# Increase LDO current with PNP

I have seen this circuit a few times. It takes a common 3 pin LDO, and adds a PNP transistor between VIN and VOUT, to increase the amount of current it can provide at the regulated voltage. It is present in the On Semi LM7805 datasheet.

At this time I'm looking at the Holtek 7133, a 30mA 3.3V fixed LDO. It also has the same circuit.

With VIN at 5V and Tr1 a common 2n3906 100mA PNP transistor, how would I calculate R1? How much does this change the quiescent current of 5 microamps? What's the minimum VIN required for this to work with this PNP (The LDO only needs a 0.2V drop out at max).

And what is the new maximum current in this configuration? Since R1 limits VIN, I'm assuming that all the load current goes through the PNP, so in this case 100mA, not the LDO's + the PNP's 30mA + 100mA.

• I'm interested because a PIR module I ordered has the 7133, and instead of adding another regulator for my microcontroller, adding a PNP would be easier, I only need a slight bit more current. – Passerby Jul 15 '15 at 6:10
• The correct name for the part is not "LDO" but "LDO regulator". – Andy aka Jul 15 '15 at 10:35
• Considering the audience of this question... what possible reason is there to type it all out? Can it be mixed up with a different type of part? Is it in anyway ambiguous what it refers to? – Passerby Jul 15 '15 at 10:45

The input voltage will need to exceed the output by the Vbe of the transistor + VCE(sat) + V(dropout)[LDO]. Under most conditions, that is about 1V plus the LDO dropout voltage. If you can live with the added V(in), then the outboard boost will operate.

The boost circuit will turn on at about 0.6V/R1. If you wanted to engage the boost at 30mA, then the resistor is 0.6/0.03 = 20 ohms. That 30 mA will (mostly) still pass through the regulator. The mostly refers to the fact that the transistor base current must be supplied from the input, so the output provided from the LDO is 30mA - Ib. All other current is being supplied from the outboard boost transistor.

The actual voltage at which the boost engages at will vary significantly over temperature due to the -2.1mV/Kelvin temperature coefficient of the base-emitter junction.

The actual output current is load dependent. As this circuit has no current limit, you will need to ensure that the transistor can handle the expected load current.

Make sure you choose a resistor with a suitable power rating (in this case 30mW minimum - I would choose a 100mW device at a minimum for this function).

HTH

• The correct name for the part is not "LDO" but "LDO regulator". – Andy aka Jul 15 '15 at 10:36

R1 would limit the current to the HT71XX. Any excess current would then pass through TR1.

$$\frac{V_{EB}}{R1} = I_{limit}$$

Because of the emitter base drop, Vin has to be a diode drop higher than what the drop out of the HT71XX is.

The circuit as it might a problem, because anything higher than what the current limit into the HT71XX is, gets directed though the pass transistor. (ie, very low loads, short circuits etc..).

Additional current limiters or fold back current might be something to look into.