Such a popular three-pin voltage regulator as LM317 (datasheet) is specified to work at minimum 3 Volts across the chip. This humble parameter is not even stated directly in the datasheet: I found it only specified as "conditions of measurements" for the reference voltage Min/Typ/Max. The rest of the document assumes even higher wasted voltage: "Unless otherwise specified, VIN − VOUT = 5V".
Something inside me protests against losing more than 3 Volts on a stupid transistor that does only that: heats up the whole design. The popular solution to this problem - switched mode power supplies - we do not consider here because of the noise they produce. Of course there are techniques to reduce the noise. But... as the ancient wisdom says: "He does not fight, therefore he is unbeatable in the world."
Roots of the idea
The main idea of the layout discussed in this article has been inspired by one of many datasheet's on TL431. Look at how National Semiconductor / TI suggest us to use this tiny jewelry:
Vo ~= Vref * (1+R1/R2)
To my humble opinion the suggested layout has no real advantages against common parts like 7805 or LM317. Minimal voltage dropout here could hardly be less than 2 Volts. Also, there's no over-current or overheating protection. The only imaginable benefit versus 3-pin regulators is: the max current can be pumped up as high as one wants to.
Evolution of the idea
Recently I faced the necessity to get stabilized 12.6V at 2A from the 12V 5A transformer's secondary. Power-wise that was a nice fit. The only problem was that the ripple voltage along with losses on rectifiers left me say a Volt or two max for the loss across the regulator.
Which active component can act as a regulator with sub-volt dropout? MOSFET: modern power devices come with the RdsON of few milliOhms max. With only few Amperes current - we are losing just few milliVolts across the device.
A straight replacement of the Darlington from the schematic cited above would not do any good for us. The threshold voltage of power MOSFET's can be 3 to 5 Volts with "usual" devices and still above 1 Volt for those "logic" ones. This voltage would dictate the lowest possible drop across our regulator.
It would have been very interesting to try either DEPLETION MOSFET or J-FET in а similar layout. Unfortunately decent power devices of these types are not available. Nothing that I know, at least. (Please, please correct me and tell me that I am wrong!)
An additional low-current power supply comes to the rescue. It should provide the potential of few Volts above the input positive rail: that would be just enough to pull the gate of the MOSFET up to open the device. Since there is virtually no current through the MOSFET's gate - the additional power supply needs to deliver only few milliAmperes of the current into the pull-up resistor.
Extreme LDO Regulator - skeleton sch.
When the potential on the TL431' sense pin gets lower than its threshold of 2.5 Volts due to any reason that lowers the output voltage - the chip conducts less current. Thus it "loosens" the MOSFET's gate that gets pulled up higher by the current through that pull-up resistor. The FET starts conducting more current and pulls the output higher - by doing that effectively restoring the balance.
In an opposite scenario, when for whatever reason the output (and its direct derivative - TL431 control pin) gets higher than needed - the layout works similarly well. TL431 starts conducting more current, pulls the MOSFET's gate down thus reducing the current through its channel. The output gets lower.
In a real-life device I wanted to have some protection in addition to the slow-blow fuse at the transformer's primary. Thus I decided to sacrifice some 0.5 Volts dropped across the regulator under the normal operation conditions - and gain SAFETY.
By the way it could be much less of a dropout voltage even with the over-current protection. But such a precision protection circuitry becomes slightly more complicated. Nevertheless should you ever need such a solution - let's have a chat 😉
MOSFET + TL431 = high grade LDO Voltage Linear Regulator
30.Jan.2012: Tested 🙂 Works great!
With load currents higher than 2A power rectifiers should be attached to some heat-sink. R8=0; C7=0.1 ... 10uF ceramic or film.
With the R5-R6-R7 values as drawn - the output voltage can be regulated between 9 and 16 Volts. Of course the real maximum is dictated by the transformer's secondaries.
Make sure R4 can withstand the full load current: PmaxR4 ~= 0.5 / R. For these particular specifications I would take R4 rated at 2W.
Why would you build this
For example: in a vacuum tube based project in order to feed tubes filaments with the DC.
Why DC for dumb heaters? Even more: such precisely stabilized direct current?
Feeding heaters by DC reduces leakage of the mains frequency or its second harmonic into the signal path. There are several ways the hum gets into the signal through the tube heaters. In fact this topic worth another thorough article...
We do want the voltage applied to heaters to stay within tight tolerances. There is data showing that exceeding heaters specified voltage by 10% could reduce tube's life-span by ten-fold. Consider some 5% deviations allowed in the power mains voltage plus some 5% to 10% tolerances in the transformers and its output variation in particular design depending on the actual load...
By the way the very same schematic could be used for feeding 6.3V filaments. Provided the transformer puts our at least 6V AC (RMS) and R5 value is decreased to 5.6KOhm.
Consider this LDO voltage regulator is used for supplying DC to the thermionic valve's filament. In such an application soft start would be desirable. The change necessary for obtaining smooth startup curve would be extremely simple: use 1000uF in place of C4 and add 1KOhm resistance between the doubler's bridge positive output and C4 "+" terminal.
Vacuum Tube Myth Busted
As I mentioned above feeding tube heaters with DC has several benefits. However there can also be very realistic reasons that explain why certain "gurus" do not like DC at heaters. These worth a note here, so that those willing to utilize the LDO regulator described in this article do not fall into these traps.
Let's consider potentially the worst scenario when one wanted to "upgrade" an existing tube amplifier by installing a rectifier and a regulator for filaments.
Believe it or not, but in most cases the transformer becomes overloaded after such vivisection of the device. In fact the secondary must be rated to supply AC current of around 1.8 times greater than the DC current at the rectifier's output. Before such an "upgrade" it most probably was Ok working at AC currents slightly below its rating. After an "upgrade" the current spikes charging the reservoir capacitor will:
overheat the transformer;
induce new and very unwanted noise into the high voltage supply, especially if the device was fed with one transformer for both heaters and B+.
In place of a conclusion
I would not claim a patent for such a basic schematic. Even though I came up with this idea on my own few years ago - later on I've seen similar layouts being used by several experienced designers elsewhere. By this article I simply want to share this useful design pattern with you, my friends.
I cannot understand how can you get stabilized 12.6V at 2A from the 12V 5A with a LDO linear regulator? How are you going to turn on the MOSFET that will require a minimum voltage of VGSthmin+Vout. I am puzzled here.. Ah, great blog btw!
Hi, this is a nice post.
I would like to use this regulator in my next design for powering up a 24Bit ADC.
current required will be about 150-200mA for the whole circuit. And this will be 12V battery powered. Can you suggest a LDO schematic with very low ripple using TL431?? The board space is limited.
thanks in advance
Also, can you write more about TL431, how it is linear and such?
TL/LM431 is a little electronic jewelry indeed :)If you consider this thing as being a simple OpAmp with a magic voltage reference and the output BJT as shown on its block diagram – you'll be fine. Just remember to give it at least 1mA to chew on.Also under no circumstances it will let the voltage on its input to go much higher than its threshold (2.5V).But… give me some time, please! And thanks for the brilliant idea for the new post to this blog 😉
Abdullah, thanks for reading and asking! :)Let me put a quick answer for now.12V AC at the transformer's secondary will give us some 15V of usable DC, once rectified:12V * 1.414 = 16.9V.Then we lose some 0.4V on one of the Schottky diodes.2A 100Hz 15kuF result in some 1.3V ripple voltage.A fraction of a volt will be lost on windings internal resistance naturally.Hence we still have quite some room for the LDO to heat up.The need for the extra voltage that would account for up to few volts of Vgs needed in order to control the MOSFET we manage here by using the voltage doubler:C2 & C3 get charged up to some 15V via the lower part of the small bridge at the moment when the corresponding trafo's output goes negative.C4 is charged to some 30V: when there's a positive peak on the secondary it sums with the C2 or C3 charge and flows into the positive terminal of C4 via the top part of the small bridge.R1 will have to drop around 12V: 30V (C4) – 12.6V (output) – 0.6 (R4) – Vgs.With say 12V across R1 we have decent current to feed TL431 and enough room to control the gate of the MOSFET.Pls do not hesitate to bug me more if anything remains open.
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Ah, so clever! I couldn't see that they form a voltage doubler!This design is great it only dissipates about 80mW at 2A. But why are the ICs with same performance are sooo much more expensive?One more question, how would a P type MOSFET perform here in contrast to an N type in terms of performance and price?
Wait… the MOSFET still requires quite some heat-sink: at 2A the thing will dissipate around 6W.Regarding P-ch FETs – indeed it seems to be more attractive to use such thing here and get rid of the voltage doubler. Unfortunately P-channel FETs are approximately 3 times worse than their N-channel counterparts. Be it price, input capacitance, max current or min_RdsON…I would be cautious using P-ch device here: the loop gain becomes much higher – that may provoke difficult to cure instabilities. Sorry, for now I've no intention to try that myself.The N-channel FET works as a follower in the original schematic. Thus it's adding no voltage gain and works calm.
Hmm, N-channel seems a really good option, then.But I do not understand how this thing dissipates 6W!
Ah, Ok, a bit more if we want to be precise :)The MOSFET's channel has to drop something around 2V : 15V+ at the reservoir capacitor – 12.6V at the output – 0.6V at R4. That gives us 4W at 2A.Add 0.4V * 2A * (~~1.8) heat on Schottky diodes. That 1.8 is a very rough approximation that accounts for the pulse character of the big cap's charging current spikes.And for the complete picture R4 gets warm too with its 1.5W to dissipate.Here I deliberately do not want to go with more precision – just remember how big the mains power voltage variations can be day/night, with or w/o elevator or other big loads. They say it should stay within 6% to 10%
http://en.wikipedia.org/wiki/Mains_electricity_by_country. I would not trust that claim 🙂
You are right; The MOSFET will have to drop the voltage on itself, I cannot just think that as I*I*Rds, because it will not be in fully conduction mode. Thanks for the answers and patience, really appreciate it 🙂
Woah this blog is wonderful i really like reading your articles. Stay up the great work! You realize, a lot of people are looking around for this info, you can aid them greatly.
MOSFET + TL431 = компенсационный стабилизатор напряжения | MyElectrons.ru
This is an interesting article, but there are a few things that bug me:
1. You don’t have an output capacitor. My experience with similar devices is that you really need it, or you may have huge problems with self-excitation and oscillations;
2. There is no point in making a current limiter, since the MOSFET is a current limiter itself. See page 3 of the datasheet. There is a certain current that can flow through the device for each value of the Gate to Source voltage. Apparently the used device is hugely over-specified, and the graph starts at 8A for 4.5V Vgs, but you can probably choose a less powerful device.
For example IRL520N is specified for drain currents below 1A for Vgs=2.5V, or 3A for Vgs=3V, and the last value is stable over temperature. You need to have a zenner of 3V apparently (this still needs to be tested!), or decrease the value of the auxiliary supply (again a simple zener diode should be fine).
Anyway, why would you use a 50A MOSFET in a circuit that is limited to 2A, and loose precious 0.6V in the process?
Dimitar, sorry I missed your comment when you posted it!
1. Fully agree, in a real implementation I always have had a not so big electrolytic at the output. The trick there is that one should not try to use an extremely low ESR there – in order to avoid possible oscillations too.
2. I’d also agree with just a humble note: limiting the MOSFET’s current by its Vgs is not as repeatable (i.e. requires thorough testing / calibration) as when using a current sense resistance. Anyway thanks a lot for bringing this up! I’ll consider this technique seriously in a power supply I’m working on right now…
Why such an over-sized device? Because that was the one I had at hand at that time 😉
Hey…I was wondering if one could use this methodology to build a negative regulated supply.
Using p-channel MOSFET for split supply audio amplifiers.
Personally I would not mess with p-channel MOSFET’s unless absolutely necessary: in general p-ch are three times worse than n-ch (in some parameters or others: Cin, Rdson, price, etc.)
For the negative rail in a split supply I’d rather be taking the same n-channel MOSFET and control it differently.
Thx a lot for this very interesting circuit, was a bit difficult to understand for me, but, with comment and answer, it seems clearer now.
I am working in a circuit to use AC from a bike dynamo hub to charge DC devices and a buffer battery. In my first circuit, I used a TL431 as High-Current Shunt Regulator as you can see in the datasheet here :
I would like to improve my regulation with a mosfet and I’m facing this problem : I’m already using voltage doubler before my regulator because my dynamo is 6V and I need 12.6V to charge my battery, so I cannot use your circuit.
Do you see another way to rise the voltage (without using a battery) to control the mosfet?
Hope this is crystal clear 🙂
good to see your interest in this little schematic 🙂
If I had to get 12.6V out of some 6V source I would try looking into integrated boost converters. These will give you all voltages and currents you need, and protection, along with the power efficiency unobtainable in a purely analog design like presented above.
If you’d like to discuss this further, I may need to know more specifics about your project requirements, such as its max and nominal current.
Thx for your answer.
The circuit I have build is from a german website. My working base was this schematic :
witch I made with AT328P instead AT8L to make my own arduino program and with Mosfet instead of relay.
You can find some of the characteristics in the simplified drawing.
Switches are commuted according to the frequency (the speed rotation) of the dynamo hub.
I have made some measurment to check witch combination of capacitor is best at each frequncy.
Now, I would like to improve the regulation to get rid of the thermal switch used to protect my transistor from overheating and to improve the efficiency.
I have think about boost regulator but I am not sure I won’t get same amount of power, good track to follow.