9v pir wiring, "T" symbol and cpc farnell

[tim_ber]

Since Nigel introduced me to the wonders of CPC Farnell I have started many small projects and spent huge sums of money.

I hope Nigel owns the company or something, because I am most grateful for the link to them and they offer great quality items, still with free delivery.

Now I have a tiny PIR. That is the only tiny thing I have; everything else is normally proportioned.

The output is only 5mA and as that is of little use I searched how to wire it up properly and found this:
http://www.etang.co.uk/datasheet/hc-sr501 body sensor/hc-sr501 body sensor datasheet.pdf

I did not know what componenet "T" was so I thought it may be where the item one wants the powered item to go and that is what I did (I think I stuck an LED there). I can now turn a small light on via the PIR. Big whoopee.

But then I looked at the spec of Farnell transistors and saw some gave out 600mA and some 100mA and some even 1A and I bought a few.

1) I am thinking the item one wants to trigger via the PIR and transistor does not go at "T", especially in light of the AC wiring diagram beneath the 12V one, so WHAT IS "T"? And what is it for? I don't understand that bit of the circuit but I know it does not work without a diode there.

2) As I have made a few circuits with the powered item at "T" is that ok even though it drains a 150mAh 9V battery quite quickly?

3) If you have told me what "T" is, then can I simply wire onto the emitter of the transistor whatever I like as long as it is withing the spec of the transistors? (continuous collector current 1Amp specified on one of the transistors I bought.

http://cpc.farnell.com/jsp/search/productdetail.jsp?SKU=SC06808
http://cpc.farnell.com/jsp/search/productdetail.jsp?SKU=SC10969

Many thanks if you answer my numpty electronic questions; the projects are providing a lot of fun and keeping my brain cells from packing up though.

 

[nigelmercier]

After a quick look, I think they are suggesting that a DC load can go where the T symbol is. Depends what you want to switch, but for anything more than 1A I would use a relay there. This will draw about 150mA through the transistor, so you need one with a gain (Hfe) of about 150 for 5mA input.

[Later] I should explain that, because if you try to switch a large load directly with a transistor it may not work. The rule of thumb is that you should have an overhead of 5X the gain when using a transistor as a switch. In other words, 5mA in, gain 150 gives 750mA out. Divide this by 5, gives switch capability of 150mA.

 

[halcyon]

Many thanks if you answer my numpty electronic questions; the projects are providing a lot of fun and keeping my brain cells from packing up though.

Relay coil, based on the diode along side.

Brian

 

[tim_ber]

Thanks.
Maybe I am getting better at this then.
I have bought some 1 A transistors and the load may not exceed that so unbelievably, I actually wired something correctly for once - I placed an LED there previously drawing maybe 100mA and the transistor did allow up to 200mA I think.

I did fry 2 different transistor before I started looking up specs of the transistors I was using.

When I started looking for relays, I started to find "power transistors", but I think the 1Amp transistors I have bought will do the job.

I was worried that the arrow pointing down from the transistor emiter on the wiring diagram was not going to ground (as I thought at first) but was going to the device to be powered, especially as I didn't know what "T" meant.

Thanks for clearing it up for me.

I am going to try and find a cheap, suitable relay now; just in case I want to power something larger. I tried Farnell but the choice was huge (not a problem for clever people) but daunting for me.
Many thanks.

 

[AngusMcDoon]

T is your load. What do you want to switch? I would use a power MOSFET with the gate connected to the trigger output of the PIR, because...

1) You can switch some pretty heft loads
2) They take virtually zero gate current - they are voltage controlled switches
3) No faffing with Hfe values, just connect it up and it works

 

[cimo]

T is your load. What do you want to switch? I would use a power MOSFET with the gate connected to the trigger output of the PIR, because...

1) You can switch some pretty heft loads
2) They take virtually zero gate current - they are voltage controlled switches
3) No faffing with Hfe values, just connect it up and it works

+1.

(T is a relay coil - with a reversed biased "freewheeling" diode shown in parallel).

 

[nigelmercier]

(T is a relay coil - with a reversed biased "freewheeling" diode shown in parallel).

I don't think you can deduce this just because there is a diode, it could easily be a simple DC load. They have shown the opto-isolated triac in the bottom diagram, so if they were implying it should be a relay, they would have drawn it properly.

 

[halcyon]

I don't think you can deduce this just because there is a diode, it could easily be a simple DC load. They have shown the opto-isolated triac in the bottom diagram, so if they were implying it should be a relay, they would have drawn it properly.

If it was a simple DC load why put a flywheel diode across it? why put a T for a DC load ?

It's immaterial what or how they have drawn it, that circuit with a relay will work, though transistor drive circuit is very basic, with a relay and the OP wanted to switch a load.

Brian