L200 and LM741 Power Supply

Hello,
I built the attached schematic. I found that it is working, the minimum output voltage is around 2.8V, the maximum is around 20V. The minimum current is 0.25A and the maximum current is 1.6A. I used an 15Vac/2A transformer and a 4A bridge rectifier. I was wondering why the output voltage and current is not falling down to 0V, and I found that the voltage on the 5v6 Zener is about 0.4-0.5V. I disconnected the GND for 741 and for L200, and I found that the voltage on the zener raised to about 4.1V. I replaced the zener with a new one, and the same thing happened again. I also tried with a new L200, but the same thing happened again.

I can't find the problem, can someone please help me with some advices ?

Hi 123mmm
Welcome to ETO

Did you look at the datasheet for the L200 ?

Have a look here .. **broken link removed**

The 2nd bullet point on pg 1 .. ??

Have fun

S

C1, D5, D7 and R1 are there to generate a negative operating voltage. It is filtered by C4 and stabilised at -5.6 volts by zener D6. The 'Ground' pins of the L200, the opamp, and R5 are connected to this -5.6 volts. They must NOT be connected to Ground.

Use of this below ground operating voltage for the negative connection of these parts is what will allow the power supply output to operate down to zero volts output.

If it's not doing that, I'd check your connections. First I'd look to at the case/tab of the L200. Even though it's listed as being GND, in this circuit, it must not be connected to GND, but must be connected to the same net as pin 3.

I checked the tab and it is connected to the -5.6V voltage, but if the tab (and pin 3) is connected to -5.6V, then the voltage drops as I specified in the first post.
I also checked the Zener diode, the 2 1N4004 diodes, I replaced L200 and LM741, but the problem persists.
I checked the connections and they seems to be good, no bridges.

What else should I check ?

123mmm said:

I checked the tab and it is connected to the -5.6V voltage, but if the tab (and pin 3) is connected to -5.6V, then the voltage drops as I specified in the first post.
I also checked the Zener diode, the 2 1N4004 diodes, I replaced L200 and LM741, but the problem persists.
I checked the connections and they seems to be good, no bridges.

What else should I check ?

Click to expand...

Hello 123mm and ETO

Today I build the same circuit with exact the same problems as you had. It seems that out of the negative help voltage net, which should provide the ability to regulate down to 0V around 40mA is drawn. This leads to a break down of the voltage. See R1 -> 680x0.04 = 27V ... that can't be stable.
But let's not hesitate and provide our own negative voltage from a bench supply: That done i get very bad behavior. If the output voltage (P2) is set to max, no current limiting with P1) is happening. This only works when you drop the output voltage down with P2. At a output voltage range (set with P2) from 20-80% the current regulation seems to work quite good. But sometimes a latch up can occur and max-current is set. That is not what i want to have for a PSU The voltage regulation characteristics are bearable. At my breadboard build there is a shift of 0,3V when loaded with 400mA. But that must not mean anything to a proper build PCB circuit.

I build this because it looked very simple and fast to make. And I wanted to know if there is anything about the advices not to do it. Besides this IC has "adjustable" in its datasheet, it is intended to run as a fixed regulator (once it is adjusted). At least all application notes have fixed current settings.

Oh and let's have a schematic we can look at without getting bad eyes. This is schematic 064 from the Book 302 Schaltungen, Elektor.

One thing: Personally I don't think this was a good design or something that works. First, no pcb layout was provided and second: who wants a bench power supply that drops to -1...-2V (when the negative bias is working) if you pull P2 in zero position? Because that is exactly what happens

For the Application-Notes-Aficionados:
A DESIGNERS GUIDE TO THE L200 VOLTAGE REGULATOR **broken link removed**

Despite I am not building this, I've done the dish-cleaning before serving the food.
Here are my design files for a PCB if someone wants to play with it.

Jedzia said:

One thing: Personally I don't think this was a good design or something that works. First, no pcb layout was provided and second: who wants a bench power supply that drops to -1...-2V (when the negative bias is working) if you pull P2 in zero position? Because that is exactly what happens

Click to expand...

Add a (about 1k) resistor to the 10k pot so the total resistance can not drop below 1k. That should help that problem.
OR
Try this part. 3080 It will go to 0 volts with out a negative supply. Have used it. There are other versions for different current levels. 3080, 3081, 3083=3A 3083

Note how one LT3080 is set to regulate current (max) and the other is set to regulate voltage.

ronsimpson said:

Add a (about 1k) resistor to the 10k pot so the total resistance can not drop below 1k. That should help that problem.

Click to expand...

Thanks, good idea. That works and must be adjusted to hit 0V.

ronsimpson said:

OR
Try this part. 3080

Click to expand...

The next I try (for fun)will be a LM317 U/I doubledecker. But this was about the L200 and especially that Elektor Design from the year 2000. The schematic spooks around the web, but you never see someone who has built that successfully. Except one on youtube, who discarded it cos its worse characteristics for a LM350 build

german, not very informative, beginner(no offense)
Oh and for me it is not vital, more a junk-parts-bin party. I just made the same observations as the OP.

123mmm said:

.................
the maximum current is 1.6A. I used an 15Vac/2A transformer

Click to expand...

Note that a 2Arms transformer should be loaded with no more that about 1Adc steady state , due to the high RMS current drawn by a rectifier-filter supply, otherwise the transformer can overheat.
The transformer derating for such a supply is about 50%.