# Sziklai Darlington Transistor Configuration

I'm building a infrared 38khz emitter and I would like to use 4 or 5 LED in multiples directions.

I found this article, where in the end it shows a Sziklai Darlington schematic and dont use resistors. I'm not sure if it is OK.

Based on the article, I made this example considering 2 LED emitters. I also included R2 and R3 to limit the current on LEDs.

It's not clear to me the current flow.

How do I calculate (or check) if it will work as I expect?

Do I need base resistors on T2 and T3?

It can be calculated with tipical values.

• That link you provided suggests a fairly bad circuit design. I'm glad to see that you had the sense to insert resistors. However, even then it still not a particularly good design approach. It's just better than what your link suggested. What is the LED type? Do you have a link or datasheet for it? (I think I already gather that you have +5 V and also +3.3 V available to you.) – jonk Oct 1 '17 at 16:52
• I'm using TSAL6100 from Vishay. And I will drive from a ESP8266, but I will have 5V and can get more current from that (I'm thinking 250mA peak) – BrunoAraujo Oct 1 '17 at 17:08
• Will you be operating them continuously at 250 mA? – jonk Oct 1 '17 at 17:10
• No. It will be used as infrared emitter for TV and Ar conditioner – BrunoAraujo Oct 1 '17 at 17:11
• Do you know what the duty cycle is likely to be? I'm asking all this because it looks like about 1.5 V at 250 mA, or almost 400 mW. However, the package says 230 C/W thermal. You need to be careful here. But I suppose you are telling me that this is probably not an issue? (If so, I accept that. Just want your attention to this for a moment.) – jonk Oct 1 '17 at 17:14

Staying with BJTs (MOSFETs might be a good option, too), the design starts out with your TSAL6100 IR LED, being operated at $250\:\textrm{mA}$.

From the datasheet, I find: And derive the following:

1. $R_{typ}=\frac{2.2\:\textrm{V}-1.35\:\textrm{V}}{1\:\textrm{A}-100\:\textrm{mA}}\approx 1\:\Omega,\quad\therefore V_{250\:\textrm{mA}}\approx 1.5\:\textrm{V}$

2. $R_{max}=\frac{3\:\textrm{V}-1.6\:\textrm{V}}{1\:\textrm{A}-100\:\textrm{mA}}\approx 1.6\:\Omega,\quad\therefore V_{250\:\textrm{mA}}\approx 1.8\:\textrm{V}$

So the worst case I might expect from the IR LEDs is $1.8\:\textrm{V}$. The difference between $5.0\:\textrm{V}$ rail and a $3.3\:\textrm{V}$ rail is $1.7\:\textrm{V}$. These details suggest to me that we might be able to operate a simple NPN in a TO-92 as an emitter follower.

simulate this circuit – Schematic created using CircuitLab

If this could work out, it has a strong argument of simplicity going for it. Let's look over the details.

Assume your I/O pin can be assumed to be from $3.2\:\textrm{V}$ to $3.3\:\textrm{V}$ and let's also assume a cheap 2N4401 in a TO-92 package with a guaranteed minimum $\beta=100$ at $I_C=150\:\textrm{mA}$ (see OnSemi 2N4401 datasheet) with a maximum $\beta=300$. We'll skate on slightly thinner ice here and assume we can "get" $\beta=100$ at $I_C=250\:\textrm{mA}$. This means your I/O pin only needs to source $2.5\:\textrm{mA}$ into the base, assuming that the BJT can stay out of saturation, with $\mid V_{CE}\mid \ge\mid V_{BE}\mid$.

You may be able to handle 5 of these with a single I/O pin. Maybe. As this would be $12.5\:\textrm{mA}$ total for 5. (You can just tie the bases together.) The ESP8266 I/O pin is limited, I think, to $12\:\textrm{mA}$ total. So... maybe.

From the datasheet, it looks like we should expect about $V_{BE}\approx 900\:\textrm{mV}$, so the emitter would be about $2.3-2.4\:\textrm{V}$. With the LED in the collector leg and requiring from $1.5-1.8\:\textrm{V}$ for its own operation, the NPN collector will be from $3.2-3.5\:\textrm{V}$. These details suggest that the NPN can indeed be kept out of saturation here, since $800\:\textrm{mV} \le V_{CE} \le 1.2\:\textrm{V}$. So that's another good mark for it.

For temperature calculations, things get more difficult. Power in the NPN will be from $200-300\:\textrm{mW}$. Which is ... significant. With a typical package to ambient thermal resistance of $200\:\frac{^\circ\textrm{C}}{\textrm{W}}$, this is an increase over ambient of from $40\:^\circ\textrm{C}$ to $60\:^\circ\textrm{C}$. Which is not at all small.

So this worries me about the TO-92 packaging.

Except that you are saying that this is very intermittent. So, I think you might consider giving the 2N4401 a try. It may work out fine for you.

If you are worried (and perhaps you should be), or just want a back-up plan while still keeping the simplicity of the design, then perhaps consider a D44H11 in the TO-220 package. These can be left open to the air with $62.5\:\frac{^\circ\textrm{C}}{\textrm{W}}$ or else bonded to a heat sink for much better figures. It also has a relatively high $\beta$ at fairly high collector currents, so it is probably a good pick on that basis, as well.

Either way, $R_1$ is easy to set. We already know the voltage at the emitter (see above) and can figure out that $R_1\approx 10\:\Omega$ would be about right, as well.

NOTES:

You will not often find this arrangement on the web, if you search. There are good reasons for that. But in your case, with the voltages available and the specific type of IR LED you are using and the limited number of them that you want, I think you can get by with something this simple. (If you take heed of the thermal considerations.) All your problem boundaries would seem to allow a design like this.

If I were you, I might buy up some D44H11 devices in TO-220 packaging just in case I need them. But I might also try some 2N4401 parts and see just how hot they get. Given the intermittent use you mentioned, they might work just fine. And the circuit certainly is simple.

As the temperature of the BJT rises, the $V_{BE}$ shrinks at a rate of around $-2\:\frac{\textrm{mV}}{^\circ\textrm{C}}$. With a $50\:^\circ\textrm{C}$ rise, this means an increase of about $100\:\textrm{mV}$ across $R_1$ -- or another $10\:\textrm{mA}$ through your LED and the NPN BJT. (About 4-5% change.) But also, $V_{CE}$ shrinks a little (about 10% change) and so in this particular case the resulting dissipation in the BJT due to a rising temperature will actually yield a slight decline in dissipation to counter that effect. Which is a good thing. (Note that this applies in this specific case and isn't a general rule about dissipation.)

This is an example of minimalist designing that takes into account all of the issues that are specific to a situation. Your I/O pin current limitations, your IR LED requirements, etc. As you can see, it cuts things quite close to the edge on several areas. However, that's taking into account worst case and the reality is likely to be better than that and probably 'good enough.' The circuit has the huge advantage of simplicity, so for hobby use I'd give it a shot.

If you want to greatly reduce the current compliance requirements from your I/O pin, you can try the following:

simulate this circuit

It will place a higher dissipation load onto each of the BJTs assigned to LEDs, since the $V_{CE}$ will increase. But since you are using these in a very intermittent fashion, that still may be just fine. And this almost entirely removes the load off of the I/O pin. The risk you take is if the IR LEDs are turned on and left on for some accidental reason, you might cook the BJTs in the TO-92 packages. But that was a bit of a risk before, too. And you can still replace the BJTs assigned to LEDs with the D44H11 devices, if you want to be safer here. If so, this circuit might be better because it also keeps the I/O pin current loading very, very light. And the D44H11 can tolerate a modest increase in dissipation. (Especially so if you mount the D44H11 BJTs on a short strip of aluminum, for example.)

• How can you assume the i/o pin is 3.3V? isn't it 5V or 0? Does the led reduce the voltage and adds resistance? that's what the first calculation is for? And how did you reach the last part for the 10 Ohm? Lastly, don't you need a resistor from the i/o pin? – shinzou Oct 3 '17 at 21:40
• @shinzou Bruno is using an ESP8266. Their I/O is always below 3.6 V, typically 3.3 V. When a BJT is used as an emitter follower, there is no need for a base resistor. The resistor is easily calculated as $\frac{3.3\:\textrm{V} - 900\:\textrm{mV}}{250\:\textrm{mA}}$. – jonk Oct 3 '17 at 22:20

Considering Vce(sat) specs are based on 10% base current one can use a lower limit of R1/R2 of 10*10=100. But considering you are using 5V rather than 3.3V, for a 1.2V LED with an ESR ~ 0.5 Ohms +/-50% , R2 will be larger than ESR of IR LED to obtain 1A. e.g. (5-1.8V-1Vce)/1A= 2.2 ohms Thus R1 must be 220 Ohms.

Now this neglects the source impedance of your 3.3V driver which can vary from 25 to 50 ohms and the precise Vce(sat) of your Darlington. But at least it gives you a method to compute how to achieve the max rated current of your LED with a duty cycle that does not exceed the average power.

## An adequate heatsink is needed

• rather than two under-rated 5mm LED's with LED's never exceed the absolute max If and Pd.
• if you checked the 1A specs on a 5mm LED you may find the Vf ~4V which now implies an ESR of ~3 ohms and R2 near 1 ohm with a duty cycle max to not exceed xxx mW. Using an LED rated for 1A such as this we see the Vf variation at 1A to be large as this reflects tolerances on the ESR not the low 5mA current Vf of 1 V