LabVIEW
Convert NI-9477 To 24V Sourcing Output
@JScherer wrote:
@RTSLVU wrote:
Here's some ideas on driving 24V devices with your DIO...
Thanks for the designs. However, these look more complicated than what the other diagram I posted looked like. Is what I posted earlier the "open collector," option? Is there a reason to use the MOSFET or Bipolar designs over this one?
Your other diagram is not equivalent and not useful for this situation. As you can see this is to define the inactive potential of a switch to a digital input. If the switch is open, a (minimal) current path is provided to pull the digital input up and make it a defined voltage. Many digital inputs when left open, will float somewhere around half of the supply voltage, which is in an area where they are not defined to operate with a defined logic level. That can make them oscillate and that will not only create a random digital signal to the rest of the circuit but will often also result in high currents in the input stage of the digital input that can even destroy the circuit (although most digital circuits are protected nowadays against that but the potential random oscillating signal is bad enough).
A pull-up on a digital OUTPUT to drive a load is useless. You would have to use a pull-up that has the same impedance as your load and this will result that your load only sees half of the supply voltage, and hence operate likely in a bad operating range. And to switch of the load you would switch on the digital output which now has to pull all the current from the pull-up resistor. This means:
state your load pull-up resistor
digital output off operates at half Vsupply also half Vsupply and dissipates the same amount of power as your load
digital output on switched off full VSupply over it and dissipates 4 times the power the load had.
So you have following very bad disadvantages:
- When your load is switched off you dissipate 4 times as much energy in the pull-up resistor as you need to operate your load
- When your load is switched on your pull-up resistor still needs to dissipate the same amount of energy as your load
- This will most likely mean that you need a VERY heavy resistor depending on the load you try to control.
- The current your digital switch will need to pull is double the current your load needs to operate
- In order to operate your load at its nominal voltage you need double the voltage supply which only makes the power dissipation even worse.
- You could reduce that doubling of the voltage supply by choosing a smaller pull-up resistor than what your load is but that will only make the ratio between the power dissipated in the pull-up resistor when your load is switched off to when it is switched on even worse.
Basically a passive logic level inverter through a pull-up resistor is a perfect heater!!!!!!!
So i have a few questions:
- What sort of load is it? What are the specs? Voltage = 24V, Current = ???, Minimum operating voltage?
- Why does this load have to be driven by a sinking output? What makes it need to have its other pole connected to Vsupply?
