I made a circuit and it worked for a while with three sets of five LEDs connected to it but I added another four sets and it stopped working pretty quickly. The LEDs are only 20 mA so this thing was only drawing like <150 mA and I think these transistors are rated at 600 mA or more, right? The 2 V signal comes from PIR sensors but that's not really important. Should these transistors without heatsinks not be okay in this situation? Which would be the most likely to break and if it does, can it take out others with it? Thanks.

Perhaps I could put two of my transistors in parallel to support the LEDs with less chance of one breaking. Perhaps I bought some crappy transistors.

 

Perhaps the transistor is close to the hot LEDs. The higher the ambient temperature, the lower the permissible power. I suggest using a transistor with a large gain and a larger dissipated power.
See

 

Since the led always light up when the sensor is triggered and not flashing, so the draw current just using its 80% as 16 mA, and the total current is I = 16 mA x6 = 96 mA, if the bjt still feel any warm or heat then you can change to 2sc1384, that is 1A, and when you use the 2sc1384 then you don't have to worry about the temperature any more.

 

I also have some 2SD468 that I savaged from old TV sets. It is better than 2n2222 for your application. If you dont have any Jap transistor, you may use 2N3019 (TO5) It is a 1A 5W transistor.

But I think ZTX690B is a good choice as it has good DC gain which lowers the Vc(sat) when operating at higher Ic current.

Allen

 

It is good transistor.

 

With the added LEDs the base current of the driver transistor may not be enough. Try reducing the value of the 4k resistor to, say, ~2k.

 

Wow, Collector current = 2A and 1.5W in TO92 footprint. Should stock some in my part drawer just in case....

Allen

 

With the added LEDs the base current of the driver transistor may not be enough. Try reducing the value of the 4k resistor to, say, ~2k.

Insufficient base drive is almost certainly the problem, but decreasing the 4 kΩ resistor alone won't quite do the trick: the 10 kΩ resistor connected to it has to be reduced also.

I'd recommend reducing the 10 kΩ to 2 kΩ, and the 4 kΩ to 1 kΩ. That will give the 2N2222 5 mA of base drive, which should be sufficient to ensure it stays in saturation despite the added collector load current.

 

I extracted the 2N2222 which takes current from the LEDs and the base and collector are now shorted.

The problem is not heat from the LEDs because they are not in physical proximity to the transistor.

I have a salvaged transistor labelled C6144 SG 1CS and a FDB6035AL. Would either of these be likely to work?

 

I have a salvaged transistor labelled C6144 SG 1CS and a FDB6035AL. Would either of these be likely to work?

Either one should work. But the slow turn off from the capacitor may still be problematic.

 

Thanks. The capacitor is meant to enable fading on and off; is that a problem? Would it need to be a different value with a different transistor?

 

Thanks. The capacitor is meant to enable fading on and off; is that a problem? Would it need to be a different value with a different transistor?

The potential problem is the extra heat dissipation in the transistor when operating in the linear region.

If you don't have a fade, and only allow the transistor to operate as a saturated switch, then you minimize power dissipation. When the LEDs are off, voltage across the transistor is high, but there's zero current, so no power dissipation (15V x 0A = 0W.) When the transistor is fully saturated, current is high, but voltage across transistor is minimized (often around 0.3V or less CE drop if I remember correctly) so power dissipation is relatively low (0.3V x 0.15A = 0.045W, or 45mW.)

However, if you use the capacitor to create a fade, or if you continuously run with inadequate base current as in your current schematic, then you have significant voltage across the transistor and significant current through it at the same time, creating much higher levels of power to dissipate. These numbers are off the cuff guesstimates, but as you fade you might see scenarios like:
2.5V x 0.10A = 0.25W (250mW)

My math here is fuzzy at best, and based on a lot of assumptions about the rest of your circuit, but the point of the examples is that a transistor can run relatively cool switching a lot of power when fully saturated, but will run much hotter in the linear region.

 

the 10 kΩ resistor connected to it has to be reduced also.

Correct. I forgot that

 

Hi,

Begin your troubleshooting by.

removing the capacitor, then wire jumper bypass the PNP, remove all input signals from the front,
then check VCE as shown. Leave out the 1K at this time.


You may have around 2 to 3 volts.

As mentioned you need to get more base current, so take a voltage reading across one of your 160 ohm resistors, solve for current through it, then multiply by the 7 branches overall. that will give you the collector current.

one branch may have around 12 to 15 mA flowing through it, so up the current through each branch, making sure not to exceed power rating of the transistor.

Try 20mA per branch, which is around 140mA collector current.
That's a lot of current, so you need to saturate the transistor, so as to keep power disp. down.
See if a 1K resistor at the base will give a VCE of 400mV or less.
If it does it may light up the whole panel of led's. If not than lower the value until VCE is in the 400mV region.
Remember to keep the transistor power rating under controle.

This way you can at least see if your transistor can drive this load. SAFELY without burning out.

Then from there begin step by step adding in the other components, when you add the PNP the 1K may need to be lowered and so on.

 

What about just replacing the whole lot with a BC337-25 transistor (which I should have used in the first place!). It's much more sensitive and with the voltage at the base at ~0.7 V, the voltage at the collector is like 0.05 V. The fading with the capacitor takes a few seconds.

 

Good design change, nice job.

 

20mA in the 160 ohm resistors causes a voltage drop of 3.2V. The LEDs have a total of 15V - 3.2V= 11.8V and since there are 5 LEDs in series then each LED is 11.8V/5= 2.36V. Then they are not 2.0V red LEDs. If they are 2.0V red LEDs then their current is 5V/160 ohms= 31.3mA and the seven strings total 219mA, not 150mA.

The datasheet for most little transistors like a 2N2222 says that the base current must be 1/10th the collector current for the transistor to be a fair switch. Then the base current should be 21.9mA. But the base current in the first circuit is only 1V/(14k)= 1mA which is MUCH TOO LOW. The 14k total must be reduced to about 636 ohms then the fade capacitor must be increased to a massive 10,350uF which is crazy. Use a power darlngton transistor instead.

 

20mA in the 160 ohm resistors causes a voltage drop of 3.2V. The LEDs have a total of 15V - 3.2V= 11.8V and since there are 5 LEDs in series then each LED is 11.8V/5= 2.36V. Then they are not 2.0V red LEDs. If they are 2.0V red LEDs then their current is 5V/160 ohms= 31.3mA and the seven strings total 219mA, not 150mA.

The datasheet for most little transistors like a 2N2222 says that the base current must be 1/10th the collector current for the transistor to be a fair switch. Then the base current should be 21.9mA. But the base current in the first circuit is only 1V/(14k)= 1mA which is MUCH TOO LOW. The 14k total must be reduced to about 636 ohms then the fade capacitor must be increased to a massive 10,350uF which is crazy. Use a power darlngton transistor instead.

Why not just use a single BC337-25 transistor?

 

Why not just use a single BC337-25 transistor?

Because its base current needs to be so high that you will need a massive and expensive 10,350uF fade capacitor. Also,the BC337 is tiny so it might melt from the heat like your 2N2222 did.

A darlington power transistor is larger than a tiny BC337 so it can withstand much more heat, and it needs much less base current so the original resistors and small capacitor can be used.