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0-30V Stabilized Power Supply


redwire
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(40V - 1.35V - 0.55V) x 5A= 191W.

can you explain me the 2 values??? 1.35V and 0.55V...

5A in R7 that is 0.27 ohms= 1.35V.
5A/3= 1.67A x 0.33 ohms (the emitter resistors for the three output transistors)= 0.55V.

And i can understand the part....
" But then its chip will be extremely hot so maybe 90- degrees C would be more reliable."

Do you want to say the it would be safer for the transistor to work in a temperature of 90 degrees C? What do you mean when you say "its chip" ?

Yes.
It is safer not to operate the transistor silicon chip that is inside the case of the transistor at its absolute maximum temperature. Some manufacturers say so on their datasheets.
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What about the opamp differential voltage? U2 has almost none. Because the gain is low. The resistance is low.

Kevin,
Didn't you learn anything about opamps? U2 has a DC gain of 200,000 times. If one input is +2.0V then the other input is +2.000005V and the differential voltage is 0.000005V (almost none). The negative feedback sets the gain of the entire amplifier to 2.7 times.

What resistance is low? The input resistance of U2 is 70M ohms as listed in its datasheet. U2 drives the fairly high input resistance of the driver transistor.
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U2 doesn't have any differential voltage at higher output voltages. At zero volts ouput, it has more differential voltage because of the -3volt power supply. I'm not sure how this can affect the voltage regulation.

Kevin, you are completely WRONG!
1) The output voltage has only a slight effect on the differential input voltage of opamp U2. It increases slightly, it is not eliminated at higher output voltages because U2 is an amplifier.
2) The power supply voltage has no effect on the input differential voltage.
Opamp U2 does not have a negative supply voltage. It did in the original project that had many errors.
Opamp U3 has a -1.3V power supply voltage so that its output can go to about -0.7V to forward-bias D9 to force the output voltage of the project to zero when the load current of the project exceeds 3A. 
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On my unit with an output of 30.0V,  U2 pin 6 has an output of 30.07, pin 2 11.46V,  Pin 3 = 11.44. No load

If U2, pin 6 has an output of 13.8V, pin 2 = 5.02,  pin 3 =5.00V

R12 is 55.6K , R11 with pot = 34.5 KOhms;  amplification = 1+ 55.6/34.5 = 2.61.  This is very close to  measured values.

Values measured with Radio Shack multimeter so digits shouldn't be taken too far.

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The MJ15003 is a good power transistor.       The thermal Resistance, junction to case is about the lowest you can get but getting that to transfer to the heat sink is more difficult.  The power derating of 1.43 W/C is high if you are running hot but when you start at 250W you've got plently of  headroom.    Have you checked out the MJ11016 it is a darlington transistor and if you don't mind a 1v drop (or less) in the total output, it has a gain of 1000 and will reduce the load on the BD139. The thermal Resistance, junction to case is close to the MJ15003.  I did a test with only one MJ11016 driving a 5A-12v  load on a large heat sink (outside a case) with no fan for about 5 minutes without failure. The BD139 only had a small heat sink and did not get that hot.     I would feel very confortable with 3 transistors no matter what type (MJ15003, 2N5886 or MJ11016), running 5-7 amps.


A have been working on this a log time ago and i have just one question for you, which element in circuit is total allowed output current adepend on??
And just one thing, is project according to your parts list work properly at all or you did some changes on project, i am talking about 5A modified project, last table i have is below.

post-82005-14279144602566_thumb.png

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The entire output current flows through the output transistors and their emitter resistors. Since they have a high voltage across them when the output voltage is low or is shorted then they get very hot when the output current is 5A then they need good heatsinking.

The MC34071 is not available in a through holes DIL package anymore so we use only the TLE2141 opamps now.

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denci,  

This is a picture of a MJ11016 (upper left hand corner of picture) I was using to test my new smd board
http://www.electronics-lab.com/forum/index.php?action=dlattach;topic=7317.0;attach=19559
At this time I was using a brake light bulb  (about 2.5 A).   The previous 5A test was done on a cold day in my garage and the heat sink was not in a project case.   These are not the worst case conditions.   When you package things up and contain the heat you are going to need at least 2-3 transistors. large heatsinks and a small fan if you want  5A and rock solid durability. Then your output will likely be determined by what size transformer you can afford.    Digikey and Mouser have a nice 6.25 A, 28 V transformer by Triad for about $32.
Note that the worst conditions are at low voltage and high current (or a short).   When the voltage was at 1 V using a 0.27 Ohm, 10W resistor just like the one on the board, you could feel the heat pouring off the MJ11016.  The same happened when I set the current for 2.5A and shorted the output.  I didn't hold the shorted conditions for an extended period because  heatsink was getting quite.  The bd139 without a heat sink was not hot.  

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Guest sonycman

Hello, audioguru!
Trying to adapt your schematics for microcontroller control and indication.
Thinking about changing 11.2 volt reference voltage to 5.0 volt and using digital potentiometers.

Which purpose C5 serves for? Should I connect it phisycally close to Q2, as on the picture, or better close to U2?
Can it be ceramic type, or better polyester?

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Thinking about changing 11.2 volt reference voltage to 5.0 volt and using digital potentiometers.

The 5.6V zener diode has very good voltage regulation (lower and higher voltage zener diodes are worse) and has zero voltage change when the temperature changes unlike other zener diode voltages. The opamp U1 that feeds this zener diode is a current source for even better voltage regulation. Another member said to use a 78L05 regulator instead.

Which purpose C5 serves for? Should I connect it phisycally close to Q2, as on the picture, or better close to U2? Can it be ceramic type, or better polyester?

C5 is a high frequency filter. Mount it close to opamp U2. It should be a polyester film type.
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Guest sonycman

zTher 5.6V zener diode has very good voltage regulation (lower and higher voltage zener diodes are worse0 and has zero voltage change when the temperature changes unlike other zener diode voltages. The opamp U1 that feeds this zener diode is a current source for even better voltage regulation.
another member said to use a 78L05 regulator instead.

What about TL431 precise voltage reference? Looks even simpler than opamp and its discrete elements.
Though TL431 might need buffering too...

C5 is a high frequency filter. Mount it close to opamp U2. It should be a polyester film type.

So it is shunt for U2. Okay, understood!

Thanks for help!
Sorry for bad english...  ::)
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sonycman,  U3 requires a voltage between 0 and 1.5V (for 3A output) and U2 requires a voltage between 0 and 11.2V to adjust the output voltage.  If you are using a microcontroller, couldn't  you delete U1, and U3 and feed a DAC signal to U2 of 0-5 V to control voltage (Yes you will need to adjust the gain for U2 by changing R12 to about  70K and R1 to about 10K so you get a gain of about 6 so that the 0-4.5V signal will provide the same amplification as the original project).  The microcontroller would also control the current by measuring  the voltage drop across R7 and  replace  U3.  If the voltage drop across R7 was greater than your set voltage, then the microcontroller could automatically drop the voltage to U2 until it matched or was lower than the set voltage.  You could also incorporate a temperature sensor that could also reduce current if a certain value is achieved.    Make sure you pick a processer that has enough pins to support these features and a LCD display.  Not sure if a digital potentiometer is necessary. 

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Guest sonycman

redwire,
Yes, using DAC instead digital potentiometer is another way to go, will see about it.
It must be the DAC with the same precision and external reference voltage as digital POT.

As to remove U3 and forcing to limit current in software by MCU - did`not agreed in that.
Current limiting must be done in hardware only by high speed opamps for reliability reasons.
MCU work is to set up references for opamps and to read voltage and current values via ADC for displaying it.

Thanks for helping!

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  • 3 weeks later...
Guest electroguy

Unfortunately, both the op-amps MC34071 and TLE2141 is not available in my country but there is MC34074 available.
Are you using the single op-amps or the duals ?
Is it possible to replace the three MC34071s with one MC34074 ? but the problem remains in U2 for the pins 1 & 5

post-108796-14279144631804_thumb.jpg

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Unfortunately, both the op-amps MC34071 and TLE2141 is not available in my country but there is MC34074 available.
Are you using the single op-amps or the duals ?
Is it possible to replace the three MC34071s with one MC34074 ? but the problem remains in U2 for the pins 1 & 5

An MC34074 will probably get too hot.
You do not say what is your country. Go to www.farnell.com and click on the flag of your country. They have warehouses all over the world called Element14.
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Guest electroguy

I am from Bangladesh,it is a country in South Asia.
Unfortunately, again,  I don't see the name of my country in Element14.
If I use MC34074 with a big heat sink and a computer fan and did you use the offset null for U2 ?
On the other hand, MC34074 do not have any offset null.
Save me!!!

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I do not know why you cannot buy ordinary parts in your country.

An MC34071 quad opamp is in an epoxy package that cannot have a heatsink. You can glue on a heatsink but the epoxy will not conduct heat to it well.

The input offset adjustment on opamp U2 sets the output to exactly 0V when the voltage-setting pot is at zero.
Without adjusting the offset then the output could be plus or minus 20mV.

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Should I try this old modified one proposed by Ivan on 02/2010.
Oh, I forget to mention that I already bought the old parts last year :(

The circuit you found uses a 7812 voltage regulator so that the original opamps do not have a supply voltage too high. Then opamp U1 will not work unless its circuit is changed to reduce its output voltage.
The maximum output from opamp U2 is about +9.5V and the maximum output voltage from the project is about +6V. If the original 24VAC transformer is used then Q2 will get extremely hot and might fail and the output transistors will get very hot.
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Unfortunately, both the op-amps MC34071 and TLE2141 is not available in my country but there is MC34074 available.
Are you using the single op-amps or the duals ?
Is it possible to replace the three MC34071s with one MC34074 ? but the problem remains in U2 for the pins 1 & 5

Audioguru wrote:  An MC34074 will probably get too hot.


U1 and U3 need to provide very little output current.  U2 is the real driver in this circuit.  If electroguy utilized a darlington power transistor instead of the 2N3055 then the BD 139, and subsequently the MC3074 would need very little current to drive the BD139.  I think the quad opamp could handle this load.  I would suggest the MJ11016.  Yes, you don't have the input offset pins, so setting the voltage to exactly zero may be impossible but a few millivolts should be a problem?  I think this would be more preferable than building the old circuit.
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