1
\$\begingroup\$

I have been working on a high-powered LED light bar. The design powers them almost directly from the mains, but limits current using a 555 and beefier NPN. Below is a simplified schematic.

Obviously the LED resistance changes with several factors, and I have blown out a few, but that was due to soldering issues on my part. Essentially I know the circuit works. The only hold-up is that I have tried two different NPN transistors (salvaged from old CRTs) and both end up with Vce voltages around 35V causing them to heat up rather quickly since the current is around 1 amp. I will link the datasheets for the two in below. I have measured the output of the timer at 2V RMS and set the resistor values for each transistor according to their Ib vs Vce graphs (around 0.2A) since neither include a Beta value. My question is: how do I bring the Vce down to around 5V or lower?

Fuji C3866 | Toshiba 2SC5386 https://drive.google.com/folderview?id=0B4n242xWd2TBOUVBZF9uQ1BDc2M&usp=sharing

EDIT: Removed terrible CAD, added hand drawn circuit These are 10W, 12V LEDs

enter image description here

\$\endgroup\$
2
\$\begingroup\$

You can't bring Vce down - you are operating it with pulse width modulation therefore the transistor is either fully-on of fully-off. What you may measure is the average Vce.

And now to the main issue - you are not controlling the current it seems from your circuit diagram - there is nothing to stabilize the PWM duty cycle - there is no current feedback hence it is open-loop voltage control.

The next issue is you need an inductor and capacitor to filter the "strong" square wave delivered by your switching transistor - the peak voltage across the LEDs will be basically rectified AC and this could kill them unless there is internal current limiting but, if there is then why both with this ornate solution?

Bad, bad circuit it seems. Take a look at this: -

enter image description here

Note the inductor L1 in series with the LED string and note Rsense to detect current and control it.

\$\endgroup\$
  • \$\begingroup\$ Very bad indeed, those poor innocent LEDs burnt to crisps :-( \$\endgroup\$ – Bimpelrekkie Dec 14 '15 at 21:59
  • \$\begingroup\$ Almost criminal LOL \$\endgroup\$ – Andy aka Dec 14 '15 at 22:32
  • \$\begingroup\$ Ok, first of all thank you for your feedback. Foremost, the inductor and feedback diode seem like a very good idea. Second, I probably should have indicated that I am used 9x 10W (12V) LEDs which I did a few tests on before hand to obtain a general range of internal resistance. As I stated, the circuit does work; the LEDs operate perfectly and with the intended PWM. The idea behind the PWM was to leave out a large resistor and maintain average current that way (which is where I seem to have forgotten the inductor) \$\endgroup\$ – Trey Carpenter Dec 15 '15 at 0:23
  • \$\begingroup\$ You are still missing 50% of the point. You need to monitor current thru the string of LEDs and adjust PWM duty cycle to keep it constant. You just cannot hope for the best that things won't drift. Maybe the LEDs themselves are better protected internally and this saves the day as you believe but I cannot check this as I don't have a data sheet or part number (I'm not going to follow ebay links btw - if it aint got a data sheet (pdf) I wouldn't buy it!) \$\endgroup\$ – Andy aka Dec 15 '15 at 8:43
1
\$\begingroup\$

You must not treat (a string of) LEDs as having a resistance ! Treating LEDs this way is a recipe for disaster (as you have noticed burning out those poor LEDs).

LEDs have a very sharp voltage-current relation. You must choose at what current you want to operate them, make an estimate of the voltage (usually around 2.5 - 3 V per LED) and then add one or more resistors in series or make a proper current source using a transistor. Your LED current is unlimited, no wonder they burn out. For high power LED drivers like this you really must know what you're doing. Be able to make power calculations etc.

Also 165 Volts is dangerous, 1 A through the transistor. Are you sure you know what you're doing ? I doubt it.

"My question is: how do I bring the Vce down to around 5V or lower?" This Vce is NOT your problem ! Your problem is much more fundamental to your design.

Please educate yourself first before proceeding. What you need to learn first cannot be explained here in a few lines.

\$\endgroup\$
  • \$\begingroup\$ I have considered and addressed your comment in the one above. I don't know what I'm doing - this is a learning experience. And Vce can be changed because I have a 25 ohm pot as R2 and it changes by +-10V. \$\endgroup\$ – Trey Carpenter Dec 15 '15 at 0:27
  • \$\begingroup\$ OK, some comments on your new schematic: Ground at bottom right: you MUST add this, without it you will be unable to control the NPN transistor. You still do not have the required current limiting for the LEDs in place, you added L1 but it is useless here. Andy's circuit has an inductor but that is because the IC that is used there is a switching converter with current feedback. You have nothing of that sort. Same for D0, not needed. You think you control Vce by changing R2 but you're not. R2 controls the on/off time of the NPN transistor. \$\endgroup\$ – Bimpelrekkie Dec 15 '15 at 6:57
  • \$\begingroup\$ When the NPN is on still too much current will flow burning out the LEDs. But I measured Vce ! Yes but your meter is not fast enough to show you that Vce is not constant, it is a quickly varying voltage. To measure it properly you need an oscilloscope or measure in a fixed position, NPN off and then NPN on. The only way you can make this design usable is to add resistors in series with the LEDs. But do take power dissipation into account. If you don't know what I'm talking about, then look it up and study it. It is basic electronics knowledge. \$\endgroup\$ – Bimpelrekkie Dec 15 '15 at 7:02
1
\$\begingroup\$

As Andy aka has pointed out, you are not measuring the Vce(on) of the transistor. You are measuring the average Vce in the presence of the PWM, and you should expect that to be much higher than Vce(on). However, since you seem to be having heat problems, I'll assume that your transistor is not being properly driven.

Your problem is most likely that you are not able to provide enough base current to turn the transistor fully on. The rule of thumb is that, when a transistor is in saturation, it has a gain of about 10 to 20. This means that, for a load current of 1 amp, you'll need about 100 mA of base current, which should be well within the capabilities of a 555. Actually, from the data sheet, at 1 amp collector current, a 100 mA base drive will only drop your Vce to 5 volts using the C3866, so I don't recommend using it. Note that hfe is the same as beta. Use the other one.

You need to keep in mind that, with a 5 volt supply, a 555 putting out .2 amps will only produce about 2.5 volts. Furthermore, the Vbe will be about 0.7 volts, leaving only about 1.8 volts across the resistor. So the resistor should be about 10 ohms.

I'll disagree with Andy about your filtering. I believe that your cap is adequate. With a 50% duty cycle your average current is about 0.5 amps, and for a capacitor $$dV/dt = \frac{i}{C} = \frac{0.5}{.001} = 500 \text{volt/sec}$$ and with a full-wave rectifier the charge period is 1/120 second, so the variation in maximum peak voltage $$\Delta V = \frac{500}{120} = \text{~}4\text{volts} $$ If your PWM frequency is high, a large capacitor may have some difficulties with it, but I don't see how this is likely to produce your transistor problem.

I'd recommend that you double-check the LED current. Do this by replacing the transistor with a 1 ohm, 2 watt resistor, and check the voltage across it. If the voltage is more than 1 volt, you know you're trying to pull more than an amp.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.