what is this part? it looks like a transformer but

Hi,
what is this part ? Anyone known this part? it looks like a transformer but in one side have 15v and in other side have 0v in the circuit, I think that must be activated with another part.

It looks like a CT (Current Transformer) to me?
I am unable to pinpoint a manufacturer from the part number/s.

Hello rezaf,

it looks like a transformer of a SMPS, an pulse transformer or current transformer.

I think that must be activated with another part.

Click to expand...

Yes, thats right. Look at the ICs in the pictures. I think this is the primary side of the transformer. Look their for pulses.

If you don't find something, please tell what ICs are use.

Regards

Rainer

Thanks Friends.
the circuit that have this transformer, is useful for amplifying output of proximity sensors (NAMUR standard) or mechanical contacts in the automation system of industries.
the module have two main part. the left side of board is for input channels and have pullup resistors, two op-amp(LM224), a MUX(4052), a PIC MCU(16C711) that give address to MUX for choosing the input channel, a 8v Regulator, and a few DIP switches for jumper settings. the right side have a supply, a CAN interface IC(PCA82C250), a PIC MCU(16C711), CAN controller chip(intel AN82527), a mosfet(IRFD24) and driver of it(UC3715), two latch IC(74HC573), LEDs for displaying of channels. also there is two opto-coupler and the transformer (the transformer that attached picture of it in the first post) between this two part of module.
I think that the operation of circuit is like this (according to datasheet and my tests):
A constant voltage of typically 8.2 VDC is applied to the terminals of an input circuit. A source impedance or internal resistor (Ri, typ. 1 k?) senses the current flow passing through the sensor at that voltage. When the proximity sensor per EN 60947-5-6 (NAMUR) is damped, the current flow in the input circuit changes. The current differential from the undamped to the damped state is used to trigger the amplifier at a defined switching point and, thus, the analogue signal is converted into a digital signal.

when I give supply voltage about 15v and connect a input circuit(shunt resistor to simulate proximity sensor according to catalog)to the input channel, the module don't give 8v supply to my input circuit and thus can't detect input, thus the status LED blink constantly. when I give 15v to the right side of board and give a 8v directly to the left side of board(after 8v regulator) the left side go activating and the right side too. with this method all two part of board have related voltage, but the transformer and mosfets do not allow communication between the two left and right sides of board to detect input circuits error or reading inputs. I think that this module must have a enable circuit for complete operation. Is this true????
(the enable pin of mosfet's driver is connected to PIC and main connector and also the input pin of driver connected to CAN controller chip)

Very Thanks for your helps.
Best Regards.

Please help me to find the way for activating output of this transformer and 8v regulator. I don't know anything about this transformers and switching circuits. Please help me.