Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

Interfacing interlocks with pic controller

Status
Not open for further replies.
Hi,

the circuit is correct now.
Hint: instead of 2 x BAT54 you may use one BAT54S

The circuit will protect the input.
But it will not suppress contact bouncing.

Both informations are already given in this thread:
* contact bouncing time
* how to calculate tau

Klaus
 

KlausST said:
Hi,

the circuit is correct now.
Hint: instead of 2 x BAT54 you may use one BAT54S

The circuit will protect the input.
But it will not suppress contact bouncing.

Both informations are already given in this thread:
* contact bouncing time
* how to calculate tau

Klaus
Click to expand...

For Tau,

10K X 100nF,then it have to divide by 1000 for milliseconds..?

- - - Updated - - -

img_5.jpg

Pull up resistor is placed in right position..?
If we using an interlock signal as +5VDC instead of GND means also shall we use this same circuit expect pullup resistor..?
 

Yes, all correct. The protection is against voltages outside of VSS and VDD so it will work when the interlock is active low or active high, just swap the pull-up resistor for a pull-down instead.

Brian.
 

Hi,

No, I don't think this is correct.
There are two series resistors, one for overvoltage protection, the other as filter.
Both corrupt V_IL to the microcontroller input, as they act as resistor divider in combination with the pullup resistor.

Use the pullup resistor at the most left side, then you are safe.
Btw: I'd go more low ohmic with the pullup. I'd say 10mA as switch current is not too much.

Klaus
 

22K pull up/down with 2K total series resistance to the switch should give a low of 0.4V or a high of 4.4V. Most 5V PICs have a low guaranteed at 0.8V and a high at 2V so technically it should be OK.

As this is an interlock and therefore not driven, I think I would drop the 1K series resistors to say 470 Ohms if there were worries about logic levels not being reached. If in doubt, move the pull up/down resistor to the interlock input instead of the PIC input. The circuit should function exactly the same.

Brian.
 

Thanks for your kindly support friends,now it working fine...
 

Hai,
There is an cooling fan in my machine,which attached with an heatsink to reduce the heat.
I have to connect the cooling as one part in interlock.
Cooling fan is 24VDC/0.164A to 0.240A.

When the fan failure to work means,total system should not run.For monitoring the fan,tried to measure the current during running condition and rotor blocked condition.
Above the values are measured.
I used ACS712 to measured the current.But its not working fine.
Is there any way to measure the current or to monitor the operation of the fan.
 

Current measurement can hardly act as a reliable fan operation monitor. Depending on the failure mode, the actual current may be any value between zero and overcurrent, including almost regular current.

Popular fan monitoring methods are
- fan with inbuilt speed sensor
- true airflow sensor (e.g. self-heated PTC)
- heat sink temperature sensor
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top