DIY Audio Projects Forum

Several posts before Suncalc gave an exsample of floating CCS circuit. "Floating" means you can use it (connected serially) in any part of your circuit, because it is unreferenced to any "ground" or "power" rails.
If you still haven't convinced there is an example from the page 16 of LM317 datasheet from National Semiconductors:

I have to be more specific.
- What will be the constant current value in LM317HV input? How do we calculate it?
- How do we make LM317HV input currrent steady resilient to voltage changes?

Hi, Within the limits of the voltage ratings for the LM317 (57 for the HV version) once you program it to be a constant current device it doesn't matter what the voltage across it actually is. As long as it is above the minimum (about 4 volts) and below the max it will be quite consistent in current flow. Well below 1% in my testing. The current is set by the sense resistor. If you put the sense resistor between the ground and output terminals (like I do in the amps) it will develop a voltage based on the current flowing through it. The LM317 will adjust its internal resistance to keep the actual voltage between the adjust pin and output pin to quite close to 1.25 volts. If the current tries to increase then it would increase the voltage from 1.25 and the IC will counter act it. A really simple application. Incidentally many 3 terminal regulators will do the same within their ratings. If you ignore the small amount of current the IC uses internally you can approximate the current setting by dividing 1.25 by the desired current in amperes. This will provide the value in ohms of the needed resistor. The accuracy of the calculation is usually limited to the precision of the resistor.

In the amplifiers I have determined what I feel is the optimum value for the current for each tube and set the LM317 accordingly. The 25 ohm control shifts a small amount of voltage in a see-saw fashion between the two cathodes. (remember that the pot will develop a voltage across it depending on the current flow through it). This way one cathode can be as much as about 2 volts different from the other under static (no signal) conditions. This will in turn change the grid to ground voltages and effectively alter the bias for the tubes. If the tubes are too mis-matched then balance can not be obtained and the tubes should not be used as they probably won't play well together and the sound will suffer.

Good listening
Bruce

_________________
Some of my DIY Tube Amplifier Projects:

• PoddWatt: 5751 / EL84 Push-Pull DIY Tube Amplifier Project
• OddBlocks: KT88 Push-Pull DIY Tube Amplifiers (Mono Blocks)
• ForeWatt: ECC802S (12AU7 / ECC82) DIY Tube Preamplifier Project
• GrooveWatt: Tube (Valve) RIAA Phono Preamplifier Project

Thanks Bruce,
So, the input current is equal to the output current + a very small amount of dissipated current inside IC. Even though dissipated current may vary with the change of potential difference between input and output of IC, since it's a very small amount we neglect it and we assume input current is also a constant current as the output current of IC ( LM317HV).

Hi, in simple terms, what goes in comes out minus about 10 microamps. The voltage across the LM317 times the current through it equals how much power in the form of heat it must dissipate. Several watts in many of my amps. In the KT120 amps that is why there are two LM317 in parallel. You would expect that they would not share the load well...but they do if purchased from the same source and lot (usually the case). Sharing was within a few percent and the difference unimportant.

Good listening
Bruce

_________________
Some of my DIY Tube Amplifier Projects:

• PoddWatt: 5751 / EL84 Push-Pull DIY Tube Amplifier Project
• OddBlocks: KT88 Push-Pull DIY Tube Amplifiers (Mono Blocks)
• ForeWatt: ECC802S (12AU7 / ECC82) DIY Tube Preamplifier Project
• GrooveWatt: Tube (Valve) RIAA Phono Preamplifier Project

Now, I got the whole picture and it feels great in terms of confidence. I must also add you do have a different but extraordinary approach by using LM317HVs this way.

It's not so much important but I want to know if I'm right. When you apply signal to up and running amplifier does used/dissipated current in IC change (10 micro amps when idle) of course if you could have ever measured it?

Hi, I never checked on the 10 ua. It is not significant when the main amount is 20,000 times bigger (apprx). Since it is only enough to provide the reference it might fluctuate wit the signal, but again insignificant. Overall dissipation will fluctuate a small amount on an instantaneous basis but the average will be unchanged at the idle setting. The number of signal swings above zero will be equal to the ones below as the amplifier is AC coupled and does not have to handle any DC signals.

Good listening
Bruce

_________________
Some of my DIY Tube Amplifier Projects:

• PoddWatt: 5751 / EL84 Push-Pull DIY Tube Amplifier Project
• OddBlocks: KT88 Push-Pull DIY Tube Amplifiers (Mono Blocks)
• ForeWatt: ECC802S (12AU7 / ECC82) DIY Tube Preamplifier Project
• GrooveWatt: Tube (Valve) RIAA Phono Preamplifier Project

I'd like to clarify several things.
First of all I'd not name the adj-pin current as "dissipated". It is more a defining current. Secondly: the current does not depend on "the change of potential difference between input and output of IC" (it does depends on temperature and there is some initial margins of the current across batches).
Small note: there is no "minus" between the "in" and the "out" currents. What goes in, that comes out. There is no "third way" the current can flow. So the question: becomes formulated not very correctly. Idle current through the IC is set by the resistor between out and adj and it can never be 10 micro amps (minimal current through the LM317HV is 10 milliamps). The current through the adj as I pointed above does not depend on the voltage between in and out.

Hi, I sure hope we don't confuse the requester on this. My comment was intending to indicate that the current in the output terminal and through the sense resistor was not perfectly the same as the input current. The actual very very small amount diverted by the IC that goes out the adjust terminal fits into the formulas for determining the set value. It can matter with low power versions of the LM317 in low current applications. For the high power application here it is completely insignificant. As such it has no bearing on the dissipation of the device. The power dissipation in the form of heat in the LM317 is equal to the current entering the LM317 times the voltage from input to output terminals.

Good listening
Bruce

_________________
Some of my DIY Tube Amplifier Projects:

• PoddWatt: 5751 / EL84 Push-Pull DIY Tube Amplifier Project
• OddBlocks: KT88 Push-Pull DIY Tube Amplifiers (Mono Blocks)
• ForeWatt: ECC802S (12AU7 / ECC82) DIY Tube Preamplifier Project
• GrooveWatt: Tube (Valve) RIAA Phono Preamplifier Project

Which one is true?

Iin = Iout + Iadj
or
Iin= Iout + Iadj + Idis

Iout: I refer to the curent on the output terminal of LM317HV.
Idis: dissipated current inside the IC.