0

I used power transistor to drive high power LED (used in streetlights)

I added resistor between the Transistor Collector and the LED to verify which part of the transistor causing too much heat, and found out that it is really on the collector part.

Originally, the LED is powered up by 42 DC Volts so I wanted to keep that source level.

I added Photoresistor to turn ON/OFF the LED

The problem is that the transistor heat much.

I also wanted to keep the luminous/power dissipated by the LED as much as possible.

UPDATE: I don't know the specs of the LED but I know the original power source of it: 24V - 40V; 0.45A - 0.8A; Maximum 42V

enter image description here

Sam San
  • 185
  • 2
  • 12
  • 2 things: - (1) You need to control the LED current (normally by a series resistor or by a current controlled supply). (2) You need to ensure your BJT is fully turned ON (enough base current) or it will get very hot. – JIm Dearden Jul 16 '17 at 15:19
  • I am very bigginer. I don't know how much BASE CURRENT is considered as enough. – Sam San Jul 16 '17 at 15:20
  • 2
    When your a beginner what you do is this: find some design made by someone else and build that or get some ready-made module to achieve your goal. Your circuit is simple, too simple actually and it has certain unwanted properties you are unaware of. If the power source is a LED driver (which I expect it is) it will not be happy with this circuit as a load. This circuit is not a "proper" design. You are a beginner. Beginners cannot make a proper design yet. You need experience for that. – Bimpelrekkie Jul 16 '17 at 15:30
  • What you should do is get a module like this: https://www.aliexpress.com/item/10pcs-Free-Shipping-1-Channel-LSR-Light-Sensitive-Sensor-Relay-Photosensitive-Optical-Relay-Module-5V-For/32695422199.html?spm=2114.search0302.4.2.ooJrPf and then switch on/off the 110/220V side of the LED driver (power source). Connect the LED directly to the output of the LED driver. That will just work, no transistors getting hot, no erratic on/off behavior. – Bimpelrekkie Jul 16 '17 at 15:34
  • Yup. I am using the original LED driver, I just add additional circuit to turn it ON/OFF. – Sam San Jul 16 '17 at 15:37
  • I have worked with relay before, and I don't like the side effect of it that's why I am staying away from relay as much as possible. – Sam San Jul 16 '17 at 15:41
  • Are you saying that you are using the original power source (which is about $42:\textrm{V}$) and the original LED light that worked with that power source? And that there used to be a relay involved, but there's a side-effect of the relay that you don't like? Did the original circuit ALSO have a photo-resistor? Are you using the same one it used? – jonk Jul 16 '17 at 17:19
  • I really think that Andy's comment about using a comparator, if with hysteresis, is going to be a huge help in making this "just work nicely" for you. (You may want to include a potentiometer to allow you to adjust things.) Also, I think you will need to select (have designed for you) an appropriate "power switch" circuit driven by the comparator to switch your LED load properly. All comparators I know about, by themselves, are just way too weak. (The entire thing could also be handled well with just three BJTs. But a comparator + output driver would be the professional's choice.) – jonk Jul 16 '17 at 17:52
  • @jonk the original circuit doesn't have relay and photo-resistor. I used the relay before in different project that has LCD and I have a garbled LCD display whenever the relay is switching. – Sam San Jul 17 '17 at 07:02
  • 1
    @JamVille Then see my answer below. Hopefully, some help. You will need to pick out a photo-resistor, though. But I think it will do fine for typical units. – jonk Jul 17 '17 at 07:30

5 Answers5

2

Oh, cripes. I'll add a BJT circuit that uses your existing TIP122. You don't specify the LDR, so I can't be sure I've got the hysteresis thresholds right. But I think these will do. I'm offering this because it's probably the "least change" to what you have now and it may work okay for you, assuming that I read you correctly when you updated your question. (But I lack enough information to know for sure.) I'm also assuming here that it is easier for you to grab a couple of jelly bean PNP BJTs than it is to grab a specific comparator IC.

Here it is, below. I'm going to stay with your drawing format which buses the power supply around. I don't like doing that, because it can distract from understanding the circuit. But I suspect it may communicate better in your case. So I'll stick with the format you are currently comfortable with.

schematic

simulate this circuit – Schematic created using CircuitLab

I kept the \$200\:\textrm{k}\$ resistor near the LDR in your circuit, because I don't know what caused you to use that value. I felt safer keeping it in place, for now. You can adjust \$R_1\$ to change the light level threshold, a bit. Also, \$R_3\$ and \$R_5\$ have useful effects when you change them. But I think I really need to know what LDR you are using (datasheet?) before I can offer much better. At least, this provides a very versatile topology that can be adapted to your needs when more information is available. The basic idea is solid and it uses a key part that I know you already have (the TIP122) adding only a minimum number of cheap parts to get the rest done for you.

jonk
  • 77,876
  • 6
  • 77
  • 188
  • with this circuit, I wonder if dimming the LED can save the energy consumption or the power just dissipated to other parts like the added resistors and transistors. What can you say? – Sam San Jul 17 '17 at 11:23
  • The circuit above solved the heating problem but the dimmer effect was gone. – Sam San Jul 17 '17 at 13:48
  • @JamVille Yes. Dimming means power dissipation in the circuit. That would be a bad thing. If you want dimming in this environment, then a more complex circuit (PWM, probably) is required. – jonk Jul 17 '17 at 15:56
  • Does PWM as a dimmer can save energy? – Sam San Jul 17 '17 at 16:35
  • 1
    @JamVille Yes. PWM means that the switch itself doesn't dissipate energy since it can stay out of being in a linear mode, dropping lots of voltage. Which means a better percentage of the average power going to the load. But not every load can tolerate PWM. So whether or not you can use the technique varies by the situation. I know nothing at all about your odd LED. So I cannot tell you how it will behave under PWM. – jonk Jul 17 '17 at 23:46
  • I wish you could help me with this one https://electronics.stackexchange.com/questions/318724/how-to-drop-the-external-source-output-voltage-using-arduino – Sam San Jul 21 '17 at 17:05
1

The LED has an exponential current-versus-voltage IV characteristic. Its a diode. The transistor has more of a constant current, with variable-collector voltage, behavior. Result? Most of the VDD appears across the transistor, ensuring the transistor overheats.

Insert 1Kohm resistor in collector, in series with LED, to limit the current.

That resistor will dissipate (42-2)*(42-2)/1,000 = 1.6 watts, if the transistor is saturated.

analogsystemsrf
  • 34,253
  • 2
  • 19
  • 48
1

UPDATE: I don't know the specs of the LED but I know the original power source of it: 24V - 40V; 0.45A - 0.8A; Maximum 42V

It seems to me that if you are trying to dim the LED by dropping 18 volts across the transistor you will get significant power loss

The LED will now have 24 volts across it and presumably want to take 0.45 amps. That current will flow through the transistor and dissipate 8.1 watts in it i.e. 18 volts at 0.45 amps is an internal power dissipation of 8.1 watts in the transistor.

If on the other hand you only want the transistor to turn on or turn off (based on the photoresistor) you need to use a comparator so that there is no in-between half state that causes the transistor to dissipate all this power. You need to go down this route if you want this to work properly.

Andy aka
  • 456,226
  • 28
  • 367
  • 807
0

For a maximum base current of only \$\frac{42{\mathrm V}}{200{\mathrm kΩ}} = \bf 0.21{\mathrm mA}\ \$ you may need another transistor stage.

The typical current gain of the TIP122 is 2500. From \$I_B = 0.21{\mathrm mA}\ \$, you get \$I_C = 525{\mathrm mA}\ \$. Not the 800mA you may require. This will result in the TIP122 only "half-opening" and dissipating a lot of power as you are using it an analogue amplifier instead of driving it into saturation.

Check the specs of your lamp.

Janka
  • 14,276
  • 1
  • 21
  • 35
0

darlingtons are poorly suited for switches as they have very high Vce at saturation.

Go with a mosfet or bjt + driver.

dannyf
  • 4,342
  • 1
  • 8
  • 9