Lm2576 adjustable psu problem

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polashd

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I’ve made a power supply circuit based on the attached diagram which I found from the net. I modified it little bit to be able to control both voltage & current. Added high value resistors parallel to high value caps to avoid accidental shock as they may hold energy for long time. This also helps switching to lower voltage faster.

Now the problem is the output voltage is very low. It rises up to little over 7v (according to formula which should be little more then 36v) and the rise is also very slow.

I can’t understand why its behaving like this.
Can any one please suggest me?
 
Connect a 100k resistors from pin-4 FDB to ground.
I think the two diode idea is not good. If you want to try it you need a pull down resistor on pin-4.
 
Which 2576 version are you using, adjustable or one of the fixed varieties? The diode from Rsc to pin 4 clamps the output voltage at 1 diode drop above the 2576's internal reference times the 27K pot adjustment. With the 27K pot wide open this works out to only 25 V max for the adjustable version, so there is an issue there. Also, the adjustable version of the 2576 does need a pull-down at the FB input.

ak
 
I think another problem could occur. If the voltage adjustment pot is set to it's maximum value then I think the output voltage will be about 25 volts. This means that the inputs of the first op amp will be sitting at half that voltage. 12.5 volts. (Due to R3 and R4) As the positive rail for the op amps is 12 volts the inputs will be just outside this and could cause a problem.

Les.
 
polashd said:
I can’t understand why its behaving like this.
Click to expand...
Perhaps the current limiter is kicking in. What is the value of your load resistor? What resistance value is R6 set at (23k is not a standard pot value)?
 
alec_t said:
Perhaps the current limiter is kicking in. What is the value of your load resistor? What resistance value is R6 set at (23k is not a standard pot value)?
Click to expand...
I tried without R6 (completely disconnecting) and again using a 22k fixed resistor as R6, in all cases result is more or less same. The pot is labeled 22k and my multi meter measures 23.5k (4.5% more). I didn't test it with a load yet. Just measuring the output with a multi meter.
 
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I did something almost the same, the biggest diffrence being there is only one diode on the feedback pin, the voltage resistor divider feedback goes direct to the feedback pin no diode, a diode isnt necessary on this section as the op amp can easily overide the resistor dividers o/p voltage, the op amp does have the diode on its o/p.
Delete the diode from the voltage feedback section, this will also improve regulation slightly as the diode will change its forward drop with temp.
If you configure the differential op amp to have gain you can use its o/p to drive an ammeter movement (in voltage mode of course).
 
polashd said:
I connected a 100k from FB to GND and now it's working. I'm getting 1.5v to 37v now. Thanks a lot.
Click to expand...
Normally the FB should see some division of the output voltage. It should be well connected to Vout.
The FB pin is high impedance.
What you have; Vout pulls up on FB, AND the current limit pulls up on FB, BUT nothing pulls down. This why your circuit does not work well. It needs some thing to pull down when Vout is low.
 
I have two points. One is that there will be noise on any switch mode power supply circuit, and the two diodes on the feedback pin will have the effect of feeding back the peak value rather than the average. That is why the 100k resistor was so important. It might benefit to put a capacitor in parallel with R7 and one in parallel with the 1.42 k resistor. Those would help to filter spikes on the current and voltage line respectively.

Secondly, you need very accurate resistors for resistors R1 - R4. IC2A's only function is to level-shift the voltage across Rsc. Now if you use 1% resistors, and you are running at 30 V and no current, if R1 and R4 are 1% high, while R2 and R3 are 1% low, pin 3 on the IC2A will be at 15.15 V. Now to get pin 2 to 15.15 V, the output of IC2 A will have to be at nearly 0.6 V. That means that 30 V output is seen as 6 A, so your current limit would change by 6 A as the output voltage changed, without adjusting R6.

Admittedly, that is worst case with the resistors being at their limits, and in the worst combination, but even if two resistors were perfect, and two were 0.5 % off, you would see around 1.5 A error. Whatever the error, it is dependent on the output voltage.

You could change the circuit so that Rsc is in the ground line. That would need the load negative not to be at your circuit ground which might not work for your application. You would need IC2A to level shift the actual output voltage, which would otherwise be reduced by the Rsc voltage, but as the Rsc voltage is probably a lot smaller than the output voltage, that is less of a problem.

You could buy really good resistors for R1 - R4. 0.1% resistors aren't a lot of money if you only have to buy four, but you could still get an error of 0.6 A. If you want better than that, the resistors will be very expensive.

You can also trim the value of one of R1 or R2 with a series resistance. With no load on the power supply, and the output voltage low, check that there is no voltage on the output of IC2A. Then put the output voltage to maximum and check the output of IC2A. Adjust which ever of R1 or R2 gets the output of IC2A to zero. I suggest you start with 1% or better resistors, and all of the same batch, as trimming will only compensate for the initial values of the resistors, and will not allow for variations with temperature, ageing or voltage. Resistors with poor accuracy will generally vary more than accurate resistors, as well as not being adjusted as accurately.

Many designs that you find on the internet don't allow for component tolerance, so they might well work, but they might not if your components have different accuracies from the originals.
 
Here is something to try. I have not tried it.

Remove the diode from the feedback path.
Connect current limit to the on/off pin. Pin should not be pulled negative. So 100k to ground and diode to only lift up.
If you used an opamp, (with the input common mode range that includes the negative supply), then you don't need the -12V.
The IC2A will not work if the Vout is too high. Watch out.
 
Hello,

Diode OR circuits are typical in regulators, in order to allow both voltage and current limit to control the output. The arrangement is usually a little different though, such that the dont affect the voltage. In this circuit they will affect the voltage a little over temperature, but it depends how accurate you need it to be over temperature. Since the feedback voltage is low, the change in diode voltage over temperature could cause a 10 fold change in output voltage when the ratio of output voltage to feedback voltage is high. For example, for a feedback voltage of 2.5v and output of 25v, if the 2.5v changes by 0.1v the output will change by 1 volt. That may or may not be acceptable.

Be careful if you add any parallel caps in any feedback path. The simplex idea is usually to "filter" the signal but they also introduce a new pole which wasnt there previously which could cause massive oscillation.

The circuit must be tested over full input range and full output load range. In particular, the current limit must be tested carefully because of the introduction of the two op amps. Op amps also introduce poles into the feedback path which can easily cause oscillation. The usual rule of thumb is that the feedback path must be faster than the fastest response of the error amplifier. Resistors are never a problem here because they dont add any poles, but caps and op amps always do, so careful testing is always a good idea.
 
This is a very practical situation. I didn't pay attention to this matter before making the circuit and used resistors of 5% tolerance. and now the result is the current limit is nowhere near the expected value. If I replace one R of each pair (R1-R2 & R2-R3) with variable resistor to make voltage division of each pair as 50% will it solve the problem?
 
MrAl said:
Diode OR circuits are typical in regulators, in order to allow both voltage and current limit to control the output.
Click to expand...
Yes but the diodes are normally after the error amp not before. In this case you can't get to the right place.

If you really need to do this:
Remove the diode in the feed back loop. Add a 1k resistor. The resistor is needed if the POT is turned to zero ohms. The resistor will not effect the loop. It will allow IC2B through a diode to lift up on "FDB" when it needs to.
 
It would be far better to have a potentiometer to trim the ratio. You only need one adjustment. You should separate the joint of R3, R4 and pin 3 of IC2A. Take a 4k7 potentiometer, connect the wiper to pin 3 of IC2A, and each end of the potentiometer to the ends of R3 and R4 that were just connected. Then all you need to do is to set the power supply to its maximum voltage, don't connect a load, and adjust the potentiometer to give zero ouput of IC2.

Variable resistors or potentiometers are usually worse than fixed resistors in terms of accuracy and stability, which is why you would better to keep all four resistors and to add a potentiometer to just adjust out the tolerance.
 
START PARTIAL QUOTE FROM MRAL:
Diode OR circuits are typical in regulators, in order to allow both voltage and current limit to control the output
END PARTIAL QUOTE


Hi,

Ha ha, i am not sure why you would quote PART of my reply and then state some sort of correction or modification when the modification to that statement already followed that first sentence. That's called taking something out of context.
The full context is that it is typical to do this, but in this circuit it doesnt work as well as usual because the arrangement (topology) is different. In particular, pay attention to the third sentence in the full quote below.
Still, i am happy to see that you provided a schematic, which is always more useful than words in electronics

Here's the full quote which maybe you should read over again:
START FULL QUOTE FROM MRAL:
Diode OR circuits are typical in regulators, in order to allow both voltage and current limit to control the output. The arrangement is usually a little different though, such that they dont affect the voltage. In this circuit they will affect the voltage a little over temperature, but it depends how accurate you need it to be over temperature. Since the feedback voltage is low, the change in diode voltage over temperature could cause a 10 fold change in output voltage when the ratio of output voltage to feedback voltage is high. For example, for a feedback voltage of 2.5v and output of 25v, if the 2.5v changes by 0.1v the output will change by 1 volt. That may or may not be acceptable.
END QUOTE
 
I've added a 5k pot between R1&R2, but It's not correcting the problem.
Then I made the circuit in bread board with all new components. Rsc is connected with a psu (V1, variable) and Tl082 is connected to another psu(V2) of 12v (Both grounds connected together). The circuit has been tested both with load (a motor which draws around 100ma at around 9V) and without load at V1 output.

I measured output of Tl082 at pin 1 and pin 7. both pins outputs near V2 (near 12v).
I tested adding a pot(10k) first between R1& R2, then between R3 & R4. In all three cases the outputs are almost same, it doesn't change with the change in pot's value.
It seems that 'no matter what the outputs shall remain close to V2'.
I don't have much experience with Op-amp. I need a solution, please.
 
I don't understand, (my English) but, by your picture.
The TL082 has a "common mode input range" of (plus supply to minus supply-3V) so the inputs will work from +12V to -9V, in this case.
Because of R4, R3 makes a 2:1 voltage divider, the V1 voltage could be as great as +24V or down to -18V. If the voltage is beyond that range the op-amp will act strange. (In my head I can not see the 10K pot)

The Job of the op-amp is to have both inputs at the same voltage.
Pin3 should be at 1/2 of V1. (normally 1% resistors or better).
Pin2 should also be at 1/2 of V1.
But you say Pin1 is at supply, near +12V. So probably Pin 2 is also very high. Why?
Please measure the inputs voltage and see that it is below 12V.
 


Hi,

If the voltage of V1 is greater than -12v it should work, as it can go even lower than that, so test it at a voltage greater than -12v and see if it works then (like -5v for example).

The circuit, as shown, should work. When you use TWO pots however, you'd have to try adjustments for both pots, first one a little then the other a little, then the first one again a little, then the second one a little more. You should not try adjusting one at a time alone.
Actually, you should remove the pots as you dont need them. Make sure all the connections are right, and the resistor values are really what you think they are. One wrong value resistor would cause a problem. Make sure the supply voltages get to the actual pin of the IC too, measuring right at the pin itself on the IC chip not from a circuit trace or socket pin.

The basic theory of this circuit also suggests that you can eliminate the second op amp (eventually). That's because you can set up the first op amp to have gain too, as long as the inputs do not exceed the limits for the op amp and that depends on the range of voltage for V1 and partly on the current in the load.

You might state what V1 actually is, and what the load can be or what the load current can be.

So right now it sounds like something very elementary is wrong, and could even be a bad op amp IC. We could test the IC itself first if you like. When playing around with these circuits sometimes the IC blows out and we dont know it right away. You might also try another IC chip, but testing one is a good idea before you use it anyway.
 
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