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0-30 Vdc Stabilized Power Supply


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Hi Pigster,
Nice going, you fixed it. ;D

How well does it perform?
If you can find a very big 10 ohms resistor or ten smaller 100 ohms resistors in parallel, use them to provide a 30V/3A load:
1) can it provide a 30V/3A output?
2) How much or how little is the AC ripple voltage at the output?
3) How much or how little does the output DC voltage change from 30V from having no load to having full load?

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The original circuit should work fine up to 15V at 1A if you replace the old opamps with the newer higher voltage ones. You probably should recalculate the resistors that set the maximum voltage and c

Hi, as promised I made an English translation of my working. Maybe there is few mistakes and I am sorry for that ! Good reading. ExplicationEN.pdf

February 23 above on this page has the latest schematic of the revised 3A lab power supply.

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Hi Gruber,
See if www.farnell.com has parts distribution in your country, click on your flag.

Quote: The OPA445AP (Texas Instruments, Burr-Brown division) has a 90V rating, a 741A opamp, an MC34071CP (used to be Motorola, then ON Semi and now maybe another name), a TLE2141CP (Texas Instruments) and a National Semi opamp have a 44V rating, but they aren't guaranteed to work so high, just withstand it for a moment without damage. "Continuous exposure to this stress level may affect product reliability".

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Hi all,

I made the power supply and it 'all' works. The supply delivers a 30V/3A supply! I can change the voltage, and I also can change the current, but not over the whole range. And the LED doesn't light on any time. I haven't tested anything (I'm not at home right now) but maybe someone can give me the solution because he is known with the problem.....

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Hi Smitjuh,
Welcome to our forum. ;D
It is good to hear that your project works. Did you make the original or the modified one?

You should be able to adjust the voltage from very close to 0V all the way to 30V.

The current regulator is supposed to reduce the output voltage and light the LED whenever the load current exceeds the setting of the current pot. So you need to check the few parts that perform the current regulation. ;D

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Hi  audioguru,

Thanx for the warm welkom  ;)
I made the power supply with parts I still had in my hobby room. I created it in one weekend!
But not fully working.. whats a shame.

Pin 6 from U3 has a 33 volt potential. When I set pin 3 of U3 to 1V. as reference, what must pin 6 and pin 2 become? which values? My LED won't light... I checked all components in currentmaes. circuit. (resistors, Q3...) they look fine.

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Hi Smitjuh,
The current regulator, U3, is used as a comparator with overall feedback through the entire voltage regulator. Its reference voltage from the pot is referenced to the rectifier's common voltage and one side of R7. Its input pin 2 is connected to the project's output's 0V terminal which is the other side of R7. When load current flows in R7 then a voltage is developed across it and when its voltage exceeds the voltage from the pot then U3's output voltage reduces and reduces the project's output voltage just enough to regulate the current at the setting of the pot.

The voltage at the input of U2 has a max of only 11.2V, so in order for U3 to reduce this voltage through D9 then the output voltage of U3 must drop from its resting voltage of about 33V down to 10.5V or less which turns on Q3 to light the LED.

When you set the pot's voltage to 1V without a load, then a load current slightly more than 1/0.47= 2.13A must flow through the load and R7 for the current regulator to become active.

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Hi , i have built the power supply and for several try it has worked but ( it has never show me current at the ampmete rconnected in series with the positive wire) when i loaded  suddenly, i  just can read  0 v in voltmeter output.
I dont know if it has to do with Q1 ant the shutdown protection or just Q1 has blown off.
Any idea?
Thanxs

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I made the circuit as it is on the web.
The thing is, that there was no reading of the ampmeter ( i have not reading of the ampmeter before ) when the power source came down, but the led wasn´t on when it happened.

The load was a stepper motor of 1.4 amps per phase and  5 v, connected by a tanslator circuit of TIP120 transistors (that the have a protection diode), that were controlled via a circuit with the parallel port.(all conections were ok)

I think that the problem has to be with Q1 as it is alreay on ,and i read a 45v betwen the colector and the emiter ,probably in the saturation region.

I dont know if Q1 has burn out or just the negative rail is still at the the level of the shut down protecction circuit.

Thanks for your support.

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Hi Zama,
Q1 is supposed to be turned off when the project is operating. It turns on only when the mains is turned off. Q1 must conduct the short-circuit current of U2 which could be 20mA if your project's U2 is strong.

If the project's output voltage is set to 30V when the mains is turned off, then Q1 must dissipate up to 30V x 20mA= 600mW which is much too much for a little BC558 transistor which causes its destruction.
That's why in the forum I recommend using a TIP31A power transistor for Q1. ;D

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Hi audioguru,

reading this forum, you're the man for all my questions ... I think  ;)

As entioned before in this topic, my current control circuit won't work properly. I will give some measurement results. On Pin 3 I can control the voltage from 0 to 1,7V (no load) witch is correct, because R18 forms together with P2 andR17 a resistor devider. At max, the next equation gives us: (10k/(56k+10K+33))*10,7=1,62V

When I put a load on the circuit (Resistor) I can control pin # from -0,2 to 1,5V... Thats is already strange for me...

The Base of Q3 has to become more negative than his emitter voltage so the trasistor can switch on, what lights the LED. To get the Base voltage far below 34V, pin 6 of U3 must get Low. This is only possible when Pin 2 of U3 becomes higher then pin3 of the op-amp. And here is the mystery.. pin 2 only gets lower when a current is flowing trough the shunt resistor R7!  ???

What am I missing?? I hope someone can help me!! thnx in advance!

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Hi Smit,
The current regulator works by using U3 as a comparator to compare the voltage across R7 caused by load current, with the setting of the current pot P2 and causing the output voltage to be reduced through D9 to maintain the set current.
Since the reference voltage is +11.2V, the output of U3 goes from being high when it is not regulating the current, to at most +10.6V during current regulation. Of course when the output of U3 is only +10.6V or less then it turns on the LED through Q3.

3A though R7 produces only 1.41V at pin 2 of U3 but pin 3 is 1.70V when the pot is at max. Why? Because pots have a wide tolerance and if your 10k pot is actually 8k then the project will produce very close to 3A.
Another member reduced the value of R18 and added a trimpot in series to adjust the max setting of P2 to be exactly 3.0A. He also added a trimpot in series with the voltage-adjusting pot to adjust the max setting of P1 to be exactly 30.0V.

The schematic of the current regulator is confusing to look at since the output 0V is not the same as the rectifier 0V with R7 is in between. It is not confusing if you look at all voltages with the rectifier 0V as a common reference.

Your 11.2V reference is low at 10.7V probably because you used a zener diode for D8 that is rated at a high current. The original circuit uses 4.7k for R4 which sets a zener diode current of only 1.2mA. I recommend using a BZX79C5V6 zener diode which is rated at 5mA, and changing R4 to 1k for a zener diode current of 5.6mA.

In my sketch the load current is 3A and the pot is max. Reduce the setting of the pot P2 and you will see how it regulates the current. ;D

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U3 is "not really" used as a comparator, at least not in a "true sense". It is true, while the output current draw is below the current set point level, the output of U3 is sitting at the positive rail, but that's where the comparator similarity ends.

It's configured as an integrator.......... as the voltage across R7 increases, and for the sake of simplicity, I'll say U3 starts to "turn-on" (go towards it's negative rail), the clamp diode D9 starts to pull down the output voltage set point votage at U2 pin 3. It will only pull it down as far as it needs to so that the current limit point is not exceeded (foldback). C8 controls the rate at which the level changes. U3 does not slam to the negative rail like a comparator would, although it could go there if it needed too. In fact, if U3 was being uesd as a true comparator, the power supply would go into a "hic-up mode" as U3 went rail-to-rail.

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Hi Indulis,
Welcome to our forum. ;D
You are correct by stating that U3 is not being used as a true comparator. It is using its inputs to compare the voltage across R7 that is caused by load current, with the voltage from the current-setting pot. Then it uses its internal gain of a few hundred thousand to very accurately reduce the output voltage so that the load current is exactly what is needed.
U3 has negative feedback and is used as an error amplifier. Comparator ICs never have negative feedback because they would oscillate.

The time constant of R21 (10k) and C8 (330pF) is pretty quick at only 3.3us so I would say that it is used for loop compensation to avoid oscillation at a high frequency, instead of being used as an integrator. ;D

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Yeah, I'm sure C8 is used as a loop compensation cap (limit/control U3 output slew, but that's what feedback is all about), but that doesn't stop it from being a integrator, which IT IS!!!!

OK.

I saw a post on another fourm that covered feedback in comparators circuits and why it is used!!!

Comparator ICs don't have built-in frequency compensation so they can have gain at high frequencies. Therefore they never use negative feedback with a resistor or integrating capacitor in order to avoid oscillation. Usually comparator circuits use a small amount of positive feedback to avoid oscillation at the threshold voltage caused by stray capacitance or noise. ;D
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Hi audioguru,

I changed R4 to a 1K value, and my Voltage value has climed from 10.7 tp 11.2V!! I did not used the BZX zener, but this one will be delivered soon I hope. I looked at you're schmatic and see that the voltage acros R7 becomes 1,4V when there is a current flow of 3A.

I tried it and what happens.... When I measure the voltage accros R7 when there is a current of 3Amp, the voltage is -1.4V!!!!!!! How can it become negative???

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I think I made an error on my previous calculation which I corrected.
It looks like the current-setting pot has scaling resistors such that the max current the project will try to give is much too high at 4.1A. It will be even higher if the 10k pot is actually higher than 10k. If the pot is 12k then the max current the project will try to produce is 4.74A!
The project will melt or smoke if the output voltage is low. The transformer will be overloaded.
I think a trimpot should be added in series with R18 to reduce the max current to 3A. A 4.7k or 5k trimpot wired as a rheostat would work well. ;D

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Hi!!

I replaced some components with the better components mentioned by audioguru. The BZX zener and the both TIP31A's also some resistors. But my LED won't light!! The output from U3 (pin6) is 34V when pin2 is in voltage value above pin3 Pin6 remains 34V. but when I turn back P2, pin2 of U3 gets below pin3 and the ouput, pin6, is decreasing til say, 26V. Is this in theory enough to light the LED?

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