Politics?
No.... I meant film in the 1 to 5 uF region. Leakage currents from electrolitics make the less suitable forom low noise and stable DC voltage . I am sorry to leave this circuit hanging, but my surgery recovery is slow and I am in a lot of pain. The voltage prgramming formula now use 1.25 volts PLUS the base to emmiter voltage of the emitter follower transistor at 1.25 mA collector current . The resistor across the 1.85 volts can be in the 10K to 50K region with the larger values showing a small error due to the addition of the base current from the follower. A jfet would not have this problem with base current but will have other considerations which make it lesss suitable than a BJT even though it would be nice to get the film cap even lower in value.
No.... I meant film in the 1 to 5 uF region. Leakage currents from electrolitics make the less suitable forom low noise and stable DC voltage . I am sorry to leave this circuit hanging, but my surgery recovery is slow and I am in a lot of pain. The voltage prgramming formula now use 1.25 volts PLUS the base to emmiter voltage of the emitter follower transistor at 1.25 mA collector current . The resistor across the 1.85 volts can be in the 10K to 50K region with the larger values showing a small error due to the addition of the base current from the follower. A jfet would not have this problem with base current but will have other considerations which make it lesss suitable than a BJT even though it would be nice to get the film cap even lower in value.
I'll try, don't hit me if i am wrong.
Vdesired = ((1,85/(R3=10k))*R4) +1,85.
Where 1.85 is the Vbe of the transistor at the given current (1.25ma according to fred but must be with a 1k R1) + the 1.25 for the 317.
These figures are ignoring the base current, i suppose you could try to include that as well but the error shouldnt be too big.
Not sure since it is not my circuit.
I tried at least.
Edit: Spelling
Vdesired = ((1,85/(R3=10k))*R4) +1,85.
Where 1.85 is the Vbe of the transistor at the given current (1.25ma according to fred but must be with a 1k R1) + the 1.25 for the 317.
These figures are ignoring the base current, i suppose you could try to include that as well but the error shouldnt be too big.
Not sure since it is not my circuit.
I tried at least.
Edit: Spelling
So with:
R1=1k
R3=10k
R4=82k
You'll have an output voltage of approx. 17,02V?
Can anyone else see if hjelm's formula is right?
Can someone simulate this circuit with the resistor values and see if Vout is correct? (I don't have any sim programs, and the free ones doesn't seem to include the Lm317)
Thanks for the help
R1=1k
R3=10k
R4=82k
You'll have an output voltage of approx. 17,02V?
Can anyone else see if hjelm's formula is right?
Can someone simulate this circuit with the resistor values and see if Vout is correct? (I don't have any sim programs, and the free ones doesn't seem to include the Lm317)
Thanks for the help
With the transistor model i have for a BC860CSIEMENS and a model i found for the lm317 from jan didden i got the values
R1 1k
R2 100
R3 10k
R4 78k
With these values i got 17.01 volts out.
The thing that was different from the theoretical values was the voltage over R3 vhich was 1.91
So therefore the formula should read
Vdesired = ((1,91/(R3=10k))*R4) +1.9mv +1,91
N.B. These values were with a reference voltage between the vout and vadj of 1.29 volts which is a high value for the LM317.
The 1.9mv is the compensation for the basecurrent.
All these figures ar approximate but should be in the area of what they should be.
R1 1k
R2 100
R3 10k
R4 78k
With these values i got 17.01 volts out.
The thing that was different from the theoretical values was the voltage over R3 vhich was 1.91
So therefore the formula should read
Vdesired = ((1,91/(R3=10k))*R4) +1.9mv +1,91
N.B. These values were with a reference voltage between the vout and vadj of 1.29 volts which is a high value for the LM317.
The 1.9mv is the compensation for the basecurrent.
All these figures ar approximate but should be in the area of what they should be.
I tried to add a follower to the circuit, but did I do it right?
How will the value of C2 affect the performance of the Fred’s circuit?
Fred recommends a 1-5uf cap. I build it but only had a 94uf. It works fine (I think) but it takes approx. 10 seconds for it to reach the final voltage (did you say low corner freq.)
The smaller cap 1-5uf performs better at higher freq, but is there anything else?
How will the value of C2 affect the performance of the Fred’s circuit?
Fred recommends a 1-5uf cap. I build it but only had a 94uf. It works fine (I think) but it takes approx. 10 seconds for it to reach the final voltage (did you say low corner freq.
) The smaller cap 1-5uf performs better at higher freq, but is there anything else?
Attachments
Flemming J P said:I tried to add a follower to the circuit, but did I do it right?
How will the value of C2 affect the performance of the Fred’s circuit?
Fred recommends a 1-5uf cap. I build it but only had a 94uf. It works fine (I think) but it takes approx. 10 seconds for it to reach the final voltage (did you say low corner freq.
The smaller cap 1-5uf performs better at higher freq, but is there anything else?
Hey Guys,
I haven't following this thread, and I'm not sure what you are trying to do, but this R7/C5 thing completely destroys any regulation that was present in the circuit. The circuit degenerates to an emitter follower for anything except the very lowest frequencies. Are you aware of this?
Jan Didden
Sounds about right
"So therefore the formula should read
Vdesired = ((1,91/(R3=10k))*R4) +1.9mv +1,91"
The voltage is the Vout - Vajd of the three terminal regulator plus the base emitter voltage of the emitter follower (the only emitter follower.........) connected to the ADJ pin. The programing formula's in the data sheet will work with this new number replacing the 1.25 volts in the data sheet. This circuit is very straight forward and seems to work as intended. Preloading is putting a resistor from Vout to ground to draw a DC current to lower the impedance of the regulator. I would start with at least 25 mA and preferably more with heat sinks on the regulators and sufficient capacitance in the unregulated supply feeding the regulator. The half ohm resistor can be 1/2 watt or more. I used three 1.5 ohm quarter watts in parallel.
I am not going to spend any more time on this since I am looking at a completely different type of IC than the standard LM317 and LT1086; which offer even better performance for a reasonably simple circuit Read the data sheet for the LM 317 a few times and it should be clear how this works. There is only one transistor and a few resistors extra. A 5 uF film cap will work fine for noise, impedance, and high frequency considerations. There is really no point in talking this thing to death when you can build it in half an hour and then go listen to it and measure it.
"So therefore the formula should read
Vdesired = ((1,91/(R3=10k))*R4) +1.9mv +1,91"
The voltage is the Vout - Vajd of the three terminal regulator plus the base emitter voltage of the emitter follower (the only emitter follower.........) connected to the ADJ pin. The programing formula's in the data sheet will work with this new number replacing the 1.25 volts in the data sheet. This circuit is very straight forward and seems to work as intended. Preloading is putting a resistor from Vout to ground to draw a DC current to lower the impedance of the regulator. I would start with at least 25 mA and preferably more with heat sinks on the regulators and sufficient capacitance in the unregulated supply feeding the regulator. The half ohm resistor can be 1/2 watt or more. I used three 1.5 ohm quarter watts in parallel.
I am not going to spend any more time on this since I am looking at a completely different type of IC than the standard LM317 and LT1086; which offer even better performance for a reasonably simple circuit Read the data sheet for the LM 317 a few times and it should be clear how this works. There is only one transistor and a few resistors extra. A 5 uF film cap will work fine for noise, impedance, and high frequency considerations. There is really no point in talking this thing to death when you can build it in half an hour and then go listen to it and measure it.
Colonel Mustard in the library with a lead pipe.
"Fred, Any clue which regulator you are eyeballing?"
Yes I am pretty sure I know what they are...... I did have to order them after all. A clue heh ?
"What speaks with one voice, yet walks on four feet in the morning, two feet at noon and three feet in the evening?" Wait a minute, that's not it!
Alright, alright ............. It has more legs than a cow but fewer legs than a spider.
"Fred, Any clue which regulator you are eyeballing?"
Yes I am pretty sure I know what they are...... I did have to order them after all. A clue heh ?
"What speaks with one voice, yet walks on four feet in the morning, two feet at noon and three feet in the evening?" Wait a minute, that's not it!
Alright, alright ............. It has more legs than a cow but fewer legs than a spider.
Attachments
one more time......... READ THE DATA SHEET!
"So I should at leaste lose C5 if I want to continue with this circuit."
No you need to lose a lot more than that.................. the negative feedback loop is in the IC and is inaccessible externally. You are degrading the performance of the regulator.
The adjustment terminal is the negative terminal of a 1.25 V voltage reference who positive terminal is connected to Vout. The programing resistor and the resistor to ground constitute positive feed back. Bypassing the resistor to ground reduces the amount of positive feedback which has the same effect as increasing the negative feedback. You have to understand what a circuit does before you go throwing parts at it to change it. I will never understand why people do this.
"So I should at leaste lose C5 if I want to continue with this circuit."
No you need to lose a lot more than that.................. the negative feedback loop is in the IC and is inaccessible externally. You are degrading the performance of the regulator.
The adjustment terminal is the negative terminal of a 1.25 V voltage reference who positive terminal is connected to Vout. The programing resistor and the resistor to ground constitute positive feed back. Bypassing the resistor to ground reduces the amount of positive feedback which has the same effect as increasing the negative feedback. You have to understand what a circuit does before you go throwing parts at it to change it. I will never understand why people do this.
Re: one more time......... READ THE DATA SHEET!
Probably because they don't understand that they don't
understand. That is always a problem, whether in electronics
or anything else; how do you know if you have understood
correctly? I think the three-terminal regulators are amongst
the most understimated components ever. They look so
deceivingly simple since they only have three pins, but
they seem to be more difficult to understand than most
people realize.
I did something similar at the age of 15 using 78xx's with
pass transistors when building a bench PSU. I also thought
then that I could add an emitter follower at the output and
also added some other interesting degrading ornaments
that were mistakenly intended to have some useful function.
Don't even ask me about the non-working current limiting
circuitry, which is highly classified information, since it could
be dangerous reading for anybody with a weak heart.
Anyway, I learnt the hard way that it didn't work, although
I didn't understand at the time why. I could adjust the voltage,
but there was no ripple reduction, and no current limiting in
any sensible way of using that term.
Fred Dieckmann said:You have to understand what a circuit does before you go throwing parts at it to change it. I will never understand why people do this.
Probably because they don't understand that they don't
understand. That is always a problem, whether in electronics
or anything else; how do you know if you have understood
correctly? I think the three-terminal regulators are amongst
the most understimated components ever. They look so
deceivingly simple since they only have three pins, but
they seem to be more difficult to understand than most
people realize.
I did something similar at the age of 15 using 78xx's with
pass transistors when building a bench PSU. I also thought
then that I could add an emitter follower at the output and
also added some other interesting degrading ornaments
that were mistakenly intended to have some useful function.
Don't even ask me about the non-working current limiting
circuitry, which is highly classified information, since it could
be dangerous reading for anybody with a weak heart.
Anyway, I learnt the hard way that it didn't work, although
I didn't understand at the time why. I could adjust the voltage,
but there was no ripple reduction, and no current limiting in
any sensible way of using that term.