# BJT Pre-Regulator

I have a circuit that has an input of 35V peak and I would like to drop the voltage down to 5V 200mA. I was recommend to use a pre-regulator in the following configuration:

The good thing is that it works. My question is what is this BJT/preregulator doing? How does it work?

• If you tell us what you really want we can tell you what you really need and how what you have really works. If you provide "simplified" questions (= randomly irrelevant) you'll get simplified answers = randomly wrong. Q1 simply implements a second series regulator that shares dissipation with U1. Vout is ~= V_CR4 - 0.8V or about 13V. The LM317 needs about Vout + 2V = 7V minimum here. As we do not know real Vin or Vin_min or anything else we do not know how low your current source goes etc or why we cannot say with certainty what the smell of the colour nine is in this case. Feb 27, 2013 at 22:12
• Voltage droop on C2+C3 in 1/2 cycle at 60 HZ is ~~~~ 1V. | Vdrop ~~~= time x current / capacitance = 0.0083 x 0.2 / 2x E-3 =0.83 ~= 1 Volt. | Make Q1 a device with 5W+ continuous dissipation capability. Pmax as shown here ~= (25*1.414 -14) * .2 = 4.4 Watts. As not shown here it depends on your actual parameters. Feb 27, 2013 at 22:16

The base voltage is set by the CR4, the emitter will be below that by 0.7V. In your case around (14.-0.7) = 13.3 V. The remaining voltage gets scrubs off across the $V_{ce}$ of Q1 as heat. $(35 - 13.3)*0.2 = 4.34 W$

Well, you're not really minimizing power dissipation here with the pre-regulator.

Your pre-regulator is a very simple linear regulator that is designed to provide a lower input voltage to your LM317 regulator. The Zener diode and R3 will provide a relatively stable 14 V. Transistor Q1 is in an emitter-follower configuration, so you get a diode drop down from the base to the emitter. This provides a coarsely regulated 13 V as the input for your LM317 regulator.

And now the down side: you're not really saving any power consumption here, you're just sharing the power dissipation between Q1 and LM317. At a 35V input and your maximum 200 mA, there will be a 22 V drop across Q1, and an 8 V drop across U1. Power is current times voltage, and Q1 is dissipating 4.4 W (HOT!!!) and U1 is now only dissipating 1.6 W. Before, U1 was dissipating all of that power (6 W), so it is definitely running cooler.

This brings up an issue with your part selection - the PZT2222AT1 is a low-power transistor. Under optimal conditions, its absolute maximum power is 0.6 W. You have now shifted the issue from your regulator to the pre-regulator transistor. I would recommend a switching regulator (buck converter) instead of a linear regulator approach, and that would dissipate much less power due to the higher efficiency.

As W5VO said, if you want to minimise energy loss then you need a switching regulator.
If you want to minimise heat dissipation in the LM315 the probably cheapest solution is to add a resistor betweem Vin and U1.

The resistance is designed to drop most of the voltage that the LM317 would otherwise dissipate.

The LM3176 has a minimum Vin-Vout of 2V at 1A - somewhat lower at 200 mA. Use at least 2V for 'safety'. Vin_min_LM317 = 5V + 2V = 7V.
Rseries_max. = V/I = (Vin_min - 7V)/Imax = say (30-7)/.2A = 115R A 100 Ohm would be about right

Power dissipation in resistor = I^2 x R = 0.2^2 x 100 = 4 Watts.
A 5W air cooled resistor is marginal. A 10W air cooled resistor is better.

At 0.2A Vdrop = IR = 0.2 x 100 = 20V.
Minimum Vin = 5V + 2V + 20V = 27V.
Should be OK. If Imax was increased then Rseries would needed to be smaller.

• I like that solution, however, I haven't been fully honest. The input is actually connected to an AC current source (which induces a voltage on the capacitors). I put 25V rms to simplify my question, as I really just wanted to know what the BJT was doing. Putting a resistor wont help. Feb 27, 2013 at 21:10