Transistor for 12 Volt circuit

[AllenPitts]

Hello ETO forum,

Working on circuit that uses a PIR (motion detector) to initiate a 555 timer to turn
on 12 volt LEDs for a couple of minutes.

R2 was changed to 100k resistor because 10k kept the LEDs on for about 6 seconds and 100k is closer to minute.
(How long do you usually spend in a closet? Your wife?)

I was showing the finished prototype to my son when I happened to touch Q2. It was very hot.

So the PCB was rebuilt with a BD437g transistor instead of the BC537. Still hot.

So was looking for a transistor that will
handle more voltage and current and found
BUL471.

Some values: (datasheets on three transistors attached herewith)

BC547 Vebo 6V Ic 500mA
BD437g Vebo 5V IC 4 A
BUL471 V(br) ebo IC 2.5 A

I have read several texts on reading transistor datasheets online.
It looks like the most sensitive value is the Vebo
which is the Emitter to Base voltage when the Collector
is open.

And I think the Ic is the max current the transistor
can handle, right?

(What is the minimum amount of voltage/current to make
the transistors come on?)

So since the the load transistor is supplying 12 volts
and the Oxilam LEDs operate at 3 watts, if supplied 12 volts they draw
3 watts/12 volts = .25 amps or 250 milliamps per LED or 2 amps total for all four.

So the BC547 at 6 volts and half an amp is way overloaded.
The BD537g could handle the current but is undervalued at 5V.
The BUL471 could handle the current at 2.5 amps but not
sure I understand the Vebo rating.

Can you provide a little adult supervision on choosing a
transistor that will take load? Maybe a MOSFET instead of a BJT?

Thanks.

Allen in Dallas

 

[audioguru]

Vebo is the reverse emitter-base breakdown voltage that never occurs in your circuit and in most other circuits.

There is no such thing as a BUL471 as in your text but the datasheet for the very high voltage BUL741 shows that for a collector current of 2A then the base current should be 600mA because it is a high voltage (why??) transistor that is a poor switch. The base current in your circuit is only 19mA.
If you use an IRF540 Mosfet then with 2A its heating will be almost nothing.

 

[sagor1]

Just because a transistor has a maximum Ic of 2.5A, does not mean you can run at what without some heatsinks. All ratings are "maximum" with all other "maximums" within limits. inproper cooling of a transistor makes the junction temps too high and can cause failure.
I agree with Audioguru, try using a MOSFET like the IRF540 or similar.

 

[crutschow]

For your application you might want to consider a retriggerable version of the 555 one-shot (LTspice simulation below):
The transistor resets the timing capacitor to 0V whenever it receives a PIR signal.
It then times out only after the end of the last PIR signal.

Timing capacitor C1 is connected to V+ instead of ground for two reasons:

1. The output doesn't trigger when power is first applied (which may not be a problem in your application).
2. Any capacitor leakage current is in parallel with R1, so it can speed up the timing, but not prevent the circuit from timing out. Thus you can increase the value of R1 to get a longer time-out without worrying about capacitor leakage preventing circuit operation.

Note that I get a turn-off delay of only about 11 seconds after the last PIR signal, not the minute you stated (the time-constant of 100k and 100µF is 10 seconds).

 

[rjenkinsgb]

(What is the minimum amount of voltage/current to make
the transistors come on?)

The collector current will be equal to the base current multiplied by the gain (hfe) of the transistor.

However, turning _fully_ on when used as a saturated switch requires the collector load current to be well under (eg. several times) the base current x gain value.

That transistor has an hfe of around 30, so the load current must be well under 30 x 20mA (the approximate base current)
That's just 600mA load current, without allowing for excess drive to give good low-loss switching.


Use a FET with a low "on" resistance, they only need gate voltage to operate and it should dissipate far less heat.
You should not need to change the circuit for that to work.

eg.

https://www.mouser.co.uk/ProductDetail/Infineon-IR/IRLB8743PBF?qs=9%252BKlkBgLFf0rfIKznQZ2TQ%3D%3D

Re. the timer part, I'd use the Q1 transistor to directly discharge the timing cap (via a low value resistor) so the timer never expires as long as the PIR keeps producing signals.

Crutschow has given one example of that setup; another option is to add a series diode and resistor (eg. 100 ohms) between the collector of Q1 and C1-R1 junction in your existing circuit. That will have the same effect without needing any great rearrangement.