need help with simple logic circuit

[rebrumm]

since i am asking this question all will know how stupid i am ,,but here goes anyway. i need a simple circuit that will let me turn a logic high to low through a mechanical limit switch.

to be more specific my driver board, 2 axis, LMD 18245 has enable ports set high and i want to put limit switchs on X and Y to set them low and shut the axis off.

a single chip would be best

russ brumm
Clairton PA

[One of Many]

since i am asking this question all will know how stupid i am ,,but here goes anyway. i need a simple circuit that will let me turn a logic high to low through a mechanical limit switch.

to be more specific my driver board, 2 axis, LMD 18245 has enable ports set high and i want to put limit switchs on X and Y to set them low and shut the axis off.

a single chip would be best

russ brumm
Clairton PA

If these are tied high to a positive voltage, you could place a single pole double throw toggle switch in the lines to the enable ports. Connect the outputs going to the enable ports on the switches common terminals and the high on one set terminals and ground on the other set. Just verify the switch orientation to the terminals so you don't create a direct short.

Another way is to use a single non-maintained push button that could trigger a flip-flop circuit to control the on/off. Somewhat more complex, but not impossible.

Do you know what the voltage sources are that tie these inputs high? This could involve associated circuits if the inputs require 24v.

DC

[gar]

051027-1950 EST USA

rebrumm:

First, go to www.national.com . Search for LMD 18245. Then find the datasheet and print it out.

The 18245 itself has a brake input. The logic inputs to this chip are TTL and CMOS compatible. A logic 1, anything from +2 to +12 v, causes braking by shorting the motor coil.

Since you indicate that your board is enabled (not braked) when the enable line is high this means there is an inverter between the enable line and the brake input to the 18245.

You have not specified the logic voltage levels at the input to you board. You need to do this.

You are describing your board as a two axis board. This may imply that you have two dc motors using two 18245s, or there are four 18245s and two stepper motors. If this one board was driving a single stepper, then you would connect both enables together.

Lets assume the board inputs are CMOS gates, and the phasing is as you indicate. Then I would put a low resistance, maybe 390 to 1000 ohms from the enable input to common (you might call it ground). This by itself would force the logic input low and thus thru the input inverter activate braking (logic 1 at the 18245 input).

Next connect a 4N35 optical isolator between your logic + supply and the enable pin. When you turn the optical coupler on this will pull the enable input high.

At the input to the optical coupler you need maybe 10 to 20 ma to over drive the coupler. Then use a normally closed limit switch that opens when the limit is made and an appropriate current limiting resistor to achieve the 10 to 20 ma current.

If you put a limit switch at advanced and another at returned, then put these two in series with the 4N35 and opening either one will stop the motor.

.

[rebrumm]

after reading the 18245 data sheet, my driver board manual and viewing the schematic .the enable lines do serve the brake on the 18245 thru the microcontroller. if i wasn't using limit switches i should have jumpered them to Vcc. now i am really confused i can run the motors in turbocnc with or without the jumpers, high or low it doesn't matter. time to email the boards designer
thanks for your time

[rebrumm]

it just occured to me. is a logic low the same as no connection? in my case logic high would be 5v so would logic low be 0v (no connection at all?

[H500]

Logic low is and no connection is not necessarily the same thing. You need to connect the pin to 0v

[CLaNZeR]

Hi Russ

The simplest way is to take your 5volts to the chip via a resistor, say a 4k7 in value. One end of the resistor to 5volts and the other end to the chip enable pin.

Then attach one end of you normally closed switch to the enable pin of chip and the other side of the switch to 0Volts.

When the switch gets activated the circuit will get broken and 5volts will appear on the chip pin rather than 0volts without creating a short.

Regards

Sean.

since i am asking this question all will know how stupid i am ,,but here goes anyway. i need a simple circuit that will let me turn a logic high to low through a mechanical limit switch.
Clairton PA

[One of Many]

it just occured to me. is a logic low the same as no connection? in my case logic high would be 5v so would logic low be 0v (no connection at all?

No, the connection would be left floating. Not a good idea in some circuits. This could leave the inputs capable of changing states at any inconvenient time due to interference. A zero should be tied to ground on most/all inputs.

DC