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DIY voltage Regulator

Lord_grezington

May 3, 2013
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Hello Everyone

Any idea why my circuit below wont work as a low cost High input Voltage regulator? Zener = 13V

Simulations show that I should be able to take around 125mA for voltages up to 60V (probably even higher if I did more playing) and keeps a stable output voltage until the input voltage drops below 18.5V but still gives me a voltage of 10.6V when the input drops to 12.5V. If this works I should be able to save around 60% of the cost of a LDO.

upload_2018-1-23_8-33-50.png

Thanks
 

(*steve*)

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It looks like it should work.

The transistor will dissipate almost 15W at your rated maximum load and input voltage. Make sure it's rated for that and that it is sufficiently heatsinked.
 

Audioguru

Sep 24, 2016
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Its output voltage is not regulated properly because it is simply an emitter follower with losses attached to a zener diode that also has losses.

The zener diode is 13V only when its current is 5mA but the 10k resistor value is too high to supply the transistor and the zener diode properly when the input voltage is less than 69V.
The base of the transistor needs about 0.5mA when its output is 60mA and the zener diode needs 5mA but will work poorly with 1mA. The total of 1.5mA in the 10k resistor needs 15V plus the 13V for the zener diode and another 0.7V for the transistor needs a minimum input of 29V for very poor voltage regulation.
 

AnalogKid

Jun 10, 2015
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I should be able to save around 60% of the cost of a LDO.
What you have is not an LDO, either in circuit configuration or in performance.

According to your numbers, you need a minimum 18.5 V input to maintain a 12.3 V output. That is a minimum dropout voltage of 5.2 V, about twice as much as with an LM317 or 7812. A true LDO would have a minimum in-out differential of about 0.5 V at your output current. Part of the problem is that R1 is too large.

One of the key circuit features of a true LDO regulator is the ability of the pass transistor to nearly saturate. For a positive output voltage, that usually means a PNP or p-channel pass transistor in a common emitter/source configuration.

Analyzing your circuit would be much easier if you provided information:
input voltage range
regulated output voltage
minimum output current
maximum output current

ak
 

WHONOES

May 20, 2017
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See circuit attached. It is an oft' forgotten simple voltage regulator update with an FET. It gives much better load regulation than the BJT version.

upload_2018-1-25_6-8-50.png
 
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Lord_grezington

May 3, 2013
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I was hoping to use this as a 12-15V supply for driving gate voltages as well as a normal LDO to supply logic circuits. The main battery supply will be a 48V lithium battery, but I think I need more current (for very brief time) to charge up the bootstraps and overcome the gate charges on the mosfets.

from this min current will be around 5.5mA

I understand I probably need more current, so I change some of the components and came up with this.



upload_2018-1-24_9-6-44.png

From this I got this data
upload_2018-1-24_9-9-40.png

I did this to see where the limits would be from a current draw of around 1.2A. With this sort of current iI will need a full buck converter which we cant afford on this. What is shows is the current starts to drop when the input voltage drops below 19.7V (which is OK for the application).
Max current through the Zener is 40mA, which is very large but is ok.

I will also simulate whonoes suggestion,

Thanks
 

(*steve*)

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Why do you want to waste 75% of the energy in your battery? A lower voltage battery will be cheaper for the same capacity in Ah, or have a higher capacity at the same price point.

Is your device just a small load compared to the main use for the battery (that might justify it)

Also LDO has a meaning different form what I think you think it has. It means Low Drop-Out, used to describe a subset of regulators that can maintain their output voltage even when the input voltage drops very close to it (e.g. with a 0.5V difference rather than the more typical 1.5 to 2.5 volts).
 

Lord_grezington

May 3, 2013
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Hello Steve

The battery is 48V 2.26KWH and the motors are rated 48V, I did not make the specs, they were just given to me along with a budget for a PCB BOM (which is around half as much as I would like it to be).
 

BobK

Jan 5, 2010
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Steve’s point is that you do not need an LDO to go from 48V to 12V. It has no advantage over a normal regulator, which will regulate to 12V from any voltage over 15V. There is not even any difference in efficiency, while your Zener regulator is actually less efficient due to the wasted current going through the zener.

Bob
 

WHONOES

May 20, 2017
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Small update on previously supplied circuit. The original works fine but the j fet is limited to about 35V.
The newer circuit is good to about 50V though you will need to use the A version of the 2n2904. Other than that, everything else is the same.
Also attached is a simulation graph showing regulation for 20V and 48V supplies and a load variation of 100:1.
Load regulation comes out at about 14mV.
 

Attachments

  • simple regulator 3a.pdf
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  • simple regulator 3a graph.pdf
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WHONOES

May 20, 2017
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Fair enough AK but, you don't learn anything by just buying stuff in.
 

AnalogKid

Jun 10, 2015
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Agree. I never use ebay modules, both because I don't have to and because I've got a ton of parts to burn through. But I recommend them when my read of the TS situation seems to call for them. "Half the budget" is a trigger. Not knowing what an LDO is speaks to a certain level of experience.

Back to the problem, even if the circuit in post #1 worked, it would not regulate nearly as well as any of the following suggestions. How tightly does the output need to be regulated? +/-xx%, ++/-xx volts. etc?

ak
 

WHONOES

May 20, 2017
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AK. I try not to buy ebay bits in but, must confess I do so occasionally out of sheer laziness. But stuff like the circuit supplied can be knocked up with bits from goodie drawers, bins, boxes wherever in no time at all. I suppose the main virtue of being able to do so means that you can build something that may be a bit off grid or may be needed instantly.
 

Gumby_Kevbo

Jan 25, 2018
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The circuits shown suffer from rather high dropout voltage, moreso the MOSFET version. They can be made to work if you have enough input voltage, and a big enough heat sink. If you can arrange to provide a bit more voltage to to the gate (or base) circuit than the main supply, then you can achieve near zero dropout voltage, and higher efficiency (assuming you can pick your input voltage).

If the input is a from a transformer, it is possible to place a voltage doubler ahead of the main rectifier and boost the voltage for the drive that way. If it is a DC supply, you can run e.g. a 555 multivibrator feeding a voltage multiplier. It may also be possible to use auxiliary windings or taps on the transformer to get the boosted supply.

Another way to get low dropout is by using a PNP or P channel output device, so the required drive is less than the input voltage. This achieves low dropout, but adds gain to the regulation loop, and can lead to poor load transient response and even instability. Better to use the N-channel/NPN which inherently attempts to stabilize load transients.
 

Heinzy

Jan 1, 2016
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For your simple FET based regulator, why use a 2V4 volt zener diode? It has a poor (positive) Temperature co-efficient compared to say 5V1 which is close to zero. Or am I missing something?
 

(*steve*)

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The circuits shown suffer from rather high dropout voltage, moreso the MOSFET version. They can be made to work if you have enough input voltage, and a big enough heat sink.

The op has an input voltage of 48V.

He doesn't need anything remotely close to a low dropout regulator.
 

BLAUJUNK

Mar 28, 2011
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It looks like it should work.

The transistor will dissipate almost 15W at your rated maximum load and input voltage. Make sure it's rated for that and that it is sufficiently heatsinked.
This is the most common regulator circuit in existence I have been using it since the early sixties for many different purposes. Have you done any voltage checks without load. The output should be zener voltage minus 0.7 Volt for the transistor threshold voltage.
 
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