IR LED with transistor

Question

I’m trying to design a IR remote controller. I tried to design IR LED driver. I designed this circuit for IR driver.

I used ESP8266 for first test for IR LED. I used though-hole version of MMBT2222A (2N2222A). I tried to design Ic current for one LED approximately 100 mA. And also I chose hfe=40 for Ib calculation and I calculated approximately 1 kΩ for base resistor. When I tried to measure with multimeter, I observed one of IR LEDs current flows about 50 mA and base current is approximately 3 mA. And LEDs voltage approximately 0.7 V. I thought BJT is not properly works on for my circuit. I tested of this circuit with 2 LED. It works about 13 meter distance (sends NEC code properly) but the voltage and current values did not come out as I calculated and also I couldn't use 5 LEDs at the same time (with 5 LEDs I used base resistors and without base resistor result is very closer).

Therefore I changed BJT to MOSFET for driving LED with voltage. And I tested this circuit IRLZ44N logic MOSFET. I tried to drive this circuit with directly 3.3 V high output. That time current flows over LED about 100 mA but when I tried with continuously NEC output signal. MOSFET not fully opened. I looked voltage with multimeter I saw 0.3 V at ESP8266 output signal. Thats the reason my guess I couldn't open MOSFET with this NEC Code. I'm open to your suggestions, thanks.

Answer 1

With \$h_{FE}=40\$, base current will have to be \$I_B=\frac{500mA}{40}=13mA\$ worst case, but there's no way you can expect the IO to source or sink that much current.

By the way, your calculation for the 1kΩ resistance is incorrect. It should be:

$$ R_B = \frac{V_{RB}}{I_B} = \frac{3.3V - 0.7V}{13mA} = 200\Omega $$

You could overcome the problem of limited current available from the IO port by connecting two transistors in a darlington pair arrangement:

^{simulate this circuit – Schematic created using CircuitLab}

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Correction: I used \$R_1=10k\Omega \$, but that was wrong. I have changed R1 as follows.

In the worst case, where \$h_{FE}=40\$, combined current gain for the pair of transistors is \${h_{FE}}^2=1600\$. To obtain collector current \$I_C=500mA\$ requires base current of at least \$I_B=\frac{500mA}{1600} = 310\mu A\$. The base will be at 1.4V instead of 0.7V (there are two base-emitter junctions in series), so maximum base resistance should be:

$$ R_B = \frac{3.3V - 1.4V}{310\mu A} = 6.1k\Omega $$

I've used \$R_1=4.7k\Omega\$ to ensure base current is well above the minimum.

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You'll lose a volt or so across Q2, so I've reduced all the LED resistances to compensate, keeping current in each at around 100mA.

In this application the 2N2222 is right on the edge of its capabilities, and will get quite warm. The IRLZ44N is a much better fit:

^{simulate this circuit}

Answer 2

With 5V power to the LEDs, 0.7V across LEDs (as hypothesized in the question), 30Ω resistors, and a saturated BJT, we'd expect like (5-0.7-0.2)/30 ≈ 140mA per LED, thus ≈ 700mA total I_C. With h_FE = 40 we'd need ≈ 18mA into the base of the BJT. But we can't get nearly that with a 1kΩ resistor and like 3V out of the ESP8266 (less a V_BE); we'll only get like 2.5mA into the base!

Drastically reducing the 1kΩ resistor (like to 100Ω) might work, but is not a recommendable solution, because

• the 12mA rating of the ESP8266 output would be significantly exceeded
• 700mA is dangerously close the I_C and I_CM of 600mA and 800mA for the MMBT2222A
• the h_FE of the MMBT2222A indeed is like 40 at high current, but that's with 1V V_CE, so we could run into thermal problems with the MMBT2222A

If we stick with a BJT and the same loads, we need a beefier BJT, with more current allowance, more dissipated power, and better gain at saturation; or a darlington (albeit at the expense of increased saturation voltage).

But as pointed in comment, we can put some LEDs in series. We can't put 4, much less 5 diodes in series, because V_FWD of the TSAL6200 is more like 1.3V @100mA, rather than 0.7V as stated in the question.

Here is a minimally modified schematic that should work, albeit with only 4 LEDs, only 100mA per LED, and only little margin for the MMBT2222A current, power dissipation and gain.

^{simulate this circuit – Schematic created using CircuitLab}

Perhaps we could use 3 LEDs in series, with like 8.2Ω series resistors, if the tolerance on the 5V is tight enough. The higher the LED supply voltage, the more LEDs we can put in series, and improve the efficiency on top of that.

The only reason I see why a properly wired IRLZ44N would not work (with the 1kΩ resistor removed or drastically reduced) is due to the high GS capacitance, which will damp things, and could be a problem in applications where the signal driving the IR LEDs is a modulated square wave (which increases selectivity).