I’m a novice. I have a fixed 12V 30 amp regulated power supply. I don’t want to modify it internally, but would like to make a separate unit to attach to the output terminals to provide a variable 0 - 11 volt 5 amp supply with a digital volt/ammeter. I have the meter and probably all the needed transistors, etc. I would to maintain or even improve the regulation. Can anyone refer me to a good circuit schematic to accomplish this?

Larry Cruzen
<SNIP>

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The LM138 series of adjustable 3-terminal positive voltage regulators is capable of supplying in excess of 5 A over a 1.2-V to 32-V output range. Don't think you will find an IC going down to 0? And what are you going to use 5A for at 0-11VDC? The LM317 can give you 1.5A.

 

Thanks for the prompt reply, Sam. 0V and 5A were rather arbitrary numbers. 1.5V min and 1.5A max would be great. Where can I get a schematic?

 

Hello,

When you have 12 Volts on the input of the LM138, the output will just reach 10 Volts at max, due to the voltage drop.
If you want a higher output voltage, have a look at some LDO regulators.

Bertus

 

The typical is on the PDF or search here for LM317 and you will find one I put together with a rectifier to use with a 120/24/12 VAC XFMR. On the schematic note the 243Ω precision resistor that is necessary and I used a small trimmer potentiometer for voltage adjust to be a bit more precise on setting output V. It does have some backlash but it will get within a few hundredths on the Vo setpoint.

http://pdf.datasheetcatalog.com/datasheets/560/275630_DS.pdf

 

Bertus is correct, note on the PDF the setoff between Vin and Vo.

 

The typical is on the PDF or search here for LM317 and you will find one I put together with a rectifier to use with a 120/24/12 VAC XFMR. On the schematic note the 243Ω precision resistor that is necessary and I used a small trimmer potentiometer for voltage adjust to be a bit more precise on setting output V. It does have some backlash but it will get within a few hundredths on the Vo setpoint.

http://pdf.datasheetcatalog.com/datasheets/560/275630_DS.pdf

Thanks. The first circuit shown looks good. Should the output be protected with a 1.5 amp fuse?

 

Good question. On mine, I depend on the fusing for the XFMR and just consider it as a module to adjunct the XFMR. The LM317 datasheet from TI is attached and is a bit more inclusive. Under section 8.3.1 Load Regulation are some notes as to what it is being used for and some capacitive protection from downstream spikes. The regulator itself is very hardy and has some internal protections. At 12VDC 1.5A there is no need for personal shock protection and as I mentioned it is a very hardy device. Your 12VDC power supply should have its own fuse or breaker.

http://www.ti.com/lit/ds/symlink/lm117.pdf

 

Hello,

If you would like to have a lower drop-out voltage, have a look at the LT1085.
That has a drop-out voltage of max 1.4 Volts at 3 A.
At 1.5 A the drop-out voltage is about 1.2 Volts.

Bertus

 

Note that a linear regulator needs to be on an appropriate heat-sink, as the power dissipated is the input voltage minus the output voltage times the load current.
The LM317 (and most other IC regulators) will limit the current to prevent over-heating if the heat-sink is not adequate.

 

The LT1083-85 series regulators may be a bit expensive.
The LM1086 is <$2 with a 1.5A output and a minimum drop of 1.5V (compared to the 3V or so for the LM317).

 

Good question. On mine, I depend on the fusing for the XFMR and just consider it as a module to adjunct the XFMR. The LM317 datasheet from TI is attached and is a bit more inclusive. Under section 8.3.1 Load Regulation are some notes as to what it is being used for and some capacitive protection from downstream spikes. The regulator itself is very hardy and has some internal protections. At 12VDC 1.5A there is no need for personal shock protection and as I mentioned it is a very hardy device. Your 12VDC power supply should have its own fuse or breaker.

http://www.ti.com/lit/ds/symlink/lm117.pdf

In looking at the Texas Instruments document provided in your link, I see that this transistor provides its own thermal and overload protection. Sounds very nice. I also note that no capacitors are needed in the circuit if the device is kept close to the source filters and rapid transient response is not needed....that makes a mighty simple circuit.

 

Indeed it is. It will probably be more than 6" from the power supply so add the input cap. I suspect you won't use less than 3V and most of the stuff I breadboard draw less than 200mA (most way less) unless you are driving a bunch of solenoid relays or motors. Have fun and Welcome to AAC!

Here is a link to a source. 10 for $0.93 delivered from china. Downside is it takes a while to get them but you will pay more for one in most places plus shipping.

https://www.aliexpress.com/af/lm317...eShip=y&SortType=price_asc&page=1&groupsort=1

 

Indeed it is. It will probably be more than 6" from the power supply so add the input cap. I suspect you won't use less than 3V and most of the stuff I breadboard draw less than 200mA (most way less) unless you are driving a bunch of solenoid relays or motors. Have fun and Welcome to AAC!

Here is a link to a source. 10 for $0.93 delivered from china. Downside is it takes a while to get them but you will pay more for one in most places plus shipping.

https://www.aliexpress.com/af/lm317...eShip=y&SortType=price_asc&page=1&groupsort=1

Thanks, SamR. Yesterday I ordered 15 LM317s from Amazon for $6.72, free shipping and 2 day delivery.

I have a comment about this circuit...out of ignorance. Remember, I’m a novice. Just an old, old mechanical engineer. In fact, when I went to college in the mid 1950’s, all was running on vacuum tubes. Transistors were new mysterious devices that carried electrons in “holes”. The sparky guys had current going in one direction and electrons in the other direction. Don’t think that has been sorted out yet. Then I spent a career designing jet engines and accepted that transistors were probably here to stay. So I recently decided that I had time to really understand these infernal devices. Bought some breadboards and a whole slew of transistors, resistors, capacitors, etc and began experimenting with some simple circuits.

Now the comment. It looks like the power is taken off this circuit across a variable resistance.Seems that would normally result in very poor voltage regulation. I guess it must be the transistor’s job to maintain the set voltage. Pretty clever if that’s the case.

 

The 3 legs on the IC are IN, REF, OUT. the potentiometer and the 234Ω resistor "tell" the IC what Voltage to set the OUT pin to. Some V is used internally by the IC so there has to be an offset between the max IN and max OUT to operate the IC. The 243 or 240Ω resistor needs to be a 1% tolerance to accurately set the OUT V. I've been messing around with Ham radios since the 70s and started out as a ME at Tenn. Tech in 69 but ended up doing Process Control Automation for a chemical refinery. Never really did any serious electronics until I retired a number of years ago to give me something to do and keep my mind occupied. I'm still learning and the guys around here have a lot more experience than me. Came from a long line of Mechanical, Marine, Railroad and Steam engineers going back to the time of James Watt. Have some wonderful old books of one of my great-great-grandfathers with etchings showing old steam equipment. Back then there were few dimensions you just took the picture and built it from scratch. Saved all the casting forms in case you decided to build another one.

 

Good to know you, Sam. I’ll put this together tomorrow and see if it will make smoke.

 

It looks like the power is taken off this circuit across a variable resistance.Seems that would normally result in very poor voltage regulation.

The resistors are just to set the output voltage.
No output current flows through them.

Below is the LTspice simulation of a typical LM317 circuit.
The LM317 is an integrated circuit (not just a transistor) that adjusts the output voltage to maintain a constant 1.25V (nominal) between the OUT and the ADJ pins.
The current through R2 is thus constant at 1.25V / 121 = 10.33mA
The output voltage is then determined by the value of R1, or Vout = (10.33 mA * R1) + 1.25V.
The output voltage is shown for three values of R1 [200Ω (blue trace), 400Ω (yellow trace), 800Ω (redt trace)] with the RLoad varying from 20Ω to 7Ω
You can see that there is no visible change in the output voltage as the load current changes.

 

Good explanation. If the LM317 establishes a constant 1.25V by varying the current thru R2, what establishes the value of R2? I’ve seen other schematics that show a 243 ohm resister in this location.

 

If the LM317 establishes a constant 1.25V by varying the current thru R2

No, the LM317 pays no attention to the current through R2, only to the voltage between the OUT and ADJ terminals.
The LM317 applies a constant 1.25V between those two terminals.
This generates a constant-current through R2 equal to 1.25V / R2.
It's the voltage across R1 due to this current that then determines the output voltage.

what establishes the value of R2?

It's somewhat arbitrary but, due to the minimum 10mA output load current required for the LM317 to stay in regulation (below), it's typically selected to be about 120 ohms which give a little over 10mA of current.

I’ve seen other schematics that show a 243 ohm resister in this location.

That's in error.
It's from the circuit for the LM117 military temperature version, which has a lower minimum load current.
It was erroneously transferred to the LM317 circuit and has been shown everywhere.

 

Understand, and thanks for the patience.