# What is the problem with my LED panel switching circuit?

I have soldered this circuit on PCB. When I switch Q1 via MCU pin-1, I see LED glowing on Q3 (not too bright though). What could be the reason for this? Even though, MCU pin 2 is pulled down to GND! It works fine on the solderless breadboard but not on the PCB.

• "It works fine on the breadboard but not on the PCB", I think you can easily conclude that there is a connection or soldering fault on the PCB. Go find it! – Bradman175 Sep 25 '16 at 3:54
• Are the Q1 LEDs all glowing brightly? – WhatRoughBeast Sep 25 '16 at 4:35
• What is the voltage at the collector of Q3? – Daniel V Sep 25 '16 at 5:00
• @DanielV voltage at the collector of Q3 is in the range of 2-3 volts. It varies. – Mat_python Sep 30 '16 at 3:34
• @Mat_python that tells me the Q3 base is at some non-zero voltage. Maybe around 1V. If you are certain that the MCU pin is true ground, somehow a current is generated across the 1k resistor at the base. A quick verification would be to replace the 1k with something smaller, say, 100 Ohms – Daniel V Sep 30 '16 at 21:51

The problem was very simple. My PCB circuit had some remnant soldering flux on it which created some conductivity. I cleaned it properly with hair dryer and it worked completely fine. Thank you all for your answers.

There must be an unwanted connection between the Q1 circuit and the Q2 circuit. If you made the pcb yourself make sure it is clean. Even an unexpected high resistance can make the led glow light.

• No. There are no connections on PCB. I tested again and re-soldered them. I think it is a faulty transistor because it showed me 0.9 K ohms resistance for reverse biased config of BE junction. – Mat_python Sep 27 '16 at 21:15
• If there are no wiring and/or silk-screen legend errors, then a logical 1 into R1 should light only the LEDs connected to Q1, so so far it seems you've made a gross error in accurately representing your circuit. – EM Fields Sep 28 '16 at 20:09

The TIP122 is a Darlington. I don't know what you are driving it with. But if it is a $5\:\textrm{V}$ Arduino or anything with $3.3\:\textrm{V}$, then you are terrorizing the Darlington with milliamps at its base!

You also are placing your strings of LEDs directly in parallel. This isn't exactly good, either. One of the pairs is likely to hog more current than the rest.

Finally, I don't know what kind of LEDs you have. But even if they are high efficiency reds, you still are only going to provide (probably) less than a milliamp through them. That $2.2\:\textrm{k}\Omega$ resistor is pretty limiting here. Even if that poor Darlington is able to sink $2\:\textrm{mA}$ (and that's not even on the spec sheet, it is so low), that's still divided into four chains of LEDs you can only hope will get the current shared among them. If equally lit, $500\:\mu\textrm{A}$ each. If not, perhaps more in one pair and less in the others.

Darlingtons are very sensitive to base currents. I'm not surprised that you are experiencing odd behaviors, given the lack of design.

The TIP122 collector is designed for $\ge 100\:\textrm{mA}$. Assuming you really are getting bright results when ON (and I'm still having a hard time with that), then try adding a $330\:\Omega$ resistor (or up to $680\:\Omega$ resistor, if that smaller value dims the LEDs too much) from the TIP122 base to ground. Do that for both TIP122 Darlingtons. Just for starters. See if that helps any. That should cut the current down to something marginally reasonable and it should provide a path for any reverse biased leakage currents, as well. I'm kind of curious how that works out.

If it does work, then I think you really need to get a decent design worked out. Write up the exact details of what you want to achieve and what you are using. This means: (1) what is driving the transistor switch; and, (2) what voltage output represents ON (and OFF, too); and, (3) what kind of LEDs are you using, what voltage do they drop when ON, and what current do they require; and, (4) what voltage rails do you have handy?

• 1.) The TIP122's absolute maximum base current spec is 120 milliamperes, so how is just a few milliamperes going to "terrorize" the part? – EM Fields Sep 25 '16 at 21:28
• 2.) I downvoted your answer because it's mostly nonsensical. You have access to online data sheets, yet post opinion instead of fact. – EM Fields Sep 25 '16 at 21:46
• @EMFields I did access the datasheet and I suppose you don't like me to speak obliquely. I didn't mean anything electronic by it. It was just a figure of speech. But a Darlington, supplying only a few mA of collector current has a large gain and does NOT need the kind of supplied drive. We'll simply have to agree to disagree since you didn't post any specific objections. – jonk Sep 25 '16 at 23:47
• First, speaking obliquely is quite different from making specious claims, of which the latter seems to apply here, since putting a few milliamps into a TIP122's base is perfectly acceptable. Second, the OP's schematic called out LTL-307EE LEDs, with a data sheet available here , so it appears you didn't even do due diligence before posting your opine. Third, the Darlington is being used as a switch, so your advice to beta bias is, at best, ludicrous. There's a lot more if you want to hear it. – EM Fields Sep 27 '16 at 0:33
• @EMFields No, I didn't believe the "LTL-307EE." Mostly because of my own experiences using CircuitLab. I think you are inflating its importance. I was speaking elliptically about the base current. Doesn't change my thoughts the drive because I was also thinking that the OP had some likely wiring problems. That's why I suggested the low valued pull-down as a test. (The lack of information flowing the OP makes me far less interested, regardless.) Have you read my other replies regarding BJTs? I am familiar with them. Have at it, though. – jonk Sep 27 '16 at 0:44

You need a resistor in each series string of LEDs in order to limit the current in each string to the required value, as shown below.

The LEDs you're using have a forward voltage drop of 2 volts with 20 milliamperes through them

You're starting with a 9 volt supply, and two LEDs in series drop 4 volts, so the ballast resistor has to drop the difference between the supply and the Led drop, which is 5 volts.

Then, from Ohm's law, $R = \frac{E}{I} = \frac{5V}{0.02A} = 250 \text{ ohms}$.

270 ohms is a standard E24 value and should easily give you good brightness from the LEDs.

• As the OP has not previously learned about using a series resistor for each series string of LEDs, it might be worth pointing out explicitly that the resistor does not have to be in the middle of each string. – Andrew Morton Sep 25 '16 at 14:05
• @AndrewMorton: That's not a bad idea, and if you think it's worthwhile, feel free to edit my post. – EM Fields Sep 25 '16 at 20:59
• I made these changes. I still have the problem with the LED brightness. I have four TIP120 connected same way for four LED panels. What could be the problem? I tested those two transistors' BE junction with 'Diode' test option on the multimeter. They showed value close to '1430' between BE (P-N). Other two working transistors have 10 K ohms of resistance between B-E. – Mat_python Sep 27 '16 at 3:06
• @Mat_python: Apples and oranges. if you tested two transistors with the diode test function of your DVM, why didn't you do the same for the other two transistors? – EM Fields Sep 27 '16 at 17:59
• Other two transistors look good. I tested both of them. I could see high resistance when connected in Reverse biased. 'faulty' transistors are brand new. What could be the problem? Should I de-solder them and replace new ones? – Mat_python Sep 27 '16 at 21:10