How the shunt works
Here it is again to refer to:
http://www.naimmods.com/2 Attack of the Redbox Preamp/0 LARS preregulation.pdf
Its not a particularly complex circuit, though there are a fair few factors you need to balance if you start tweaking it. For example, reduce the gain of the shunt transistors too much and you can see how noise elimination collapses in sims and noise creeps into the ouput waveform.
***And note another benefit of using a shunt regulator- it has current limiting, not as an add-on protection circuit that detracts from overall performance, but as an inherent part of its functionality and actually part of its high performance. The preamp and regulator are protected even in case of a short, as long as the relevant parts are rated and heatsinked to handle worst case scenarios- full on and off load.***
in operation
Overall its an error amp composed of long tailed pair which 'compares' the output voltage as measured across voltage divider (R53/R54) on one side against a filtered reference voltage (U3) on the other.
If the output voltage goes high, the darlington shunt transistors (Q30/Q31) conduct more current to ground until the output voltage drops again such that voltage across reference U3 and R53 are equal again.
If voltage goes low, the shunt transistors will conduct less, allowing more current to flow to the load instead of ground and so increasing voltage again.
Voltage drops and rises because available current is limited to that set by the CCS.
Should the load impedance drop such that load current demand exceeds that supplied by the CCS, the circuit ceases to regulate- the shunt transistors no longer pass any current and the output voltage will continue to drop until a point is reached where the load current demand has reduced to that which can be supplied by the CCS.
Offload, the full current of the CCS shunts to ground.
Pretty high gain is required of the shunt darlington pair as the drive current is also an error current to the error amp. I've chosen composite hfe of 10,000.
Overall, I've kept the shunt simple and avoided peppering it with current sources, because I'm not convinced its needed, and the LARS 6 board is complex enough as it. In a direct a-b comparison on the PRS6 it clearly beat an ALW superreg as a prereg and that's all I was looking for.
As an aside, In fact this shunt is in some ways a more elegant implementation of a superreg anyway- In the ALW/Jung, bootstrapping (using the low noise output to regulate the error amp) is achieved via a quite temperamental bolt-on current source and messy level shifting with decoupled zeners. As I remember, pre-regulation on the ALW/Jung made a huge difference to the ALW- due to the poor current regulation of that current source IIRC? Here it's intrinsic to the shunt circuit, and the current source is not a bolt on but improves isolation. I digress.
LARS5 Operating points and component values
So at this point, lets look at some operating points, component values etc. for the regulator boards and how to set them up.
a. set the shunt output voltage
Shunt output voltage is set by the voltage divider R53/54.
The voltage reference (U3) is a 6.9v buried zener (very low noise, low dynamic impedance).
That means you get approx 6.9v between the base of Q27 and V+out- the base sits 6.9v below V+out.
As the Vbe of Q27 is about 0.7v you get approx 6.9 - 0.7=6.2V below Vout at the emitter of Q27 and therefore across R51, whatever the output voltage might be.
R51 therefore behaves as a current source, and is chosen to give 1.3ma of current.- One doesn't need to build an extra full CCS with transistors, biasing etc. as the current source for the LTP. Very elegant circuit design. But means you can't just plop in jfets willy nilly instead for example.
As the emitter of Q27 and Q29 are directly connected, they must be at the same voltage and therefore their bases must also be at the same voltage. So the voltage across R53 will always be 6.9V - the same as the voltage ref U3. That's the equilibrium state for the error amp, and why the regulator works; it always tries to maintain this equivalence.
6.9V across R53 generates a current. As the base of Q29 effectively gobbles no current that same current will pass through R54 and generate a voltage across it also. So by choosing the value of R54, one can determine the voltage across it is and therefore what V+out will be.
For example, if both R53 and R54 are 1k, then the output voltage across R54 must also be the same, 6.9V, so the total voltage across both resistors must be 2 x 6.9= 13.8V.
If R54 is 2x R53, so R54 is 2k then the voltage across R54 must be 2x 6.9V so the output voltage will be 3x 6.9V=20.7v.
However, DC impedances of both sides should be matched as this is a bipolar error amp so significant currents flow through the bases of Q27/Q29. Imbalances add to the 'error budget'.
Impedance of R48+R49 should equal that of R53//R54. R49 is a damping resistor to ensure no oscillation of Q27. Its probably not necessary as C24 is polyester which I've not had troubles with, and is in some ways undesirable, but rather be safe.
b. What output current should one set the CCS for?
The max that the CCS can provide is determined by the limits of the transistors in the current path.
The gyrator pass transistor is a BC550 which has a max Ic of 100ma if you look at the datasheet.
http://www.datasheetcatalog.org/datasheet/fairchild/BC550.pdf
The CCS is composed of another BC550; and a 2sc2705 cascode transistor which has a max Ic of 50ma from the datasheet.
http://www.datasheetcatalog.org/datasheet/toshiba/3203.pdf
So the max current the CCS can supply is limited by the 2sc2705 -50ma. Obviously its always sensible to go below max ratings so Ive set an upper limit of 45ma.
Is this enough?
To answer that we need to know that the max load current demand of the preamp board will be. Ill look at that later in detail, but the answer is nuclear worst case scenario- the LPA5 preamp uses around 30ma each for the +ve and ve rails.
[Overkill worst case scenario; +/- 5v output swing out of the preamp driving the 18k input impedance of a normal Naim poweramp e.g Nap250- see page E37:
http://www.naimaudio.com/sites/default/files/products/downloads/files/amplifiers_reference-manual_english_issue5_0.pdf
Obviously that is way more than is required but good to know there is plenty of margin. Drive current is always useful to have in reserve.
Back of a fag packet: Naim poweramps cant really take much more than +/- 50v rails without almost certain thermal runaway or just plain blowing the output devices. Done that a few times so I know.
45V are really the max Id be comfortable with and NAP250s run at 40V rails. Voltage gain of the poweramps is about 27x IIRC (have to check that) so max input voltage the amp would sanely need is 45/27=1.66 rounded up to 2V]
more shortly.
ced
Here it is again to refer to:
http://www.naimmods.com/2 Attack of the Redbox Preamp/0 LARS preregulation.pdf
Its not a particularly complex circuit, though there are a fair few factors you need to balance if you start tweaking it. For example, reduce the gain of the shunt transistors too much and you can see how noise elimination collapses in sims and noise creeps into the ouput waveform.
***And note another benefit of using a shunt regulator- it has current limiting, not as an add-on protection circuit that detracts from overall performance, but as an inherent part of its functionality and actually part of its high performance. The preamp and regulator are protected even in case of a short, as long as the relevant parts are rated and heatsinked to handle worst case scenarios- full on and off load.***
in operation
Overall its an error amp composed of long tailed pair which 'compares' the output voltage as measured across voltage divider (R53/R54) on one side against a filtered reference voltage (U3) on the other.
If the output voltage goes high, the darlington shunt transistors (Q30/Q31) conduct more current to ground until the output voltage drops again such that voltage across reference U3 and R53 are equal again.
If voltage goes low, the shunt transistors will conduct less, allowing more current to flow to the load instead of ground and so increasing voltage again.
Voltage drops and rises because available current is limited to that set by the CCS.
Should the load impedance drop such that load current demand exceeds that supplied by the CCS, the circuit ceases to regulate- the shunt transistors no longer pass any current and the output voltage will continue to drop until a point is reached where the load current demand has reduced to that which can be supplied by the CCS.
Offload, the full current of the CCS shunts to ground.
Pretty high gain is required of the shunt darlington pair as the drive current is also an error current to the error amp. I've chosen composite hfe of 10,000.
Overall, I've kept the shunt simple and avoided peppering it with current sources, because I'm not convinced its needed, and the LARS 6 board is complex enough as it. In a direct a-b comparison on the PRS6 it clearly beat an ALW superreg as a prereg and that's all I was looking for.
As an aside, In fact this shunt is in some ways a more elegant implementation of a superreg anyway- In the ALW/Jung, bootstrapping (using the low noise output to regulate the error amp) is achieved via a quite temperamental bolt-on current source and messy level shifting with decoupled zeners. As I remember, pre-regulation on the ALW/Jung made a huge difference to the ALW- due to the poor current regulation of that current source IIRC? Here it's intrinsic to the shunt circuit, and the current source is not a bolt on but improves isolation. I digress.
LARS5 Operating points and component values
So at this point, lets look at some operating points, component values etc. for the regulator boards and how to set them up.
a. set the shunt output voltage
Shunt output voltage is set by the voltage divider R53/54.
The voltage reference (U3) is a 6.9v buried zener (very low noise, low dynamic impedance).
That means you get approx 6.9v between the base of Q27 and V+out- the base sits 6.9v below V+out.
As the Vbe of Q27 is about 0.7v you get approx 6.9 - 0.7=6.2V below Vout at the emitter of Q27 and therefore across R51, whatever the output voltage might be.
R51 therefore behaves as a current source, and is chosen to give 1.3ma of current.- One doesn't need to build an extra full CCS with transistors, biasing etc. as the current source for the LTP. Very elegant circuit design. But means you can't just plop in jfets willy nilly instead for example.
As the emitter of Q27 and Q29 are directly connected, they must be at the same voltage and therefore their bases must also be at the same voltage. So the voltage across R53 will always be 6.9V - the same as the voltage ref U3. That's the equilibrium state for the error amp, and why the regulator works; it always tries to maintain this equivalence.
6.9V across R53 generates a current. As the base of Q29 effectively gobbles no current that same current will pass through R54 and generate a voltage across it also. So by choosing the value of R54, one can determine the voltage across it is and therefore what V+out will be.
For example, if both R53 and R54 are 1k, then the output voltage across R54 must also be the same, 6.9V, so the total voltage across both resistors must be 2 x 6.9= 13.8V.
If R54 is 2x R53, so R54 is 2k then the voltage across R54 must be 2x 6.9V so the output voltage will be 3x 6.9V=20.7v.
However, DC impedances of both sides should be matched as this is a bipolar error amp so significant currents flow through the bases of Q27/Q29. Imbalances add to the 'error budget'.
Impedance of R48+R49 should equal that of R53//R54. R49 is a damping resistor to ensure no oscillation of Q27. Its probably not necessary as C24 is polyester which I've not had troubles with, and is in some ways undesirable, but rather be safe.
b. What output current should one set the CCS for?
The max that the CCS can provide is determined by the limits of the transistors in the current path.
The gyrator pass transistor is a BC550 which has a max Ic of 100ma if you look at the datasheet.
http://www.datasheetcatalog.org/datasheet/fairchild/BC550.pdf
The CCS is composed of another BC550; and a 2sc2705 cascode transistor which has a max Ic of 50ma from the datasheet.
http://www.datasheetcatalog.org/datasheet/toshiba/3203.pdf
So the max current the CCS can supply is limited by the 2sc2705 -50ma. Obviously its always sensible to go below max ratings so Ive set an upper limit of 45ma.
Is this enough?
To answer that we need to know that the max load current demand of the preamp board will be. Ill look at that later in detail, but the answer is nuclear worst case scenario- the LPA5 preamp uses around 30ma each for the +ve and ve rails.
[Overkill worst case scenario; +/- 5v output swing out of the preamp driving the 18k input impedance of a normal Naim poweramp e.g Nap250- see page E37:
http://www.naimaudio.com/sites/default/files/products/downloads/files/amplifiers_reference-manual_english_issue5_0.pdf
Obviously that is way more than is required but good to know there is plenty of margin. Drive current is always useful to have in reserve.
Back of a fag packet: Naim poweramps cant really take much more than +/- 50v rails without almost certain thermal runaway or just plain blowing the output devices. Done that a few times so I know.
45V are really the max Id be comfortable with and NAP250s run at 40V rails. Voltage gain of the poweramps is about 27x IIRC (have to check that) so max input voltage the amp would sanely need is 45/27=1.66 rounded up to 2V]
more shortly.
ced