Hi again,
ronv:
I think you are right there. If the ripple can get to the reference, then the reference adds ripple to the reference for the top regulator. I guess you would want to add some capacitance then? This sounds like one of the ideas i'd like to see come out in this thread.
Diver300:
You seem to be right on the money
There's another small detail which i'll get to in a minute, but you've covered it pretty well for the usual applications of this regulator.
FATAL FLAW PROBLEM
The "fatal flaw" was that the top regulator sources current into the bottom regulator, so the bottom regulator must be made able to handle the top current plus it's own current, which as Diver said, is about twice the normal current. So with this idea in mind, the bottom regulators R1 should be 60 ohms or add another resistor in parallel to the whole bottom regulator.
VOLTAGE DROP PROBLEM
One advantage is the voltage drop due to series resistance Rs is much less with the external reference. There is another advantage which i'll address far below.
To illustrate the "Rs" drop problem (Rs is the resistance between the very output of the regulator and the top bias resistor R1 which is 120 ohms but sometimes is made as high as 220 ohms) we can first calculate the voltage using the 'normal' equation i provided in the previous thread:
Vout=(Vref*(R2+R1)*RL)/(R1*RL+Rs*RL+Rs*R2+Rs*R1)
where Vref is the internal reference voltage usually taken to be 1.25v,
R1 is the top bias resistor usually 120 to 220 ohms,
R2 is the bottom bias resistor (the voltage set resistor),
RL is the load resistance loaded somewhere after the Rs resistor so that the load current flows through Rs also,
Rs is the pesky series resistance at the output of the regulator.
That equation is for the single IC version, and calculating the output with:
R1=120
R2=1243.2,
Rs=0.000,
RL=14.2,
Vref=1.25,
we get:
Vout=14.2 volts
No surprise there since R2 was calculated R2 to provide us with this voltage.
Next, raise Rs up to 0.020 ohms (the wire resistance) and we get now:
Vout=13.974
So we see that small wire resistance cost us 0.2 volts, and that's a lot for a circuit that is supposed to be regulating the voltage.
Lets see what happens with the new circuit, where the second LM317 is used as an external voltage reference...
Because the circuit topology is different now, we have a new equation:
Vout=((Vref*R1+Vext*R1+Rs*Vext)*RL)/(R1*RL+Rs*RL+Rs*R1)
One thing that stands out already if we compare the two equations: the first equation has Rs in the denominator only, while the second equation has Rs in the numerator as well. So for the first equation when Rs increases, the voltage must drop, but in the second equation when Rs increases there is a compensating term in the numerator which means it should not go down as far. Lets see what happens when we evaluate.
With this new equation the output at the very output of the regulator will be whatever is at the ADJ terminal plus the reference, so we need Vext to be:
Vext=14.2-Vref=14.2-1.25=12.95 volts.
So plugging in:
R1=120,
RL=14.2,
Vref=1.25,
Vext=12.95,
Rs=0.000,
we get:
Vout=14.2 volts.
Again this is what we set it for and since Rs=0 there's no drop at all.
Now lets set Rs=0.020 again and we get:
Vout=14.18 volts.
Wow, the error before was 0.2v and now it is only 0.02 volts, so that's quite an improvement.
So we see a big advantage here. Be back in a minute with the second advantage which will be interesting for bench power supplies.
ADVANTAGE #2, TEMPERATURE COMPENSATION
First let me say that this second advantage is applicable mainly to bench power supplies or power supplies which operate in ambient air temperatures which are relatively stable. It is amazing how much difference this can make with a bench power supply made from an LM317 though.
One of the problems with the LM317 is that the voltage reference is located on the same chip as its drive circuit, so when the drive circuit heats up so does the voltage reference. This results in significant output voltage drift for a bench supply. By removing the reference from the same package as the driver circuit at least in part, we can improve this quite a bit too. With the bulk of the reference voltage external to the chip, the external reference behavior dominates over the internal reference behavior.
Starting with the old equation:
Vout=(Vref*(R2+R1)*RL)/(R1*RL+Rs*RL+Rs*R2+Rs*R1)
with Rs=0 (we dont care about that resistance anymore) we get:
Vout=Vref*(R2+R1)/R1
Now with values selected as before:
R1=120,
R2=1243.2
we get:
Vout=14.2 volts
of course, but now let Vref vary by -1 percent and we get:
Vout=14.058
Not very nice is it? For a standard power supply this might not be a problem, but for a bench supply it's a nightmare.
And this does not come about because the ambient temperature suddenly became equal to that on Mars in the sunlight, it's because the heat sink had heated up due to the load current of 1 amp and an input output differential of maybe 8 volts or so and a decent heat sink.
Ok, so back to the new equation:
Vout=((Vref*R1+Vext*R1+Rs*Vext)*RL)/(R1*RL+Rs*RL+Rs*R1)
We can work with Rs=0 again because we are no longer concerned about that resistances effect so we have:
Vout=Vref+Vext
Raise your hand if you expected such a simple result
Too simple? Well that's what it comes down to. With the internal gain assumed to be high (and it is) that's what we get.
Right away we can see now that if the reference voltage changes it is not going to be comparable to Vout if Vext is a few times larger than the initial Vref.
With our output of 14.2v and Vref=1.25 and Vext=12.95 we of course again get 14.2 volts output, but now when we change Vref by -1 percent we get:
Vout=14.1875
and that is almost 14.19 volts, so now we've only lost about 10mv instead of 140mv, a significant improvement.
So there you have it. Two advantages to using the second regulator. BTW, the lower regulator can be an LM317L device so it is much smaller, but it MUST NOT be located on or near the same heatsink as the top regulator as it must have free air flow from the ambient air temperature that should be somewhat stable.