Help with power on transistor

Discussion in 'The Projects Forum' started by Dmm, Oct 20, 2015.

  1. Dmm

    Thread Starter Member

    Apr 13, 2015
    I would like to setup a circuit shown in the attached. I need help figuring the power on the transistor. I have some calculations shown in the attached. And I get 2.7 w on transistor. Can I simply add R5 shown to bring down the power? Am I running the calculations correctly?

    I am a complete newbie to electronics and circuits. I understand a few things but not all of it yet.

  2. ronv

    AAC Fanatic!

    Nov 12, 2008
    The total current is only .180 amps (6X.03). You used 12 volts instead of 9.5 to calculate the total current.
    I'm not sure what PIR is, but I will assume it can fully turn on the transistor. If that is the case the power in the transistor is lower as the voltage across it will only be a few 10ths of a volt.
    Known as saturation voltage.
    So you don't need to drop the current thru the transistor.
  3. ScottWang


    Aug 23, 2012
    1. You have to make sure how many volts of PIR output and calculate the values of R3.
    2. The led limiting resistors are better in series for each led, so re-calculate the R values.
    3. Make sure the power voltage of PIR, if it is under 50 mA then you can using 78L05,78L06, 78L09 to be the voltage regulator to provide the power voltage for PIR, used the voltage divider to supply power for PIR maybe is not a good idea.
  4. GS3

    Senior Member

    Sep 21, 2007
    I have no idea what PIR is but let us assume it provides enough current to take the transistor into saturation ( 5 mA?).

    Some mistakes I see.

    Ic is calculated as 12 / 53 which we can use roughly as an upper bound. In reality it is going to be less due to voltage drop in LEDs and Tr. But, OK, we can use 0.2 A for a nice round number.

    Power dissipated in the transistor is Vce * I = 0.3 * 0.2 = 0.060 W. You made the mistake of using 12 V which has nothing to do here.

    You could not insert a resistor there to decrease power without affecting the current through the LEDs so, in effect, it would be just like increasing the value of the individual LED resistors.
  5. GopherT

    AAC Fanatic!

    Nov 23, 2012
    Do you have a PIR module (mounted on a small PCB with three wires) or a bare PIR chip?

    If you have a sensor mounted on a PCB, do not use the bias resistors. Simply connect red wire to your 12V, black to ground and yellow will be at ground when nothing is sensed and 3.3V when something is sensed. That means you need a 150 ohm resistor between the sensor and the 2N4401 base.

    Otherwise, the power PN the transistor is about 0.5V drop across the C to E x (0.030A x 6) = 0.5V x 0.180 = 0.09 W. Nothing to worry about.
  6. Dmm

    Thread Starter Member

    Apr 13, 2015
    Thank you all for the comments/replies!

    Sorry, didn't think the PIR would get as much attention on this. It is a Parallax Rev B infrared sensor, product page. Data Sheet. The PIR (+) or pin 2 is the supply voltage and the sheet lists 3 to 6 VDC, so that's why I have the R1 & R2 voltage divider. I don't remember off hand, but I think the voltage on PIR + (pin 2) is around 3 or 4 volts in my setup.

    Total Current: This is where I get confused a little. I know the LEDs are rated for 30mA and I have 6 of them in parallel, so adding these up gets the 180 mA that @ronv and @GopherT reference in their reply. But What about calculating the current through each of the R4 resistors before the LED? Doesn't that get I = V/R = 12v / 316 ohm = 0.0379 A? ... *6 = 0.2274 A? But then if I account for the forward voltage of the LED then I have I = (12v - 2.5v ) / 316 ohm = 0.03A or 30mA each, *6 = 180mA. So after talking through that myself, that I see where the total current comes from. But just confusing to me, a newbie, when first looking at a simple circuit with resistors in parallel WITHOUT any LEDs. Now that I think about it a little bit it makes more sense. You have a voltage drop across the R4 and the current on R4 would be the 0.2274 A then? But then you have another voltage drop across the LED so in calculating the current on the LED you need to account for the R4 voltage drop first.

    So then onto the power on the transistor. 2N4401 also has "052" printed on the transistor if that makes any difference. It is from a bag of NPN transistors from radioshack. The packaging has printed on it Typical hFE=200, Vceo = 40V max, Ic = 600mA max, power dissipation 625 mW max. @ronv I looked at your link for the datasheet. The hFE in the datasheet lists at condition Ic=150mA, VCE = 1V, min=100 max = 300. Is that how you find the gain of the transistor? Look at the conditions and see which IC you are closest to? My total current is 180mA so close enough to the we are in the 100 to 300 range?

    So @GS3 in calculating the power dissipated in the transistor P = Vce * I, I assume this is not the "VCE" listed above in the "conditions" column. But further down the datasheet there is a VCE(sat) that lists max of 0.4V under conditions IC=150mA, IB = 15mA. So my power would be 0.4v * 180mA = 72 mW which is well under the datasheet PD max of 625mW.

    So theoretically the maximum current I can have on my circuit would be 625mW / 0.4V = 1562.5mA or 1.5A. And if using 30mA rated LEDs then the maximum number of LEDs I could connect in parallel would be 1562.5/30 = 52 LEDs. Is that logic correct?

    What is a safe amount to run these small transistors...maybe half of the maximum power dissipation? Just curious, don't think my circuit will grow to 52 LEDs.

    My last comments on @ScottWang reply. I just looked at the datasheet for the 78L05, these are new to me (most everything is!) I assume it's more efficient/reliable to use one of those? I did notice when the PIR is "off" the voltage at the voltage divider was higher than when the PIR was "on" and sending an output voltage. Maybe 1/2 to 1 v difference. I will stay with the voltage divider for now since I don't have any voltage regulators. I'll have to pick up a few of them sometime. Anyone have a list of advantages over using the voltage divider?
  7. StayatHomeElectronics

    Well-Known Member

    Sep 25, 2008
    The 78L05 is a voltage regulator which will maintain the voltage output at 5V over a large range of currents being drawn into the load, in your case the PIR and transistor. The voltage at the voltage divider will change as a result of a changing load, PIR "on" or "off".

    The voltage regulator is used to maintain a stable voltage.
  8. GopherT

    AAC Fanatic!

    Nov 23, 2012

    When calculating power on a device (transistor in this case), you DON"T measure power based on the power supply voltage. You measure power based on the voltage difference across the device. Likewise, you measure current flowing through the device.

    When the Transitor is in its ON state, you will measure between 0.2 to 0.5 volts across the Emitter to Collector. Power = (voltage across the device) x (current through the device) = 0.5 V x 0.180 A = 0.09Watts.

    Go back to your parallel & Series circuits lessons. Each of your 6 strings of LED + resisistor you will measure voltage as follows...

    2 V across the LED
    0.03A x 316 ohms = 9.5 V across each resistor

    Then you will have 0.5 across the transistor

    2V(LED) + 9.5V(Resistor) + 0.5V(Transistor) = 12V total power supply


    In theory, yes, 52 LEDs can be handled. That is assuming component in your circuit is running at a "Typical" value. Unfortunately, you cannot ask for Typical trnaistors, resistors and LEDs. You get what you get. Therefore, you can either calculate an error budget and determine how many your transistor can handle, or, you can just say about 50% (about 25).

    Also, you could cut down on the LED power quite a bit - 30mA is the max those little indicators should be run. Most of use 5 to 20 mA - depending on your application.
    Last edited: Oct 21, 2015
  9. ScottWang


    Aug 23, 2012
    Another way is to using the resistor and zener diode for the power of PIR, but the problem is that you have to know how many volts for the PIR, 5.1 V zener diode and 330 Ω for 5V power, 6.2V zener diode and 300Ω for 6V power, above are providing about 10~15 mA for PIR.