Adjusting the input voltage of a linear regulator

Thread Starter

cmartinez

Joined Jan 17, 2007
7,394
Here's my problem: I need to use a linear 3.3V regulator with an extremely low power consumption for a project that I've been working on.

The regulator in question is the XC6215B332MR-G‎ which is a tiny marvel that draws only 0.8µA when idle. All in all, my entire circuit (which includes a PIC16LF1823 MCU) ends up drawing less than 14µA when fully working.

The thing is, that until now, my circuit has been working using a power supply of 6V, which is in the very upper limit allowed as input for the XC6215B332MR-G, according to its datasheet. But now I need to draw power from a 15V source, which I'm sure will blow said regulator out of this world fairly quickly if connected directly.

The closest match I could find to replace that chip with another one that could tolerate that high of an input voltage is the AP2205-33W5-7, but the problem is that its datasheet states that it draws around 35µA when idle. Which is more than twice what my complete circuit is already drawing.

Question, is there a practical way to drop the 15V input voltage down to about 6V whilst wasting as little power as possible? The first thought that popped into my head was to use a brute force approach and place an array of many diodes in series between the power supply and the regulator's input so that they could drop the 11V extra volts from the supply.... but I suspect that things cannot be that simple...
 

Papabravo

Joined Feb 24, 2006
16,783
I haven't worked out the details, but a resistor, zener diode, and a pass transistor used as a pre-regulator might do the job. For such a low load current the drop from 15V to 6V, that's 126 μwatts. It'll take me a few minutes to whip up a schematic. Here you go. I don't know if this comes anywhere close to meeting your requirements, but I have used this in the past for Industrial Automation equipment that ran off of 24VDC. Depending on the transistor used it is robust and reliable.

1627772817102.png
 
Last edited:

Ian0

Joined Aug 7, 2020
3,232
Here's my problem: I need to use a linear 3.3V regulator with an extremely low power consumption for a project that I've been working on.

The regulator in question is the XC6215B332MR-G‎ which is a tiny marvel that draws only 0.8µA when idle. All in all, my entire circuit (which includes a PIC16LF1823 MCU) ends up drawing less than 14µA when fully working.

The thing is, that until now, my circuit has been working using a power supply of 6V, which is in the very upper limit allowed as input for the XC6215B332MR-G, according to its datasheet. But now I need to draw power from a 15V source, which I'm sure will blow said regulator out of this world fairly quickly if connected directly.

The closest match I could find to replace that chip with another one that could tolerate that high of an input voltage is the AP2205-33W5-7, but the problem is that its datasheet states that it draws around 35µA when idle. Which is more than twice what my complete circuit is already drawing.

Question, is there a practical way to drop the 15V input voltage down to about 6V whilst wasting as little power as possible? The first thought that popped into my head was to use a brute force approach and place an array of many diodes in series between the power supply and the regulator's input so that they could drop the 11V extra volts from the supply.... but I suspect that things cannot be that simple...
Your solution is in Figure 9 of this application note
http://ww1.microchip.com/downloads/en/AppNotes/AN-D66.pdf
 

mlsirkis

Joined Aug 11, 2010
32
I hadn't thought of that... simple as this? Or is another component (such as a series resistor) needed?

I hadn't thought of that... simple as this? Or is another component (such as a series resistor) needed?

The zener goes in series with the regulator Vin pin, not the regulator GND pin. The regulator GND pin ties directly to GND. Add a cap from Vin to GND, and another cap from Vout to GND. See regulator data sheet for cap recommendations.
 

Ian0

Joined Aug 7, 2020
3,232
I hadn't thought of that... simple as this? Or is another component (such as a series resistor) needed?

Zener between Vin and +15V, otherwise your output will be 13.3V!
Two possible problems with the zener
1. If your power supply drops below 13.3V, your regulator stops working - so OK if your 15V is regulated
2. Zeners drop rather less than their rated voltage at very low currents. In theory, you would have 5V going in (15V - 10V) but you may have more than that at low currents, and your reg will only stand 6V.
The Supertex circuit (post #6) solves both these problems.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
7,394
I haven't worked out the details, but a resistor, zener diode, and a pass transistor used as a pre-regulator might do the job. For such a low load current the drop from 15V to 6V, that's 126 μwatts. It'll take me a few minutes to whip up a schematic. Here you go. I don't know if this comes anywhere close to meeting your requirements, but I have used this in the past for Industrial Automation equipment that ran off of 24VDC. Depending on the transistor used it is robust and reliable.

View attachment 244837
Many, many thanks for your help, Papabravo... I'm going to sim your circuit myself and play with it a bit ... see where it takes me.
 

Ian0

Joined Aug 7, 2020
3,232
You can also make a pre-regulator with the depletion FET, in which case the FET bias resistor goes to the source of the FET not to the positive power supply. At first glance it would seem that it would give better PSRR, but as the FET has a larger Cds I'm not so sure. . .
The problem with the zener/transistor pre-regulator on a very low current supply is the zener current. You need at least a few hundred microamps, perhaps a milliamp to be sure of the zener voltage.
 

Papabravo

Joined Feb 24, 2006
16,783
Should work. Notice the difference in the symbol, the line between the drain and the source is solid, not broken. I don't know what it means that the simulation blows up when the capacitors have a non-zero ESR. The number of available depletion mode symbols and subcircuit files appears to be limited.

1627846256219.png
 

eetech00

Joined Jun 8, 2013
2,571
The closest match I could find to replace that chip with another one that could tolerate that high of an input voltage is the AP2205-33W5-7, but the problem is that its datasheet states that it draws around 35µA when idle. Which is more than twice what my complete circuit is already drawing.
AP7370,AP7380, AP7381, AP7383, AP7384.

All of the above draw 2.5uA or less and have Max Vin of 18v or greater and Vout of 3.3v available.
 
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