LM324 Power Supply with variable voltage and current

[xavier60]

 What side of the shunt(R26) are you using as ground?
The LM324 supply pins see the 12V regulator voltage plus any voltage drop across the shunt.

[mike_mike]

I measured the voltage on the LM324 on pins 4 and 11 and the output of the 7812 across the 10uF capacitor...

Aren't the voltages on pins 5 and 6 too high ? (Please see my previous post)

[xavier60]

The voltages on pins 5 and 6 seem normal. The high value resistors should protect the inputs pins from effects of high input difference voltage. I haven't looked at the data sheet.

[xavier60]

The TI data sheet says 'Differential Input Voltage Range Equal to the Power Supply Voltage"
Keep in mind that for input pins on any device, there might be some significant current draw at extreme voltages which will cause voltage drops across input resistors.  Voltages should be also measured at their source.

[floobydust]

I would not worry about overloading that LED op-amp. The 100k resistors limit any input current.

The CC/CV mode LED circuit is typically looking at the CC and CV op-amp's outputs.
So R13, instead of connecting to the pot would go to IC2A's output pin 1. A 1MEG resistor from IC2B's output to (+) input for a little hysteresis as it's acting as a comparator.
This could get rid of the very sensitive R7, if this is your problem.

[mike_mike]

The actual configuration of the led op amp should work good if it remains as it is in the last schematic ?

I am not worried about the 2k2 pot... I would let it as it is, if the led op amp is ok.

[xavier60]

The actual configuration of the led op amp should work good if it remains as it is in the last schematic ?

I am not worried about the 2k2 pot... I would let it as it is, if the led op amp is ok.

Read the responses and also the LM324's data sheet, then decide.

[mike_mike]

I know that there is an op amp in comparator configuration. If the in+ voltage is higher than in-, then the output of the opamp will be Vcc. If the in- input is higher than in+, then the output of the op amp will be 0V.
I have also had a look in the LM324 datasheet, where differential input voltage is specified = +-32V.
Also in the same datasheet, it is specified Common-mode input voltage range = V+ - 1.5V. Is there any connection between those 2 paramenters ? Which of them should be took into consideration in this circuit ?

[xavier60]

Also in the same datasheet, it is specified Common-mode input voltage range = V+ - 1.5V. Is there any connection between those 2 paramenters ? Which of them should be took into consideration in this circuit ?

The "Common-mode input voltage range = V+ - 1.5V" means that it will operate as an op-amp so long as neither input exceeds 10.5V for a 12V supply.
I believe that it will still function properly as a comparator if only one of the inputs exceeds 10.5V. Meaning that the output state will correctly correspond to the input difference polarity.

[mike_mike]

I decided to make some modification on the led signalization circuit. I will modify the values of R6, R7 and R8 and I will connect a resistor between the pin 5 and GND.
I am wondering to which GND should I connect the resistor ? To the GND on the left or on the right of the shunt resistor (0.22R) ?

[xavier60]

You want the voltage on pin 6 to be roughly at the center of the voltage swing range of the CC op-amp's output.
Ideally the resistor current should avoid flowing through the shunt. Because it's a constant current, it doesn't really matter especially if the Pot and resistors can be replaced with something 10 times higher in value.
Also, the Pot isn't really needed.

[mike_mike]

I removed the pot, I used R6=27K, R8=12K, I connected a 100k resistor between the pin 5 and the right side of the shunt resistor.
The voltages on the pin 5 and 6 are as follows:
1. With SlowCC active:
pin 5=2.53V
pin 6=3.67V
2. With 2.5A load without SlowCC active:
pin 5=5.55V
pin 6=3.74V
3. Without load:
pin 5=5.08V
pin 6=3.31V

Regarding what you said in the previous post, the 100k resistor can be connected on any of the GND (left or right of the 0.22 R shunt), but if connected to the right side of the shunt, then it will consume current through the shunt, but it can be connected on either one of the GND's ?

[xavier60]

Forget about the other post, I forgot about the extra resistor between pin 5 and ground.

[mike_mike]

In the schematic from reply #536, the 100k resistor needs to be connected to the ground on the right side of the 0.22R shunt resistor ?

[xavier60]

Ok, it's best to return the R6/R8 divider to the right side of the shunt and reference all measurements to the right side of the shunt.
This is mainly because some of the voltage changes you are seeing are due to changing voltage drop across the shunt making things confusing.

Also connect the extra 100K resistor to the right side of the shunt.

[xavier60]

You mainly want two readings on pin 5.
While there is no load.
Where the PSU is in CC mode at full current setting.

[mike_mike]

So I have the attached schematic, with the R6/R8 voltage divider connected to the right side of the shunt and with the 100k resistor connected also to the right side of the shunt.
Is this the correct version of the schematic ?

With max output current, and with an overload, the voltage on pin 5 is 3V and on pin 6 is 3.92V.

[xavier60]

So I have the attached schematic, with the R6/R8 voltage divider connected to the right side of the shunt and with the 100k resistor connected also to the right side of the shunt.
Is this the correct version of the schematic ?

With max output current, and with an overload, the voltage on pin 5 is 3V and on pin 6 is 3.92V.

"Is this the correct version of the schematic ?" I think so.
That all looks about right. Pin 5 swings from 5.5V to below pin 6 with some margin. Pin 6 could be a bit higher for extra margin,
Also, I calculate 3.7V for the R6/R8 divider?

[mike_mike]

Also, I calculate 3.7V for the R6/R8 divider?

Yes, I also calculated 3.7V for R6/R8 divider. Probably it is 3.9V because of the resistors, which have 5% tolerances (I think).
Or probably it is higher than 3.7V because of the reason you said in reply #525.

LE: last version of schematic attached

[xavier60]

With measurements referenced to the ground symbol which connects to the right side of the shunt, the voltage readings should not  be affected by the shunt voltage drop.
I would like to know the voltage range on the Base of Q1 from no load to full load. It will tell me the gm of the output path.

[xavier60]

The purpose of C15 is not clear. R21 isn't needed.

[mike_mike]

With measurements referenced to the ground symbol which connects to the right side of the shunt, the voltage readings should not  be affected by the shunt voltage drop.
I would like to know the voltage range on the Base of Q1 from no load to full load. It will tell me the gm of the output path.

I have 3.28-3.30V on the pin 6, with respect to the right side of the shunt GND. This voltage is always the same, even if there is load or there is not load at the output and even if there is active the current limit or not.

The voltage at the base of Q1, with respect to the right side of the shunt GND is: 4.58V with no load and 5.40V with about 2.5A load.

[xavier60]

I have 3.28-3.30V on the pin 6, with respect to the right side of the shunt GND. This voltage is always the same, even if there is load or there is not load at the output and even if there is active the current limit or not.

The voltage at the base of Q1, with respect to the right side of the shunt GND is: 4.58V with no load and 5.40V with about 2.5A load.

There will be some meter loading pulling the voltage down. The voltage at pin 6 should be close to  R6/R8 divider voltage when there is nothing loading it down. Is the divider still 3.9V?
The gm of the output path is 3.    1V change at the Base of Q1 causes 3A output change.

[mike_mike]

On the R6/R8 divider, there is 3.60V, while on pin 6 there is 3.29V. Is that good ?

[iMo]

The purpose of C15 is not clear. R21 isn't needed.

C15 is the blocking cap at the Vcc/Vss pins of the LM324.
Its Vss is wired at the left hand side of the shunt.