how to understand the functionality of circuit

[xavier60]

sir,

            Why Transient analysis is done in Power supplies, even though it is a DC System?(attached Document)
            Normally why capacitors are connected in parallel with the Load?

Thank you sir.

As well as seeing how the PSU's output voltage dips then recovers when there is a sudden load increase, the load transient test will show how stable the feedback loop is.
Here is another member testing a PSU with various compensating component values, https://www.eevblog.com/forum/beginners/lm324-power-supply-with-variable-voltage-and-current/msg3893150/#msg3893150
It is clear which is the most preferable response.

[audio]

Sir,

1) what is the need of bc557 T1 transistor?

2) why ic1 is present in the circuit? I guess it would make output down to zero, but how?


Thank you.

[xavier60]

Sir,

1) what is the need of bc557 T1 transistor?

2) why ic1 is present in the circuit? I guess it would make output down to zero, but how?


Thank you.

T1 is just a driver for the relay coil because the opamp might not have enough output current capability to drive the coil directly.
The opamp is configured as  voltage detector with some hysteresis, its purpose being to control the transformer's secondary tap switching.
Do you understand opamp function and application?

[audio]

Sir,

            when the output is 20V @ 5A, Then Switch should be switched to top end of the coil. But How the System Will sense and Switch?
Sorry I cant Understand the operation of the switch and why we have employed opamp in the circuit?

[xavier60]

Sir,

            when the output is 20V @ 5A, Then Switch should be switched to top end of the coil. But How the System Will sense and Switch?
Sorry I cant Understand the operation of the switch and why we have employed opamp in the circuit?

The opamp is being used as a voltage comparator. Positive feedback through R32 causes it to have 2 threshold voltages, or to have hysteresis. Search for "schmitt trigger"
Doing some rough calculations with the 10K VR set half way, the opamp's output will swing low as the PSU's output increases above about 13.5V, turning On the relay. The opamp's output will swing high again when the PSU's output decreases below about 12.5V.

[audio]

thank You..

[audio]

Hi to all,

              I Found this Circuit in internet. What i have Understood from the circuit.
1)The Transistor T1 Is parallel transistor or Darlington Transistor to supply current The Load(Connected to High Current source V_L). The Circuit has two OPAMP and forms two loop for Voltage and Current Loop to control.

2) OA1 is Inverting Amplifier and output depends on ratio of (R2/R1)*Vref

3)OA2 is doing positive feedback, could Not Understand how Current Limitation occur in the Circuit??

Thank you all

[xavier60]

The voltage at the left of the CS resistor R5 goes negative WRT GND with increasing load current. So the feedback polarity is correct, negative.

[audio]

Sir,,,

     How these arrangement Could do Current Limiting Function??

      Why There is A vref2 source in OA2?


      Thank You

[xavier60]

With no load, Vref2, R3 and R4 cause an adjustable positive voltage on the non-inverting input of AO2. The output of OA2 will be at the full supply voltage, 33V. D2 will be reverse biased having no control of T1.
If load current is increased to the point that causes OA2's non-inverting input to drop slightly below its inverting input. OA2's output will swing low, first causing D2 to forward bias, taking control of T1 from OA1. OA2 might need to continue to pull the Base of T1 lower until the output current settles to a value that causes OA2's inputs to be very close to equal, 0V.
The purpose of D1 and D2 is to cause whichever opamp that has the lowest output voltage to take control of T1. They can be called "ORing" diodes, although not strictly correct.

BTW, it's actually the current from the Constant Current source that supplies drive current to T1.

[audio]

Sir,,

            "it's actually the current from the Constant Current source that supplies drive current to T1"

1)Why current source is in between 33v source and Output of OPamp ?
2)Does that mean Current is going to come from source and opamp(milli amps) , Summed and supplied to Transistor?
3)If i Construct the circuit, how to select
             (a)opamp
             (b) Resistance Vales and Wattage.
4) Whether This RPS is Practically Usable in Day today use, Like charging the battery(high Current 10 A)

thank you

[xavier60]

Sir,,

            "it's actually the current from the Constant Current source that supplies drive current to T1"

1)Why current source is in between 33v source and Output of OPamp ?
2)Does that mean Current is going to come from source and opamp(milli amps) , Summed and supplied to Transistor?
3)If i Construct the circuit, how to select
             (a)opamp
             (b) Resistance Vales and Wattage.
4) Whether This RPS is Practically Usable in Day today use, Like charging the battery(high Current 10 A)

thank you

Keep it simple by thinking of the opamps controlling only their output voltages and so the Base voltage of T1. The Emitter of T1 follows its Base voltage. The  load determines the Emitter current. T1 determines the required Base current which is drawn from the CC source. All surplus CC current is sunk to GND by either one of the opamps.
 I dont think I want to calculate practical values.

[audio]

Sir,

           How to construct Constant Current Source? marked in image.

Thank you sir.

[xavier60]

There are a few simple ways, https://www.nexperia.com/applications/sub-systems/constant-current-source.html
Search for "Constant current source"

[richard.cs]

This is a perfect example of a situation where the schematic becomes hugely more readable with an IC symbol that reflects the internal functionality, example attached. It may help to re-draw your schematic in this way.

[audio]

Sir,

      1)Why There is a current Source in the circuit, OA1 itself can Provide few milliamps of current or we could use Darlington Pair ?
      2)What is the need of current source between unregulated supply and output of OA1?

Still not clear sir.

[xavier60]

Sir,

      1)Why There is a current Source in the circuit, OA1 itself can Provide few milliamps of current or we could use Darlington Pair ?
      2)What is the need of current source between unregulated supply and output of OA1?

Still not clear sir.

The CC source doesn't connect to the output of OA1, it connects to the junction of the 2 diodes anodes.
The CC source isn't strictly needed. It could be replaced by a pull-up resistor or one of the diodes could be replaced by a resistor.
Replacing the CC source with a pull-up resistors forces a compromise. If the value is made low enough to supply enough current at  high output voltage settings, it's dissipation might be too high at low output voltage settings.
Another advantage of the CC  source is that it will inject almost no ripple current into the control path.

[audio]

sir,,

            how current limit LED circuit is operating?

                    T1 is a pnp transistor, I have applied KVL in Base emitter loop and Current is nearly 2mA. How it will going to operate as current limit indicator?

[xavier60]

sir,,

            how current limit LED circuit is operating?

                    T1 is a pnp transistor, I have applied KVL in Base emitter loop and Current is nearly 2mA. How it will going to operate as current limit indicator?

With a red LED used, that CC source should supply about 1mA. It's not critical anyway. This current mainly flows through P3 to cause a voltage drop that's proportional to its set resistance. This adjustable voltage becomes added to any voltage drop across R8 and is applied to the current limit BJT inside of the LM723. The whole idea is so that the PSU has a current limit range that can adjust down zero.

[xavier60]

I should have also mentioned that the LED function is as a crude voltage reference for the CC source circuit.
Two or more Silicon diodes in series are often used for the same purpose.

[audio]

Sir,,
     

             The 0.33ohm at 2 ampere load Current will develop .7V .Hence it will be 0.7V needed to turn current limiting transistor.

1)how this p3 variable resistor is going to be used control current of 1A?
2)how 1mA emmitter current of T1 is going to be used as Indicator?
3)Still Current limiting Function Not Clear Sir.

[xavier60]

Sir,,
     

             The 0.33ohm at 2 ampere load Current will develop .7V .Hence it will be 0.7V needed to turn current limiting transistor.

1)how this p3 variable resistor is going to be used control current of 1A?
2)how 1mA emmitter current of T1 is going to be used as Indicator?
3)Still Current limiting Function Not Clear Sir.

Ill try to answer some of that now.
Assuming the CC source is really supplying 1mA.
At 1A, the voltage across R8 will be 0.33V.
If P3 is set to 270Ω, it will have 0.27V across it which will up to 0.6V applied to the CL BJT in the LM723 and cause PSU output current limiting.
What are you planning to do eventually? Build something?

I don't fully understand question 2. The 1mA constant current always normally stays the same.

[audio]

Sir,

                          If Variable resistance P3 is at 1K and 1mA Current Flows will Devolop 1V and already o.33V across Current sense resistor because Load current is 1A. Then voltage developed across CL and CS terminal is 1.33V.so Excess base voltage results more Ib,then more Ic which reduce base current of series pass transistor of LM723 IC. REduces Load Current and succesfully Reduces load voltage. This will be the action.

                         But as per your Explanation when P3 is 270ohm, current limiting will begin?
                        1)why p3 should be 1K,why not 270 ohm?
                        2)Actually circuit does current limiting up to 1Ampere, How Sir?
                        3) i cannot correlate question 1 and 2?

thank you sir

 

[xavier60]

Sir,

                          If Variable resistance P3 is at 1K and 1mA Current Flows will Devolop 1V and already o.33V across Current sense resistor because Load current is 1A. Then voltage developed across CL and CS terminal is 1.33V.so Excess base voltage results more Ib,then more Ic which reduce base current of series pass transistor of LM723 IC. REduces Load Current and succesfully Reduces load voltage. This will be the action.

                         But as per your Explanation when P3 is 270ohm, current limiting will begin?
                        1)why p3 should be 1K,why not 270 ohm?
                        2)Actually circuit does current limiting up to 1Ampere, How Sir?
                        3) i cannot correlate question 1 and 2?

thank you sir

 

Current limiting occurs when the voltage between CL and CS reaches about 0.6V. With P3 set to 0Ω, current limit will be about 1.8A.
With P3 set to 600Ω there will be 0.6V applied between CL and CS causing the current limit to be near 0A.
 All this assumes that the CC source is really 1mA. Maybe it's not.
Do you have a link to the whole project?

 

[xavier60]

I found this, https://learnorgdata.s3-ap-southeast-2.amazonaws.com/Complex+Power+Sup/Week+7/723+reg+main+sm.pdf
It says that the output is up to 1A. Maybe I have made a mistake. The current limiting still works as I explained.
Has someone built this PSU design?

When P3 is 270Ω, the voltage drop across it at 1mA will 0.27V. So the current limiting will begin at about 1A.