Maida style Reg

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Ok lads my mistake corrected  voltages in black

cheers

skal
 

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just did a quick measurement between the base and emitter of tip50 value is 0.594vdc , the lm317 voltage between adj and output is 1.247vdc.

hope this helps

cheers

skal
 
Turning the pot full cw , the the lm317 voltage between adj pin and output is 1.047vdc and the voltage between  base and emitter of tip50 value is 0.630vdc, if i am understanding this correctly , the lm317 has drop out of regulation?

skal
 
If I understand correctly, the dropout voltage (for good regulation) is something like 25 to 30 V (in to out difference), so you can't expect more than ~360 V at the output.

Samuel
 
Samuel Groner said:
If I understand correctly, the dropout voltage (for good regulation) is something like 25 to 30 V (in to out difference), so you can't expect more than ~360 V at the output.

Samuel

i think i understand ,the pot can only change voltage over a  limited within the scope of the lm317 reg range, any thing out of that range can not be  regulated . so the range  is set by the zener?

cheers

skal
 
Samuel Groner said:
If I understand correctly, the dropout voltage (for good regulation) is something like 25 to 30 V (in to out difference), so you can't expect more than ~360 V at the output.

Samuel
No;  LM317HV guarantees good regulation for anything between 3 and 60V. But the preregulator (TIP50+associated resistors) introduces a voltage drop that's load-dependant (about 15V at 100mA). Since there is a minimum current of 12.5mA (due to the voltage-reference across the 100r resistor), the minimum drop in the prereg is about 3V. So, in theory, the maximum no-load voltage is about Vin-6V.
At 100mA, maximum output voltage would be about Vin-23V.
 
The voltage on the 317 reg is defined with Vz-Vbe-47Ic , what's about 19 Volts for Ic=12,5mA and  14V for Ic=100mA.

If there is no external load, the circuit can't work for Vin=400 and Vout=300 because the current thru base resistor is 8mA and it goes (mostly) thru Zener into output.  The darlington transistor here should be used with much higher base resistor's value, or output should be all the time loaded with minimum 10mA (27kOhm, 5 or 7W resistor on output), IMO.
 
moamps said:
The voltage on the 317 reg is defined with Vz-Vbe-47Ic ,
That's true if the Zener conducts, but there are cases where the voltage between base and output is too low. If the output voltage is set at 300V, the Zener does not conduct but the regulator works perfectly.
The darlington transistor here should be used with much higher base resistor's value, or output should be all the time loaded with minimum 10mA (27kOhm, 5 or 7W resistor on output), IMO.
With the actual 100r resistor at the adjust pin, the no-load current is 12.5mA. No need for a bleeder resistor.
 
abbey road d enfer said:
That's true if the Zener conducts, but there are cases where the voltage between base and output is too low. If the output voltage is set at 300V, the Zener does not conduct but the regulator works perfectly.

The Zener doesn't conduct only if Vin-Vout is smaller than Vz . Please reread original article.
http://www.ti.com/lit/an/snoa648/snoa648.pdf
 
moamps said:
abbey road d enfer said:
That's true if the Zener conducts, but there are cases where the voltage between base and output is too low. If the output voltage is set at 300V, the Zener does not conduct but the regulator works perfectly.

The Zener doesn't conduct only if Vin-Vout is smaller than Vz .
Exactly what I wrote...
 
abbey road d enfer said:
moamps said:
abbey road d enfer said:
That's true if the Zener conducts, but there are cases where the voltage between base and output is too low. If the output voltage is set at 300V, the Zener does not conduct but the regulator works perfectly.

The Zener doesn't conduct only if Vin-Vout is smaller than Vz .
Exactly what I wrote...

Nope. In this circuit (contrary to the simple reg without pre-regulation, where protection Zener doesn't conduct in normal operation mode) Vin-Vout must be higher than Vz  to ensure normal operation, and the Zener conducts all the time.  If the Zener doesn't conduct, the circuit doesn't work, IMO.
 
moamps said:
abbey road d enfer said:
moamps said:
abbey road d enfer said:
That's true if the Zener conducts, but there are cases where the voltage between base and output is too low. If the output voltage is set at 300V, the Zener does not conduct but the regulator works perfectly.

The Zener doesn't conduct only if Vin-Vout is smaller than Vz .
Exactly what I wrote...
Nope. In this circuit (contrary to the simple reg without pre-regulation, where protection Zener doesn't conduct in normal operation mode) Vin-Vout must be higher than Vz  to ensure normal operation, and the Zener conducts all the time.  If the Zener doesn't conduct, the circuit doesn't work, IMO.
Why? The zener is there just to make sure that Vin-Vout stays within reasonable limits and the TIP50 takes its share of the dissipated heat, but I have simulated the circuit, it works even with with only 3V across the prereg. But the available current is nil. At 6V across the prereg, there is a total of 35mA available, so 22.5mA for the load.
 
abbey road d enfer said:
moamps said:
Nope. In this circuit (contrary to the simple reg without pre-regulation, where protection Zener doesn't conduct in normal operation mode) Vin-Vout must be higher than Vz  to ensure normal operation, and the Zener conducts all the time.  If the Zener doesn't conduct, the circuit doesn't work, IMO.
Why? The zener is there just to make sure that Vin-Vout stays within reasonable limits

The Zener is there to keep Vin(reg)-Vout constant, independent of Vin-Vout voltage. It is essential in this circuit.

and the TIP50 takes its share of the dissipated heat, but I have simulated the circuit, it works even with with only 3V across the prereg. But the available current is nil. At 6V across the prereg, there is a total of 35mA available, so 22.5mA for the load.

Have you simulate the circuit with  transistor's hfe=30 or so? Your math stands only if the transistor is in saturation.
This circuit is not intended to work in Vin-Vout≤Vz regime.
 
moamps said:
abbey road d enfer said:
moamps said:
Nope. In this circuit (contrary to the simple reg without pre-regulation, where protection Zener doesn't conduct in normal operation mode) Vin-Vout must be higher than Vz  to ensure normal operation, and the Zener conducts all the time.  If the Zener doesn't conduct, the circuit doesn't work, IMO.
Why? The zener is there just to make sure that Vin-Vout stays within reasonable limits

The Zener is there to keep Vin(reg)-Vout constant, independent of Vin-Vout voltage. It is essential in this circuit.
Why? The regulator can work with anything between 3 and 60V. The choice of Zener voltage is dictated by the targetted max current and associated dissipation in the LM317. [/quote]

and the TIP50 takes its share of the dissipated heat, but I have simulated the circuit, it works even with with only 3V across the prereg. But the available current is nil. At 6V across the prereg, there is a total of 35mA available, so 22.5mA for the load.

Have you simulate the circuit with  transistor's hfe=30 or so? Your math stands only if the transistor is in saturation. [/quote] No. With 3V across the prereg @ 12.5mA, the transistor has about 800mV Vce; admittedly, it's close to saturation, but not completely.
In fact some combinations of Vin-Vout and current put the transistor in saturation; but it does not prevent the regulator to do its job.
 
This circuit is not intended to work in Vin-Vout≤Vz regime.
  I know the idea is to maintain a reasonable voltage across the LM317, and the TIP50 to take its share, which is the case when the output voltage is more than 20V below unreg B+, but it does not mean that the circuit does not regulate  when Vin-Vout is lower than 20V. In that case, dissipation in the LM317 is limited and the TIP50 does not need to take a share.
 
The Zener is there to keep Vin(reg)-Vout constant, independent of Vin-Vout voltage. It is essential in this circuit.
Why? The regulator can work with anything between 3 and 60V. The choice of Zener voltage is dictated by the targetted max current and associated dissipation in the LM317.

You answered your own question.  Usually the Zener voltage is small, about 6V, just enough to for sure have Vin(reg)-Vout ≥3V and to have maximum current and  power available.

Have you simulate the circuit with  transistor's hfe=30 or so? Your math stands only if the transistor is in saturation.
No. With 3V across the prereg @ 12.5mA, the transistor has about 800mV Vce; admittedly, it's close to saturation, but not completely.
In fact some combinations of Vin-Vout and current put the transistor in saturation; but it does not prevent the regulator to do its job.

If Vin-Vin(reg)=3V and I=12,5mA then:

Vre=47I=0,6V, Vrc=100I=1,25V
Vrb=Vin-Vin(reg)-Vbe-Vre=3-0,7-0,6=1,7V
Irb=Vrb/Rb=1,7/10000=0,17mA
for hfe=50, Ic=Ie=Irb x hfe=0,17x50=8,5mA !!!

So you haven't enough base current to drive transistor to get minimum needed current, without any load. 
 
This circuit is not intended to work in Vin-Vout≤Vz regime.
  I know the idea is to maintain a reasonable voltage across the LM317, and the TIP50 to take its share, which is the case when the output voltage is more than 20V below unreg B+, but it does not mean that the circuit does not regulate  when Vin-Vout is lower than 20V...

Practically, you don't know that for sure.
For example:
Vin-Vin(reg)=10V, I =35mA
Vre=1,65V, Vrb=Vin-Vin(reg)-Vbe-Vre=10-0,6-1,65=7,75V
Irb=Vrb/Rb=7,75/10000=0,775mA
to get Ic=35mA at Vce=5V, hfe must be 50 what for power transistors at so small currents isn't so usual.

If the loading changes, you can be partially in unregulated regime.

So I proposed earlier use of  darlington transistor for prereg transistor, where higher value of Rb can be chosen (the bleeding current to output via Zener is small) without problems with driving the prereg transistor.

I have used this topology few times, and have nothing more to add here.

 
moamps said:
The Zener is there to keep Vin(reg)-Vout constant, independent of Vin-Vout voltage. It is essential in this circuit.
Why? The regulator can work with anything between 3 and 60V. The choice of Zener voltage is dictated by the targetted max current and associated dissipation in the LM317.

You answered your own question. 
No, I did not. Ther is no need for this voltage to be constant. It just needs to be clamped at a reasonable voltage.
Have you simulate the circuit with  transistor's hfe=30 or so? Your math stands only if the transistor is in saturation.
No. With 3V across the prereg @ 12.5mA, the transistor has about 800mV Vce; admittedly, it's close to saturation, but not completely.
In fact some combinations of Vin-Vout and current put the transistor in saturation; but it does not prevent the regulator to do its job.

If Vin-Vin(reg)=3V and I=12,5mA then:

Vre=47I=0,6V, Vrc=100I=1,25V
Vrb=Vin-Vin(reg)-Vbe-Vre=3-0,7-0,6=1,7V
Irb=Vrb/Rb=1,7/10000=0,17mA
for hfe=50, Ic=Ie=Irb x hfe=0,17x50=8,5mA !!!

So you haven't enough base current to drive transistor to get minimum needed current, without any load.
 
This circuit is not intended to work in Vin-Vout≤Vz regime.
  I know the idea is to maintain a reasonable voltage across the LM317, and the TIP50 to take its share, which is the case when the output voltage is more than 20V below unreg B+, but it does not mean that the circuit does not regulate  when Vin-Vout is lower than 20V...

Practically, you don't know that for sure.
For example:
Vin-Vin(reg)=10V, I =35mA
Vre=1,65V, Vrb=Vin-Vin(reg)-Vbe-Vre=10-0,6-1,65=7,75V
Irb=Vrb/Rb=7,75/10000=0,775mA
to get Ic=35mA at Vce=5V, hfe must be 50 what for power transistors at so small currents isn't so usual.

If the loading changes, you can be partially in unregulated regime.

So I proposed earlier use of  darlington transistor for prereg transistor, where higher value of Rb can be chosen (the bleeding current to output via Zener is small) without problems with driving the prereg transistor.

I have used this topology few times, and have nothing more to add here.
What you don't understand is that the transistor, being essentially in voltage-follower mode, is self-regulating. Whenever the current demand increases, the voltage across it increases too and the base current increases in such a way that the transistor never runs out of juice.
Also you based your comments on the fact thet hfe is 50; the datasheet for TIP50 shows that it can vary between 20 and 150.
I agree that in the extreme case of hfe=30, operation with 3V across will not be sustainable, but for hfe=50, operation at about 4.5 V is stable. What is 1.5V difference compared to 390?
This is just silly argumentation.
 
What you don't understand is that the transistor, being essentially in voltage-follower mode, is self-regulating. Whenever the current demand increases, the voltage across it increases too and the base current increases in such a way that the transistor never runs out of juice.

Yeah, there is no enough voltage across the transistor and no enough  base current you can play with.

Also you based your comments on the fact thet hfe is 50; the datasheet for TIP50 shows that it can vary between 20 and 150. I agree that in the extree case of hfe=30, operation with 3V across will not be sustainable, but for hfe=50, operation at about 4.5 V is stable. What is 1.5V difference compared to 390?
This is just silly argumentation.

The silly argumentation here is advocating something what's working only if you are lucky when you pick the transistor from the bin. 
 
moamps said:
The silly argumentation here is advocating something what's working only if you are lucky when you pick the transistor from the bin.
This is turning into a perfectly useless dick contest. You are trying to prove that something does not work at all past some limits, and I'm just saying that operation in degraded mode is not a total failure.
I'm not advocating anything, I'm just saying that there's a grey area where one should express opinions with caution. And certainly the area where Vin-Vout is comprised between 6 and 20V is such.
 

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