24V DC to 5V DC Power converter - Is it OK to use?

[mntlvr]

mikes
Yes you can use resistors if the Pulse and Enable and Direction all draw a constant current but you better make sure that is the gospel and the wattage is high enough to dissipate the heat , but if these Drivers do not all draw the same current then according to ohms law the value of the resistors for one may not be the same for the others and you have to check this current under no load and full load so It is safer to just use Buck Converter which cost about 3 or 4 dollars which is a lot cheaper than what the drivers cost of the input voltage varies by more than +/- .5 VDC
So it was much safer in my thinking. R=E/I so you must know the exact Voltage each driver will drop and the total current each driver consumes. Remember the TTL or CMOS logic is only tolerant of a few millivolts of change above 5VDC

[martyscncgarage]

OR if all you are trying to do is provide signals to your drivers (enable, direction, step) you can use the 24v supply and some current limiting resistors. I'm still not sure why the prevailing opinion is that is is simpler to have an additional power supply for 5v

Given that you can get the power supplies for less than $20, vs. the time (I understand the experience can just whip in the resistors) and simplicity, (read no brainer) seems easier to me to buy the dual output power supply. It would be nice if Centroid provided the dual output but there is fear the inexperienced newbies will miswire. (It has happened....)

Marty

[mikes]

Please do not forget that the driver inputs are all using optoisolators. They already have a current limiting resistor on the driver, and is relatively simple to get or determine the current draw via a quick test with a 5v supply, from the specs of the driver, or just look at the opto being used and get the current draw off it spec. Many of the drivers I have looked at, tell you the value resistor needed for 24v operation.

The LEDs in the optoisolators need something like 10-18mA, so I don't know what you mean by "wattage is high enough to dissipate the heat". You will be safe with a 1/2watt resistor, (.018A x (24v - 5v) = .33watts max, but I think in most cases you can be closer to 10mA and go with a 1/4watt resistor. I don't understand what you are saying regarding testing under load and no load. The driver's and acorn's load have no play on the value or wattage of the resistor its simply the current needed by the optoisolator (with its internal limiting resistor) and the Vcc.

[mikes]

Given that you can get the power supplies for less than $20

To each their own, Marty. I am not saying you are wrong, just that there is a simple alternative to adding another PSU. $20 can buy a lot of resistors.

Of course, the 5v logic output of the DB25 is yet another option, but no enable output there.

[martyscncgarage]

Given that you can get the power supplies for less than $20

To each their own, Marty. I am not saying you are wrong, just that there is a simple alternative to adding another PSU. $20 can buy a lot of resistors.

Of course, the 5v logic output of the DB25 is yet another option, but no enable output there.

You clearly know what you are doing. Don't assume others have your skills and abilities though...
Cheers,
Marty

[mntlvr]

Please do not forget that the driver inputs are all using optoisolators. They already have a current limiting resistor on the driver, and is relatively simple to get or determine the current draw via a quick test with a 5v supply, from the specs of the driver, or just look at the opto being used and get the current draw off it spec. Many of the drivers I have looked at, tell you the value resistor needed for 24v operation.

The LEDs in the optoisolators need something like 10-18mA, so I don't know what you mean by "wattage is high enough to dissipate the heat". You will be safe with a 1/2watt resistor, (.018A x (24v - 5v) = .33watts max, but I think in most cases you can be closer to 10mA and go with a 1/4watt resistor. I don't understand what you are saying regarding testing under load and no load. The driver's and acorn's load have no play on the value or wattage of the resistor its simply the current needed by the optoisolator (with its internal limiting resistor) and the Vcc.

Well what I mean by wattage is P=E*I so if the voltage drop on the resistor is 19 v and the current is .018 amp then the resistor must dissipate .342 watts so each resistor must be at least 0.5 watts in size not 1/4 watt and that does not leave much safety margin so I would use 1 watt rated resistor.