# LM317 voltage regulator with high current boost by 3 npn transistors

I am building a rectifier (~230V+-10%/50Hz) with voltage regulator from scratch. My output needs to be 3.3V with a maximum current(before short-circuit protection trigger) of 16A. I have chosen three npn transistors in parallel to boost my current (2N3055) up to 16 amps and calculated respective resistors, powers, temperatures in order to work properly with real elements. However I experience problems in my simulations and the output is nowhere near what I expect. Vin is what I will have after my transformer, rectifier and LC filter- around 7V with 80mV pulses. Ic across each transistors should be~5.33A.

However during a simulation with a load of 0.22 Ohms, a current of 15A is expected but this is what I get.

Vout is around 2.1V and Iout is 9.45A. Why so?

I also want to add a short-circuit protection that is triggered by Iout greater than 16A.

I guess it should be with a pnp transistor and a resistor but I am not sure how to calculate and where to connect it.

Here is a 2N3055 datasheet 2N3055 and LM317 datasheet LM317

• Why would you think current will flow through the transistors when you have the base-emitter junctions shorted out? – Trevor_G Feb 5 '18 at 15:32
• Try taking the feedback from the output. But you will have no short circuit protection regardless, other than the ballast resistors. – Spehro Pefhany Feb 5 '18 at 16:29
• @ThePhoton isn't that 1.25 * (309 +240)/240 = 3.28? – Trevor_G Feb 5 '18 at 16:35
• @Trevor_G, you're right. Brain freeze. – The Photon Feb 5 '18 at 16:36
• @Kristin Why don't you apply and extend the example provided in the datasheet for the LM317? You've already figured out the emitter degeneration to help share currents between the external BJTs. – jonk Feb 5 '18 at 16:36

Vout is around 2.1V and Iout is 9.45A. Why so?

Your feedback is being taken from the OUT pin of the regulator chip.

The voltage there is about 3.3 V.

But from there to the load, there is a Vbe drop (probably 0.8-1.0 V with these kind of currents), and a drop across the 0.11 ohm resistors (0.5-0.6 V when you get to the load current you want).

$3.3-1.5 \approx 1.8$, but the output will be a bit higher since reducing the output voltage also reduces the load current. So your result is roughly in the range you should expect.

Try taking feedback from the actual node where you want to regulate the voltage (it would be easier to say which one if you labelled your nodes).

• +1 top of C1 for feedback would be my guess, no idea about the stability, and how well three BJTs will share the load though. – Trevor_G Feb 5 '18 at 16:30
• @Trevor_G, 2N3055 is rated for 15 A max and 1.25 C/W thermal resistance, so I think OP is okay even if current sharing isn't perfect. Particularly if the transistors share a heat sink (pretty common arrangement for TO-3 packages). – The Photon Feb 5 '18 at 16:35
• I suppose...... – Trevor_G Feb 5 '18 at 16:36
• Provided you don't get one of the fake 2N3055s that has a tiny die inside. – Spehro Pefhany Feb 5 '18 at 16:38

Although I've not specifically attempted a $15\:\text{A}$ boosted LM317 before, this is along the lines of what I'd try out first. This is roughly taken from the Figure 23 you mentioned:

simulate this circuit – Schematic created using CircuitLab

In this case, I went for the D44/D45 series devices. (The PNP version has simply HORRIBLE Early Effect, but it's not a big deal here.)

The values of $R_6$, $R_8$, and $R_9$ are set to drop somewhere from $150-200\:\text{mV}$ at full load. They will need to be rated for at least $1\:\text{W}$, but I would not feel comfortable with less than $2\:\text{W}$ resistors there. If you adjust those values, please keep in mind the dissipation question. You are talking about a lot of current.

To reduce the oscillation, you really want some ESR in $C_2$ to add a nice 'zero'. If you see oscillation in the output, try adding a small series resistor to $C_2$. $15-39\:\text{m}\Omega$ (as shown with $R_{10}$) should put a crimp in the oscillation. You might just make provisions for it and jumper it, without using a resistor, if your output seems fine with the output capacitor you selected. But here is one of those cases where output capacitor ESR is actually a good thing.

Your schematic shows an AC input. That's not good. I hope your schematic was just mistaken, there.

Since the minimum specification for the LM317 is $3\:\text{V}$ from input terminal to output terminal, the externally added circuit will always have more than enough headroom to operate so long as you supply that difference.

Keep in mind this is a linear power supply. With $\approx 3.3\:\text{V}$ output and $\approx 3\:\text{V}$ overhead, you will have little better than 50% efficiency. At full load, you will have $\ge 45\:\text{W}$ wasted dissipation, not counting the load's dissipation. And more than that, likely, because this ignores whatever you have supplying the unregulated input DC voltage -- where it is likely you have still more dissipation in diode rectifiers from AC, etc.

While perhaps $3\:\text{W}$ dissipation might occur in the emitter resistors, that still leaves a pretty much all the rest with the bypass BJTs. Getting rid of $15\:\text{W}$ each will be the challenge. Note that if you want to allow a maximum junction temperature of say, $100\:^\circ\text{C}$, and the worst case ambient temperature you care about supporting is $45\:^\circ\text{C}$, then this means you need $\frac{100^\circ\:\text{C}-45^\circ\:\text{C}}{15\:\text{W}}\approx 3.7\:\frac{^\circ\text{C}}{\text{W}}$. For the parts I mentioned, junction to case is already $1.8\:\frac{^\circ\text{C}}{\text{W}}$. That leaves you only $1.9\:\frac{^\circ\text{C}}{\text{W}}$ for whatever you use as a heatsink plus the bonding interface between the BJTs and that heatsink. That's not a lot to work with.

You might consider putting more of the dissipation into the emitter resistors, I suppose. More degeneration won't hurt you. I chose to set them at about a minimum resistance for the circuit, so increasing their values will be fine. (Don't decrease them much, though.) You need to work out this balancing act on your own.

• Could you explain the way of calculating R4, R5 and R7 and if I can somehow limit the output current to 16 amps at worst. – atkristin Feb 5 '18 at 20:39
• @Kristin There's no current limiter here and I wasn't interested in sitting down to think through how the LM317 operates in order to construct a fold-back limiter (which would probably be the right next step.) For $R_4$ I just assumed about $50-60\:\text{mA}$ through the LM317 and about one VBE drop for Q1. $R_5$ isn't critical -- a larger value will put some more current through the LM317, which is fine. – jonk Feb 5 '18 at 21:18