Transistor switching circuit

Discussion in 'The Projects Forum' started by Ian_Glids, Jun 22, 2015.

1. Ian_Glids Thread Starter New Member

Feb 1, 2014
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0
Hi Guys,
I am after a little help with a switching circuit. I understand it can be done simply with a transistor, and I can build the circuit once it is designed/drawn, but I am at a loss to find a circuit that does exactly what I need.

I can't provide a circuit of what I have, but I expect an explanation should suffice...

I am constructing a remote door sensor system for an Art Gallery for my brother-in-law, so that he can go inside the house of shed and know when someone enters the Gallery. We will use an IR Beam sensor from Jaycar, item: [http://www.jaycar.com.au/Access-Con...ators/Commercial-Grade-Doorway-Beam/p/LA5193] to detect the person entering, and I want to use the output from that to trigger a 433MHz transmitter / receiver [http://www.jaycar.com.au/PRODUCTS/K...Automation/433MHz-Remote-Switch-Kit/p/KC5473] to sound a piezo buzzer. The transmitter can be triggered by several means, such as the built-in push button switch, or a rising voltage spike of >2VDC, or the closing contacts of a switch.
The output of the IR Beam unit (on the side of the speaker) is via a 3.5mm stereo plug/socket, with terminals I will refer to as C, L and R (Common, Left and Right)
In the 'un-triggered' state of the IR Beam (ie; no person present to cut the beam) the voltages are as follows (using a 12V plugpack to power the IR Beam unit):
C to L = 11.5V
C to R = 5mV
L to R = 11.5V
When the IR Beam is triggered, the voltages are:
C to L = 11.5V with no obvious spike or change
C to R temporarily drops to 0V, then returns to 5mV
L to R temporarily drops to 0V, then returns to 11.5V
I'm not keen to measure the Ohms across C to L or L to R, but resistance across C to R is 36 ohms in the un-triggered state, and is open circuit whenever and while the IR Beam is cut (ie not momentary open circuit).

Could anyone provide a circuit that can either:
• swap the change from 36 ohms to open circuit to provide a positive closing switch action (which can be temporary/momentary as the normal push button switch on the transmitter board is a momentary contact switch), or
• provide a +ve voltage spike of > 2V from the temporary C to R drop to 0V.

Ian

2. ScottWang Moderator

Aug 23, 2012
4,933
777
What you linked that two pages was wrong.

3. Alec_t AAC Fanatic!

Sep 17, 2013
5,979
1,138
I'm having trouble making sense of those figures. L is a fixed 11.5V, R changes by only 5mV, and yet L-R changes by 11.5V??

4. AnalogKid Distinguished Member

Aug 1, 2013
4,704
1,300
Seems like overkill. The same site sells magnetic door switches for \$5. Wire one directly to the transmitter. Or go to ebay for wireless door switches and doorbells.

ak

5. Kermit2 AAC Fanatic!

Feb 5, 2010
3,852
968
you say the transmitter is triggered by closing contacts?
use a relay with a 12 volt coil. wire the transmitter acrosd the contacts that are open when the relay is active. use the output that is 11.5 volts and drops to zero when beam is broken. when that voltage gors to zero the relay de-energizes and your contacts will close triggering your transmitter.

6. Ian_Glids Thread Starter New Member

Feb 1, 2014
4
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Sorry Scott, maybe the closing bracket is the problem.

7. Ian_Glids Thread Starter New Member

Feb 1, 2014
4
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Sorry, can't do that... the door needs to remain open to 'invite' customers into the Gallery.

8. Ian_Glids Thread Starter New Member

Feb 1, 2014
4
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Thanks Kermit, that looks like a good option.
Cheers,
Ian