state relays to activate.

If the power supply can only put out 2.4 watts then I wouldn't be able to trigger the relays with this supply which is supplying what they called for. This makes me think I misread the diagram and/or am over thinking this?

Thanks again for the replies.

 

[Compositepro]

A multi-meter on resistance setting cannot accurately measure the resistance of a solid state device.

 

[Oaklandishh]

Thanks for the heads up, How should I go about figuring out if I am going to fry this device then?
Do I have to either pop it open or talk to the manufacturer?

Is it safe to assume since they called for 24v and 100mA that it is safe to put that in even if my application is not the normal one?

 

[LiteYear]

I measured with a digital multi-meter between points 1 and 2 and I assumed the item in the box was a resistor, was this not a fair assumption to make?

The assumption is fine, the measurement is not. The resistor is in series with a diode. Your measurement includes the forward bias voltage of the diode, which is almost impossible to discount accurately.

Is it safe to assume since they called for 24v and 100mA that it is safe to put that in even if my application is not the normal one?

Yes. The resistor is selected by the designers to provide exactly the right current when supplied by a low impedance, 24V DC source. As long as you're only supplying 24V, you wont blow it up. If your power supply is not powerful enough to supply 100mA (ie. at least 2.4W per input), it might not work, but it definitely wont blow it up.

 

[LiteYear]

It worried me that your numbers imply 2.4W per pin. That's 8 x 2.4 = 19.2W just to drive some SSRs! So I had a look - as far as I can see the rating of 100mA applies to all input pins! So when all SSRs are on (in parallel), they are drawing 2.4W total, or about 300mW each, which is far more reasonable.

 

[Oaklandishh]

Thank you! your responses are very helpful to me.

So I guess I can use the one power source to activate one pin at a time at 100mA or multiple at lower current and the SSRs should still trigger.

Thank you for all the help and corrections everyone!

 

[MacGyverS2000]

100mA would be with ALL pins activated. For 10 pins, that's 10mA EACH. If you have 3 pins on, they will pull a combined 30mA.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[Oaklandishh]

I was going to wire the circuit as attached, If I do this, it will be supplying 100mA regardless of how many pins I have on correct? How would I wire it to not do this? Do I even want to?

If the device is rated for 100mA then it doesnt make sense to over complicate it?

Thanks,

-Eric

http://i.imgur.com/f0JJyVV.png

 

[LiteYear]

I was going to wire the circuit as attached, If I do this, it will be supplying 100mA regardless of how many pins I have on correct?

No, the internal resistors will limit each pin as necessary.

How would I wire it to not do this? Do I even want to?

Take your 250R resistor out. The internal resistors already do what you're trying to do.

If the device is rated for 100mA then it doesnt make sense to over complicate it?

That's it. The subtle aspect you're struggling with is that the device is rated for a supply with a maximum of 24V and a minimum of 100mA. As long as your supply doesn't present any more than 24V, and can deliver at least 100mA, it will work. The resistors will take care of restricting the current to the necessary level (something less than 100mA, depending on how many pins are on). We (electrical people) are so used to dealing with voltage sources that this subtlety barely enters the conscious.

Let me give you the tired old water and hose analogy to give you a mechanical perspective. The device is a spinning sprinkler with a configurable number of spouts. It is rated at 24kPa of pressure and 100mL/minute of flow. If you supply it 24kPa and don't turn any of the spouts on, it will draw no flow. If you turn one spout on, it will draw some flow but definitely not 100mL/minute. If you turn on all spouts it will draw up to 100mL/minute. If your supply can't provide the required 100mL/minute, the water will only dribble out of the spouts and it might fail to spin. If your supply can provide 10000mL/minute, the sprinkler will still only draw 100mL/minute provided there's 24kPa of pressure. If there's 50kPa of pressure, you might force 200mL/minute through the spouts, spinning the sprinkler to destruction. So to correct operate the sprinkler, you need to set the pressure tap so you get close to but not more than 24kPa and at least 100mL/minute.