Adjustable power supply questions

[jackinnj]

If you take the rectified and filtered output, create an artificial ground at the junction of two resistors (like 2k2). use the + and - rails to drive LM317L, LM337L regulators -- these cost about $0.50 each. The "adjust resistors and tantalum bypass caps get attached to the artificial ground. For +/- 6 volts the adjust resistor should be 820R, the sense resistor 220R for 5.90 Vout.

for the ap you are describing an LM317L (the 100ma version) will do fine since the heavy lifting is done by the pass transistor.

[rendisha]

Jack, I knew how to get +/- 6V, but now I am concerned about Jan's and UrSv's posts that Op-Amp supplies should be referenced to the output voltage. Does it mean that + supply should be 6V above output voltage and - supply 6V below?

...Still waiting for an answer to my last question

[halojoy]

Originally posted by UrSv


I think they the TO-3 because it is an application note and that is the regulator in question. The regulator is NOT used as the main current provider but as a regulator (suprise?). I have downloaded the datasheet again and looked closely and as far as I can see the resistor is 33 Ohms just as clearly as in the picture posted earlier.

As for #6 I would not be the right person to tell you how to do it and if I did then it would take a long time and most likely be incorrect. I do have a basic knowledge of electronics but I have clear limitations...

You can fool some people, but not me.

FACT remains: you do not use TO3 to transfer 20 mA
If you recommend this, you should do it for your self,
and not fool other people into it.

33 ohms will run 2 Ampere , which is exactly what the IC
can take, plus 3 Ampere in Transistor
making that 5 Ampere Power supply.

UrSv why do you want to RUN all 5 AMPERE
through The power Transistor.

Why uses Nelson Pass Several power transistors in parallell,
if he runs 20mA in one, and 5 AMPERE in other.

Please, somebody out there!
God save us from bad Teachers!

/halojoy is in dispare- from the sillyness going round here

[halojoy]

Originally posted by halojoy

You can fool some people, but not me.

FACT remains: you do not use TO3 to transfer 20 mA
If you recommend this, you should do it for your self,
and not fool other people into it.

33 ohms will run 2 Ampere , which is exactly what the IC
can take, plus 3 Ampere in Transistor
making that 5 Ampere Power supply.

UrSv why do you want to RUN all 5 AMPERE
through The power Transistor.

Why uses Nelson Pass Several power transistors in parallell,
if he runs 20mA in one, and 5 AMPERE in other.

Please, somebody out there!
God save us from bad Teachers!

/halojoy is in dispare- from the sillyness going round here

so if there is a mistake in a schematic
someones are going to build that, inspite of that
the circuit does not make sense.
- I do my own thinking, my own check up - if something is not
reasonable - I have to correct that -

blind followers of faulty Schematics - I did not think they were here at
www.diyAudio.com

[UrSv]

Originally posted by halojoy

You can fool some people, but not me.

FACT remains: you do not use TO3 to transfer 20 mA
If you recommend this, you should do it for your self,
and not fool other people into it.

33 ohms will run 2 Ampere , which is exactly what the IC
can take, plus 3 Ampere in Transistor
making that 5 Ampere Power supply.

UrSv why do you want to RUN all 5 AMPERE
through The power Transistor.

Why uses Nelson Pass Several power transistors in parallell,
if he runs 20mA in one, and 5 AMPERE in other.

Please, somebody out there!
God save us from bad Teachers!

/halojoy is in dispare- from the sillyness going round here

Everyone is entitled to their own view on things. I just don't see how that would make sense or even be reasonable. I would not build a regulator that would run 2 A through a 33 Ohm resistor giving 132 W (through an unspecified resistor as opposed to the 0.2 Ohm 5 W power resistor which passes most of the current IMHO) of dissipation and meaning that I would need 66 V more input than I want output. NOT including the dissipation in the pass transistor. It seems that this 66 V drop is not possible with only 35 V input, especially when output is 30 V. IMHO it does not make sense to run the regulator at or above it's limit when it is solely used as a regulator. It is not a FACT that you don't use TO-3 to transfer 20 mA it is simply so that you might if you want to. The reason to run as much current through the pass transistor is because that is why it is there. Besides that it is not me who wants to. It is National. I have however done it and it works just fine.

As for why Pass used several transistors I think the reason is that he wants to share the current between several transistors to increase safety (SOA et al) and improve various aspects of the amplifier. It would simply not be possible to run an Aleph 0 with only one transistor. You know that.

As for why you would use this or that for what reason my favourite is when using 2N3055 (I seem to recall that was it) with the case top , which is indeed TO-3, taken off using light to produce a voltage (apparently it does when exposed to light) to drive an MC amp. How is that for TO-3 transistor non-standard application?

Finally, I never claimed to be a good teacher and most likely I am not.

[roadkill]

I believe halojoy is right about the resistor being 0.33 ohm, this fits in line with the usual "current dumping outboard transistor" attachment to 3-terminal regulators. However, his calculations show that an LM317K should be used, since only this will allow the rather large power dissipation caused by 2A of current. A 33ohm resistor would allow only a small current to flow through the LM317, which would cause the circuit to oscillate if sudden transient loads appear at the output, as the LM317 and the pass transistor try to figure out who's doing what.
There may be one way to settle this: we could mail maybe Bob Pease or someone else at National and hope they have the time to reply.
rendisha, I assume you want a negative regulator so that you can get a split supply where both split voltages are adjusted together (that is, one knob controls both voltages). If so, it's probably better to use an opamp+2N3055/2955 to produce a dual tracking regulator. The only part of the circuit you'll have to be careful about is the reference voltage. If that's not well regulated and noise-free, so will the output. It's thus not good to use a zener, an LED + constant current source would be better.

[UrSv]

Originally posted by halojoy


so if there is a mistake in a schematic
someones are going to build that, inspite of that
the circuit does not make sense.
- I do my own thinking, my own check up - if something is not
reasonable - I have to correct that -

blind followers of faulty Schematics - I did not think they were here at
www.diyAudio.com

Halo,

Maybe you should take it upon you to inform National, Elektor, ELW and some others about the error in this schematic. I have seen maybe 25 different versions of this circuit and all have been working the same way. This must mean the error is spreading as you fear.

Some examples today:

http://www.gmweb.btinternet.co.uk/jlhlm3x7cb.htm
http://www.repairfaq.org/sam/samschem.htm#schslp2

[Jeff R]

I believe UrSv is correct.

The LM317 is not meant to supply much output current - it is there only to set a voltage reference to drive the pass transistor. High load currents from it would negatively impact load regulation of both it and the pass transistor's output voltage.

To help show in a simple way that the LM317 is not supply any load current, consider the current sense resistor, R3. Note that it is in series with the pass transistor, but not the LM317. If the LM317 were putting out up to 40% of the current, as Halojoy claims, then none of its up to 2 A current is being monitored by the current sensor. This does not make sense! Since the LM317's output does not flow through the current sensor, the logical conclusion is that it does not supply any significant load current at all.

There are other reasons to show the circuit, as shown, is correct, but I feel this one reason is sufficient to show that the regulator is not intended to supply load current.

Of course, the regulator does not have to be a T0-3 as the current demand is very low. We have to remember that this circuit has been around for some 25 years. Back then, TO-3 packages were nearly as cheap as TO-220's and even TO-39's (remember them?). Even a TO-220 is overkill here, but that is still what I would probably use.

[roadkill]

OK, Jeff and UrSv, maybe the regulator does serve as a reference. I've never managed to understand how outboard pass transistors work, could one of you explain? From my understanding, the outboard device works as the second half of a complementary feedback pair. Is this right? My previous post was based on a similar circuit I had seen in a "circuit cookbook".
Shouldn't it also be possible to get the output to go all the way to zero if you connected R8 (adjust pot) to -6V in place of ground? Maybe there should be another resistor b/w the pot and -6V because the lowest voltage at the ADJ terminal is -1.2V to make it output 0V.
Another question: I've not used opamps that require compensation, so what's with D2? C5's function is also a bit of a mystery to me.
Addressing rendisha's question about the opamp's supply voltage, it can be the output voltage only if you use an opamp which tolerates its supply voltage at its inputs. LM301 seems OK (from the datasheet), so you can connect it to the output voltage. Anyway, the opamp's function here is to rob the LM317's ADJ pin of current when an overcurrent condition occurs, through the diode and LED. It's probably best to have an auxilliary +/- 6V supply for the opamp alone.

[jan.didden]

Hi Roady,

To understand the outboard pass, assume that initially the pass transitor does not conduct. Now you put on a load, and the 317 starts to supply current to the load to keep the output voltage up. This current flows through the (in)famous R1 (33 Ohms). When the current reaches about 18 mA, the Vbe of the pass transistor gets high enough to cause Ic to flow, starting to add to the load current. If this current is not enough for the set output voltage, the 317 will start to conduct more, which in turn will turn on the pass transistor more etc: the output will be regulated.

BTW, I agree with the analysis that the 317 only suppllies a very small part of the total current. Why it is a K-version, beats me. The small price difference between TO3 and TO220 in those days may indeed have something to do with it.

Jan Didden