Positive and Negative regulators sections are different

[Davewalker5]

I don't understand what Q21 , 2N6315 on the Voutput pin#6 is going for the Positive +15 volts supply supply

What is VR2, Q19, Q20 doing for the Negative supply on Voutput pin#6 ? what kind of circuit is V2, Q19, Q20? and why doesn't the positive regulator not have a VR2 or Q19, Q20 only the negative regulator?

I also don't understand why they would connect the Positives Regulators CS output as the Input of the Negative supplys regulator? any reason why? the CS is the current sensing , so they are using the current sensing input voltage as the negative regulators input

What technique is this called?

 

[KrisBlueNZ]

The differences are due to the fact that the µA723 regulator IC is designed to operate as a positive regulator.

The +15V regulator shows how the 723 is supposed to be used. The pass transistor (Q21) is driven directly from the VOUT pin, and the load current is sensed across R57 and passed back to the 723 by direct connection to the CL and CS pins.

The -15V regulator uses another 723 but the designer has had to jump through a few hoops to make it work. The base drive to the pass transistor (Q19) is offset by 6.2V using VR2 (which is just a zener diode) in conjunction with R56, because the output voltage range of the 723 is not suitable for directly driving a pass transitor for negative supply regulation, and the load current is sensed across R52 but the voltages are again not suitable for direct connection to the 723's CL and CS inputs so current limiting is done externally by Q20.

The connection involving R54 and R53 forms a "tracking regulator" where the negative regulator tracks the voltage that is set by the positive regulator. R54 doesn't have a direct relationship to the CS pin of the positive regulator IC; it monitors the +15V rail voltage, which happens to be connected to that pin.

 

[Davewalker5]

The base dive to the pass transistor (Q19) is offset by 6.2V using VR2 (which is just a zener diode) in conjunction with R56, because the output voltage range of the 723 is not suitable for directly driving a pass transitor for negative supply regulation,

- So the zener diode is used to create an offset voltage? or bias voltage for the ass transistor Q19?

If they didn't use the zener diode , the negative voltage out of the negative supply regulators output would damage Q19 and Q20?

 

[Davewalker5]

Why didn't they use the Current Sensing input CS on the Negative regulator? only on the Positive Regulator

The connection involving R54 and R53 forms a "tracking regulator" where the negative regulator tracks the voltage that is set by the positive regulator. R54 doesn't have a direct relationship to the CS pin of the positive regulator IC; it monitors the +15V rail voltage, which happens to be connected to that pin.

The Tracking Regulator R54 and R53 looks like a voltage divider, how is it a tracking regulator?

 

[KrisBlueNZ]

It creates an offset voltage, yes. Without it, the circuit wouldn't work because the voltage at the VOUT pin doesn't go far enough negative. I haven't looked into the design in much detail and I'm not very familiar with the 723. Have a look at the data sheet and see if you can figure out how it's being used to regulate the negative supply.

Why didn't they use the Current Sensing input CS on the Negative regulator? only on the Positive Regulator

I explained that. It's because the voltages on the ends of the current shunt resistor (R52) are outside the usable range for the CR and CS pins of the 723. So on the negative regulator they have to do the current limiting externally, using the second transistor.

The Tracking Regulator R54 and R53 looks like a voltage divider, how is it a tracking regulator?

It is a voltage divider. The negative regulator adjusts the base voltage on the negative pass transistor (Q19) to produce the right voltage at the centre of the voltage divider. The voltage divider divides the voltage difference between the +15V rail and the -15V rail and feeds the result into the -15V regulator, which adjusts Q19's base voltage so that the voltage divider's output voltage is correct when the rails are at +15V and -15V. This is how tracking regulators work.

 

[Davewalker5]

It creates an offset voltage, yes. Without it, the circuit wouldn't work because the voltage at the VOUT pin doesn't go far enough negative

So VR2 zener diode in series makes the regulators output go more negative or have a lower or higher negative swing? how i don't get it

Ok, so they are using the resistors and transistors as an external current limiting circuit and external tracking regulator

But My Main question is how did they use the positive regulators CS current sensing input pin to be connected to the Input pin of the negative regulator? that is weird to me

They Daisy chained the positive regulators CS input to the Negative regulators input Vinput for some reason , what is this technique called?

 

[KrisBlueNZ]

So VR2 zener diode in series makes the regulators output go more negative or have a lower or higher negative swing? how i don't get it

Yes, it provides a fixed voltage offset. This makes the transistor's base more negative than the output of the regulator IC while still allowing the regulator IC to control the transistor. Read the 723 data sheet and draw up the negative regulator circuit from the point of view of how the 723 (a positive regulator controller) will see it.

But My Main question is how did they use the positive regulators CS current sensing input pin to be connected to the Input pin of the negative regulator? that is weird to me

I answered that in post #2. Last paragraph.

 

[Davewalker5]

Yes, it provides a fixed voltage offset. This makes the transistor's base more negative than the output of the regulator IC while still allowing the regulator IC to control the transistor

.

I can't see how a series zener diode can set a fixed voltage offset, i can understand a parallel zener diode to set a fixed voltage offset but not a series

If the output of the regulator is negative -10 volts, how can the series diode make the -10 volts to -12 volts?

But My Main question is how did they use the positive regulators CS current sensing input pin to be connected to the Input pin of the negative regulator? that is weird to me
I answered that in post #2. Last paragraph.

Because of the external tracking circuit? but it still doesn't make sense to me why would they use the current sensing input as the negative regulators input, that's weird and not normal

The designer is trying to make a micky mouse tracking regulator and current sensing circuits externally

 

[KrisBlueNZ]

I can't see how a series zener diode can set a fixed voltage offset, i can understand a parallel zener diode to set a fixed voltage offset but not a series. If the output of the regulator is negative -10 volts, how can the series diode make the -10 volts to -12 volts?

The voltage offset is about 6.2V. R56 causes current to flow in the zener diode. This causes a fixed voltage to appear across it. This produces the voltage offset between VOUT and the transistor's base.

Because of the external tracking circuit? but it still doesn't make sense to me why would they use the current sensing input as the negative regulators input, that's weird and not normal

They are not using the current sensing input as the negative regulator's input. They are using the +15V rail as one end of the voltage divider that feeds the negative regulator's input. It just happens that the CS pin is connected to the +15V rail.

The designer is trying to make a micky mouse tracking regulator and current sensing circuits externally

Thanks for your opinion.

 

[davenn]

.
Because of the external tracking circuit? but it still doesn't make sense to me why would they use the current sensing input as the negative regulators input, that's weird and not normal

Read again what Kris said ....

The connection involving R54 and R53 forms a "tracking regulator" where the negative regulator tracks the voltage that is set by the positive regulator. R54 doesn't have a direct relationship to the CS pin of the positive regulator IC; it monitors the +15V rail voltage, which happens to be connected to that pin.

particularly the bold bit and even more so the bold red bit

Dave

 

[davenn]

lol Kris you just beat me

 

[Davewalker5]

The voltage offset is about 6.2V.

You mean a Negative -6.2 voltage offset? how can the output of the negative regulators outputting a negative voltage and the diodes offset voltage is a Positive 6.2 vdc?

They are using the +15V rail as one end of the voltage divider that feeds the negative regulator's input.

The design did this because they put in an external tracking regulator circuit inbetween the positive regulators output to the negative regulators input

 

[davenn]

Dave
no one uses these chips nowadays they are a very old and pretty much obsolete type

the 78xx (79xx) series or the LM317 and its negative version are so much easier to use
with only a small number of external components

Dave

 

[(*steve*)]

They're a great bit of history and I expect that @Davewalker5 would find a lot if detailed information if he looked up the uA723 datasheet. I would recommend nothing less than the original Fairchild datasheet from the time that this device was most commonly available in a 10 pin metal can package.

The key features of this device are a 150mA current rating, a maximum of 800mW power dissipation (with a following breeze), and a complete lack of any form of overcurrent or thermal shutdown.

 

[Davewalker5]

here is another schematic that has the negative regulator that is different compared to the positive

They use the inverting input of a 741 to invert the voltage

But what is Q6 and Q17 doing?

Why doesn't the positive regulator has this type of circuit like Q6 and Q17

It seems like Q5 just add more current to the output pin of the positive regulator

 

[davenn]

basically

Q5 is the current pass transistor for the positive regulator
Q6 is the current pass transistor for the negative regulator

some one else can describe in greater detail if needed

 

[Davewalker5]

I thought regulators had a current pass transistors internally inside the IC chip

they have the current pass transistors externally

A lot of times I don't even see current pass transistors in regulators circuits, any advantages of using current pass transistors? because mostly they leave them out

 

[davenn]

I thought regulators had a current pass transistors internally inside the IC chip

what made you think that ?

they have the current pass transistors externally

yes ... for any given regulator, it has a max current it can supply
reread what steve typed above for the 723 regulator ....

The key features of this device are a 150mA current rating, a maximum of 800mW power dissipation (with a following breeze), and a complete lack of any form of overcurrent or thermal shutdown.

so the most it can pass is 150mA ... if you need to use it in a PSU that you need more current on the output then you must have current pass transistors

A lot of times I don't even see current pass transistors in regulators circuits, any advantages of using current pass transistors? because mostly they leave them out

which means that the maximum required output of the PSU is within the ratings of the regulator
78xx series ~ 1 Amp
LM338 ~ 5 Amps

if you need more current than that you either use pass transistors, or more efficiently, use switching regulators

Dave

 

[Davewalker5]

oh ok thanks for clearing that up

Q5 is the current pass transistor for the positive regulator
Q6 is the current pass transistor for the negative regulator

Ok i get it

But what about Q17? what does it do

Q6 and Q17 is a current pass transistor circuit it seems that outputs more current only in the negative supply compared to the positive supply

 

[davenn]

someone else can describe the use of Q17

but I'm guessing it controls the feedback for the op-amp and therefore controls the regulation