Adjusting the input voltage of a linear regulator

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cmartinez

Joined Jan 17, 2007
8,220
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.
Just looked them up, thank you. But I found that the "current-supply" of the above devices is about 300 µA:

1627864423820.png
What does that mean?
 

eetech00

Joined Jun 8, 2013
3,859
Just looked them up, thank you. But I found that the "current-supply" of the above devices is about 300 µA:

What does that mean?
That doesn't tell the whole story.

In the datasheet, the test is done with Vin at 1v or 2.7v above VOut. But Vin of your circuit will be operating well above that.
See the charts on page 6 of 17 "Ground current Vs Input Voltage". For your application it is really constant at about 1.5-2uA.
 

Ian0

Joined Aug 7, 2020
9,677
I think (and I'll have to check this) a JFET with a large Vgs(off) such as a J111 would over-volt the input of the regulator, at low output currents.
 

Papabravo

Joined Feb 24, 2006
21,159
The major difference between a JFET and a MOSFET is the gate current. The gate current in a JFET is comparable to that of a MOSFET, despite having a junction, and still less than the base current of a bipolar transistor.
 

dcbingaman

Joined Jun 30, 2021
1,065
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
An improvement on this might be to replace the 2.21K resistor and Zener with a 3 terminal voltage reference. You can get versions of it with very low quiescent current, down in the microamp range. But still good enough to drive the transistor.
 

Ian0

Joined Aug 7, 2020
9,677
As I thought, there are some inadvisable choices of JFET:
J111 and J310 will give an over-voltage situation at the input terminal
J112 and J113 won't give enough voltage at the input terminal to make it work.
I couldn't actually find any JFETS that would make it work (That doesn't mean that there aren't any)
Screenshot at 2021-08-03 18-07-07.jpg
 

HIPE

Joined Nov 11, 2020
1
All the previous replies are valid, there are many way to drop the Voltage.

But, they will all consume power while doing it.

None of the above solutions will drop the Voltage with zero power loss, so if your power budget is (35uA x 6V = 210uW) then you will lose another ((15V-6V) x 35uA = 315uW) just in the pre-regulator element, giving a total power consumption of 525uW which is more than double the power consumption at 6V.

Even if you did find a linear regulator that could cope with the 15V input, the same rule would apply. A higher input Voltage means that the regulator would have to work harder and waste more to give the 3.3V output that you need.
 

AnalogKid

Joined Aug 1, 2013
10,987
Solution without power waste is a JFET, like shown here:
No.

In a linear, series circuit, 100% of the current goes through 100% of the components 100% of the time.

JFET, MOSFET, zener, resistor, whatever ... The total power dissipation in the voltage-dropping components will be exactly the same - the voltage drop across the components times the current through the regulator to the load. In this application, that is (Vin - 3.3 V) x Iload. The voltage drop across the pre-regulator circuit might vary per the device used, pushing more or less power dissipation onto the 3.3 V regulator chip, but the total power loss is the same for any method that does not require some static current of its own (such as post #2).

I prefer the zener method, with the zener in the source leg rather the return leg, plus one or two bypass capacitors across the 3.3 V regulator input.

ak
 
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