# Design of voltage regulator and operation

I'm designing a simple voltage regulator with input voltage (Vin) from 120 to 375 Vdc, output voltage (Vout) between 3.5 and 4.5 V (more or less), and output current (Iout) 5 mA maximum.

I found the voltage regulator schematic from the image here: https://www.circuitbasics.com/voltage-regulators/

Even if there are some clues in the post on how to pick some of the values, it doesn't explain in detail how it works. For example, I'm not sure how much reverse current needs D8, how to pick R15, R16 and R17, etc.

Where could I find some more detailed information about this circuit? Can I improve the design by replacing some BJTs by MOSFETs (e.g. Q8)? Are there other circuits that would suit better my application?

• If you are trying to just get something working the On Semiconductor NCP785AH50T1G is a 450V 10mA LDO with a 5V output, does most of what you need, and isn't too expensive. Try searching for "400V LDO" to get other options May 13, 2023 at 15:06
• @Graham: I'm trying to figure out what part of 120VDC to 4VDC requires a low drop out regulator. Not all linear regulators are low drop out regulators.
– JRE
May 13, 2023 at 15:55
• You are quite right, got the wrong word. This application is quite the opposite of low dropout! May 13, 2023 at 16:04
• R19 and Q9 comprise a current limiter. Q9 will conduct when the voltage on R19 exceeds about 0.7 V, so for 5 mA, R19 would be 140 ohms. This will start to turn off Q8. May 13, 2023 at 21:32

The first thing to choose in a discrete regulator design like this is the pass transistor - most other values fall out of that initial choice.

Assuming that we stick with BJTs for now, we are looking for a transistor with a Vce of at least 375V, although adding a safety margin isn't a bad idea (400V-450V).

R15 provides a small amount of current needed for turning on Q8 and also used in biasing D8. We need to choose a value of R15 that provides the sum of these two currents. The base current of Q8 will be approximately beta times smaller than the output current. Look for beta/Hfe in the datasheet of your chosen transistor; as an example if Beta = 100, then the Q8 base current would be 5mA/100 = 50uA i.e. not much. We will get to the zener diode current later.

As an example, lets aim for an output of 4V, in the middle of your specification.

The base of Q10 acts like the feedback pin on regular LDOs. Here we are comparing the output Vout, divided down by a potential divider (R16, R17 and R18) with a reference. The reference is the zener diode voltage + 0.6V diode drop from Q10.

If we choose a 1.2V zener and add 0.6V ( = 1.8V), then we need to divide Vout from 4V to 1.8V. If you ignore R18, and just use R16 and R17 that is just a simple potential divider question.

The reverse current in D8 will depend on the actual component chosen, look for the recommended currents in the datasheet. Once you have a value (say 1mA) then R15 can be calculated, using ohms law.

R15 = (Vout + 0.6) / (Iref + Ibase_q8)

Can I improve the design by replacing some BJTs by MOSFETs (e.g. Q8)?

I don't think there is much to be gained from MOSFETs in this application but others may disagree. MOSFETS don't have any gate current which helps with efficiencies at high loads, but unlikely to make much of a difference at such low loads.

• I agree that MOSFETs won't help much, since most of this design relies on the consistent and small 0.7V $V_{BE}$ common to all BJTs. MOSFETs' $V_{GS(ON)}$ isn't so consistent between models, even from the same batch. You could build a one-off MOSFET-based regulator with resistances and diodes hand-picked for it, but performance would differ somewhat between mass-produced ones. May 14, 2023 at 3:21