[SOLVED] reason of reverse current in circuit

Status
Not open for further replies.
Hi,

Please take a photo of your circuit, where we can see the wiring.

Klaus
 

KlausST said:
Hi,

Please take a photo of your circuit, where we can see the wiring.

Klaus
Click to expand...

please see the below pic. green led is on transmitter side while yellow is on receiver. let me know if you need focus on something.







a little more closure look of above pic


 

I think you have answered your own question with that photograph.
Where are the resistors on the schematics you gave us? You only showed the IC but clearly there are some other components on the board and I strongly suspect one of those resistors is a 'pull-up' on the DI pin and that is the source of your current.

Brian.
 

Hi,

That's why I like the photos. They show the reality.

Klaus
 


what are other components you are referring to ? i have rs485, led,registers and power supply only. And I have not kept any input pin floating.
how come DI pin has a pull up register? the connected register with DI is attached to GND which should be pull down.
please bear with me, i might be asking silly and basic questions. consider me as a beginner who wants to make some good quality circuits buy experimenting things and learning fundamentals wherever is needed.
 

Hi,

On the PCBs you bought ... there is not only the bare MAX485 IC, there additionally are at least 7 resistors, 2 capacitors and a LED.

Klaus
 

Even on the photo, you can clearly recognize a pull-up resistor R4 at terminal DI. Measure the resistor value or read the marking, post it here and mark the thread as solved...
 

KlausST said:
Hi,

On the PCBs you bought ... there is not only the bare MAX485 IC, there additionally are at least 7 resistors, 2 capacitors and a LED.

Klaus
Click to expand...

are you referring below schematics from the data sheet? so what is the solution for this? all my pins are occupied.






 

The schematics in the data sheet are correct, I'm afraid you are confusing the MAX485 IC with the board you are using. The board has additional components, the pins marked on the PCB also go to other components. There IS a MAX485 on the board of course and I'm sure it is working as it should.

Without seeing the schematic of the board we can't be sure but it appears there is a resistor connected between DI and the +5V pin, probably R4 and it is that causing the current you are seeing from the DI pin to ground, not the IC itself. I would guess the board manufacturer added it so the pin doesn't float if DI isn't driven externally. I am surprised it is low enough in value to light an LED like that, personally I would use a value that would only allow a few uA to flow but on the board it must be quite low in value.

Brian.
 

FvM said:
Don't refer to the datasheet, look at your real hardware.

View attachment 165376
Click to expand...

I have read the values on the h/w this is as below.
initially my connection to PIR was not working as expected hence i diverted myself to this direction (thinking of this reverse current is main cause)
as a user of this MAX485 do i need to dig more on these values ? even simply connecting PIR sensor is not enough as per instruction. provided that I have not kept any pins floating.
what should be my next step?






showing my old circuit with PIR sensor back


 

O.K., you have 10k pull-up at DI. But what's the problem with it? Doesn't HC-SR501 provide a push-pull output?
 

FvM said:
O.K., you have 10k pull-up at DI. But what's the problem with it? Doesn't HC-SR501 provide a push-pull output?
Click to expand...
yes it seems like that. i have added a LED on PIR output and one on receiver end. if i understand correctly both should turn on and off in sync based on the motion detected. but in pic if you see only receiver LED is on. just to let you know if i test PIR as a standalone unit attached with LED (using TTL signal) it works properly which mean my PIR unit is proper.



 

Hi,

Inputs must not kept floating.
Unused outputs: leave unconnected

I assume your LED connection is not correct - but as so often - your photo does not match your given schematic.
This makes helping difficult.
It seems your LED is connected in series. This is wrong. And every LED needs a current limiting resistor. I don´t see it.
Just connect the LED to DI as you connect the LED to DO.

Klaus
 


I have not kept any input pin floating.
i thought current limiting register was not needed as led is getting power from data pin of PIR sensor which is in mA. any way i have added this.
still no effect. JFYI i have kept led on transmitter side just for troubleshooting. mapping signals on the receiver side's led. one led is off and other is always on.
below is my current circuit wiring and schematics.




 

Connecting R2 and LED this way makes no sense. In combination with pull-up on RS-485 module, the input will be permanently high. No way to signal a state change of PIR sensor.
 

FvM said:
Connecting R2 and LED this way makes no sense. In combination with pull-up on RS-485 module, the input will be permanently high. No way to signal a state change of PIR sensor.
Click to expand...
so what should be my correct schematics? please advice.
and if it is always high then led should turn always on
 

The LED and it's current limiting resistor are to monitor for the presence of a voltage. If a voltage above the forward voltage of the LED is present it will let current flow and it will light up. To check for the voltage it needs to be across that voltage, you have it wired in series and probably the wrong way around too.

It should be wired with the sensor output directly connected to DI and the resistor and LED wired in series from that connection to ground. The cathode end of the LED (the end the arrow points towards) should be the grounded end.

If your schematic is accurate and the sensor produces a 3.3V output, check what its maximum supply voltage should be. It is unusual for a 5v operated device to have a 3.3V output, usually it is the same as the supply.

Brian.
 


if i can read the PIR datasheet correctly this is what i understand. operating voltage 5 to 20V and output voltage 3.3V
i will correct the wiring and try again.

◦ Voltage: 5V – 20V
◦ Power Consumption: 65mA
◦ TTL output: 3.3V, 0V ◦
◦ Lock time: 0.2 sec
◦ Trigger methods: L – disable repeat trigger, H enable repeat trigger
◦ Sensing range: less than 120 degree, within 7 meters
◦ Temperature: – 15 ~ +70
◦ Dimension: 32*24 mm, distance between screw 28mm, M2, Lens dimension in diameter: 23mm

 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…