# Voltage regulator circuit, efficient, low necessary drop

I want to get a 4v supply from >=5v. (I.e. as little as 1v drop)

The load current will vary by a factor of 10,000. (From 5uA to 50mA).

I need the circuit to be really efficient when there's a small load.

I wanted a darlington pair set up as emitter-follower following a zener diode regulator. The problem being fairly obvious, the 1.4v drop exceeds the 1v I have available, so it's not usable.

Instead, using a single NPN follower following a ZD in isolation would give an hfe of about 500 say, meaning the current taken by the ZD with the supply at 5v would have to be 20 times greater than the load is using in standby, in order for it to be able to supply the load's peak current! And significantly greater when the supply is say 8v, the current taken through the ZD would be 200 times the load's standby current!

Is there a better way?

• The secret of LDO regulators is to use a FET instead of a BJT. – Mike DeSimone Jul 4 '11 at 15:29

A zener diode is absolutely no good for very low power (for the 1N4732A the zener voltage is specified at 53mA), and even LDOs often have ground currents 10 times your 5$\mu$A load. You want an LDO with a < 1$\mu$A ground current, like Seiko S-812C40. Output voltage is 2.0 to 6.0 V, selectable in 0.1 V steps, so there's also a 4V type. You get 65mA out. Dropout as low as 120mV, and stable even without output cap. The S-812C is available in SOT-23.

edit
I just discovered that there's also a version with a shutdown. That S-812C is even cooler than I thought! :-)

• I would never have thought of Seiko as an IC manufacturer. Would that mean that they use this voltage regulator in their watches? In that case it makes sense that it's so low-power. Are these parts commonly available through distribution? – Federico Russo Jun 17 '11 at 12:52
• @Federico - Frankly, I don't see a reason to have a voltage regulator in a battery operated watch. And even the 1$\mu$A ground current will significantly decrease battery life; a watch uses less. I used this LDO on a project for 50k/year. No problem getting it, though I don't recall which distri. It may be more difficult for low quantities. – stevenvh Jun 17 '11 at 13:14
• "Due to restrictions from the manufacturer, Mouser is not authorized to market Seiko Instruments product in your region." Bummer :-( – Federico Russo Jun 17 '11 at 13:58
• @Federico - Now available from Digikey – stevenvh Jul 12 '12 at 6:39

The already mentioned MCP1703 is about as good a fit as you could hope for. Part number you want is MCP1703T-4002... where

• 40 indicates 4 Volts and next
• .. 0 indicates fixed voltage.
• ... 2 = 2% = std availability)

BUT best of all it's in stock at Digikey for $US0.77 in ones ($0.47/100) as MCP1703T-4002E/DBCT-ND

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=MCP1703T-4002E/DBCT-ND

• Vinmax = 16V <- good for Microchip parts of this sort.
• Iout max = 250 mA for the 4V version.
• Lower Vout versions are available with lower Iout ratings.

There are 5 or more other 4V LDOs available even from Digikey but most have a zero load quiescent current in the 70 to 200 uA range.

MCP1703 Ground current (quiescent current) is 5uA max at zero load (2 uA typical). Worst case should always be used for design, 5uA is equal to your specified minimum load current, but would be hard to improve in a "roll your own" alternative on without vast effort. In particular, the availability of a 2% bandgap reference at only 5 uA quiescent current (max) places it in very select company.

For non zero loads the ground current is about 0.5 uA per ma, or a "gain" of about 2000, which is amazingly good. Overall this appears to be a superb part and good value for money as well.

Also available as std in 0.1V Vout increments from 1.2V to 5.5V and in 50 mV increments to special order.

Note that the various parameters highlighted at the top of the data sheet and on the product selection page are (as is usually the case) the best case value in each case and you usually don't get all at the same time. eg Iout max is only for >= 4V versions.

For completeness - may meet some other need better, other 4V LDOs from Digikey, none as good for this role, include:

Micrel MIC5206 Seiko 812C ROHM BA000LBSG NJR NJU241 Toshiba TA48MxxF TI (NatSemi) LP2985 TC1014 Microchip (far higher Iq)

1V is reasonable amount of headroom for a LDO (low dropout regulator). I haven't run accross a fixed voltage LDO that does 4V though, and they don't tend to come in adjustable versions either but it might be worth looking around.

You didn't say how accurate you need it, so maybe a 3.3V LDO with a diode in the ground leg will be good enough. Otherwise you could use a low power opamp to drive a transistor to make the 4V.

As for the voltage reference, there are a lot of options beyond zener diodes. There are three terminal voltage references, which are basically accurate low current voltage regulators. There are also accurate low current shunts for this purpose.

The MCP1703 has a typical ground current of 2 µA, drop-out of 625 mV and is available in a 4 V version.

• I haven't looked at the Seiko part, but a lot of other parts require a minimum resistance in series with the output cap. The MCP1700 and '1702 don't if I remember right. I think the '1702 doesn't either, but you'd better check. – Olin Lathrop Jun 18 '11 at 22:45
• @Olin The MCP1703 works with ceramic capacitors which have a tiny ESR. – Thomas O Jun 18 '11 at 22:56
• @Olin - The Seiko S-812C is stable without output cap, but when used can handle a low ESR ceramic. No series resistor. Drop-out is also better than MCP170x (120mV) – stevenvh Jun 18 '11 at 23:23
• @Stevenh Looks like a good part, but far exceeding required specs. Although it may as well be diamond encrusted if you can't get it! (I haven't checked though.) – Thomas O Jun 19 '11 at 0:43
• @Thomas - A closer look reveals that the 2 uA is quiescent current, without load. Ground current is 30 uA at a 50 mA load. – stevenvh Jul 12 '12 at 6:45

http://www.fairchildsemi.com/ds/LP/LP2951.pdf

The dropout at 100 mA is 600 mV, so you will be OK at 50 mA. You will probably need to add a load resistor on the output to get it to regulate properly with a 5 uA load, as it's only specified down to 100 uA out.

• Ouch Leon, add 140 $\mu$A ground current and you have an efficiency of 2% :-/ – stevenvh Jun 17 '11 at 11:45
• I didn't notice that. Perhaps it isn't such a good idea. :) – Leon Heller Jun 17 '11 at 12:03

Maybe try something inspired by the below. Pick a low power CMOS Op-Amp with JFET inputs. 'C' can be small (0.01uF should do). R2 can be large (on the order of 200K Ohm) and would need to be sized to get the correct Vref at the inverting terminal of the Op-Amp. Bias current through R2 could be on the order of 5uA for a workable device.

The only really tricky part of this circuit is the voltage reference. To keep this simple and fast I drew up a really crummy Vref circuit. Setting up low-power, temperature-compensated, precision voltage references (e.g., band gap reference) is beyond the scope of this reply (but if you need one, it can be built or bought). You might need seven diodes in the stack instead of six.

Here is how the quick and dirty V-reference in the pic works: stack diodes and tweak the current through the stack to get 4V at the inverting input at the device operating temperature. Delta-V / Delta-temperature will ~KILL~ the accuracy of your Vref if temps fluctuate a lot.

The rest of the circuit is straight-forward: If the voltage to the load drops below Vref, then the op-amp output goes high, pulling the MOSFET gate high and thereby throttling current to the load. If the load voltage drops, the op-amp pulls the gate low and the MOSFET lets more current through. The drain-source voltage of the MOSFET can be very small (tenths of a volt), so getting +4V off of a +5 volt rail will be no problem. The capacitor will smooth things out, make your load think it's being driven by a low impedance source, and prevent any ringing in the feedback loop.

• at low currents like 5$\mu$A the forward voltage of a general purpose diode (think 1N4148) is very sensitive to current variations, meaning that line regulation will be poor. And if you divide the output voltage to 0.6V before feeding it back to the opamp you only need 1 diode. – stevenvh Jun 18 '11 at 8:33