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What is "floated" linear voltage regulator?
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xavier60:

--- Quote from: 001 on November 21, 2018, 08:02:16 am ---
--- Quote from: spec on November 21, 2018, 07:54:59 am ---

The term floating is really a misnomer, because all voltage regulators are connected one way or another to 0V, they must be to function. Having said this, it is possible to have the regulator actually isolated from the 0V line by using an opto isolator for example.

Getting back to the 200V supply, you would connect a 100R resistor between the output and sense terminals and a 15k9 resistor from the sense terminal to 0V.

--- End quote ---

Thanx a lot! Can You tell me what reference is used in floating designs? How it "feels" ground potential?

--- End quote ---
You have been getting answers. It would help if we know how much knowledge you have. Do you understand op-amps and resistor dividers?
Zero999:

--- Quote from: ArthurDent on November 21, 2018, 04:26:01 am ---"So how does a floating regulator create/have a low impedance to ground if it never references to it?"

There is a difference between impedance and resistance. A supply's output could have an extremely high resistance to ground at D.C. but with a capacitor from the output to ground could basically be a short circuit at R.F.. A lot of supplies I have say the output can be floated +/- XXX volts.

--- End quote ---
That's true, but with a DC power supply we always need a very low DC resistance. The the linear regulator is too slow, then the output impedance will rise with increasing frequency, which is why a decoupling capacitor is used.


--- Quote from: 001 on November 21, 2018, 08:02:16 am ---
--- Quote from: spec on November 21, 2018, 07:54:59 am ---

The term floating is really a misnomer, because all voltage regulators are connected one way or another to 0V, they must be to function. Having said this, it is possible to have the regulator actually isolated from the 0V line by using an opto isolator for example.

Getting back to the 200V supply, you would connect a 100R resistor between the output and sense terminals and a 15k9 resistor from the sense terminal to 0V.

--- End quote ---

Thanx a lot! Can You tell me what reference is used in floating designs? How it "feels" ground potential?

--- End quote ---
It doesn't feel ground potential, because no terminal in the floating regulator is directly connected to ground.

Look at the simplified schematic of an LM317. It consists as an op-amp with it's negative supply rail connected to the output terminal. The inverting input is also connected to the output and the non-inverting input is connected to the adjust pin, via a 1.25V voltage reference. A tiny 50µA current source biases the reference and op-amp's non-inverting input.

As with any op-amp circuit, the op-amp adjusts its output, turning the output transistor more on or off to ensure the voltage at both its inputs is the same. Because the non-inverting input is connected to the adjust pin via 1.25V reference, the voltage at the adjust pin will be held at 1.25V below the output.



Now let's look at the basic application circuit for the LM317.


The output is connected to the adjust pin via a R1. Because the voltage between the output and adjust pin is fixed 1.25V, the current through R1 will also be constant. R2 is in series with R1, thus the current through it will also be constant. A tiny 50µA bias current flows out of the adjust pin and through R2, but it's normally small enough to ignore. Hopefully from this statement, you can derived the formula for calculating the output voltage, using Ohm's law.

The main advantage of this configuration is the op-amp's power supply is equal to the difference between the input and output voltages. This means the input-output differential limits the maximum supply voltage, rather than the op-amp or pass transistor's voltage rating. Note that if the protection circuitry (over current/temperature) is triggered, the output transistor will start to turn off, thus lowering the output voltage and increasing the total power supply voltage to the op-amp, which will be damaged if the supply voltage exceeds its maximum rating. Thus, if the protection circuitry is required the total supply voltage must be within the regulator's maximum rating, unless additional parts are added to limit it.

The downsides are the 50µA bias current flows out of the adjust pin which will cause the voltage across R2 and therefore the output voltage to be slightly higher, than the basic calculation suggests but that can be easily be calculated by adding R2*50*10-6. The op-amp's supply current flows also through the load, so a minimum current is required to ensure the op-amp has sufficient power to operate, but that can be ensured by making R1 low enough to pass the minimum current with no load.

It's possible to build your own floating regulator, from discrete parts, but it's not practical. You'll need to design a discrete op-amp with: a common mode range which includes the negative rail, a wide supply voltage (<3V to ensure the dropout voltage isn't too high, up to the maximum expected input-output voltage) and capable of outputting sufficient current to the driver transistors.  Discrete transistors aren't matched, so the performance will be inferior to an IC design and over-temperature protection is virtually impossible with a discrete design.
schmitt trigger:

--- Quote from: xavier60 on November 21, 2018, 02:51:53 am ---

It needs to be mentioned that this design uses the mains switch and transformer's primary winding to discharge the large capacitor.
Very hazardous!

--- End quote ---

 :o
I hadn't noticed that!, thanks for pointing it out.
They saved the cost of a discharge resistor. But the more complex DPDT switch instead of a DPST would offset that, don't you think?
Zero999:

--- Quote from: schmitt trigger on November 21, 2018, 03:14:36 pm ---
--- Quote from: xavier60 on November 21, 2018, 02:51:53 am ---

It needs to be mentioned that this design uses the mains switch and transformer's primary winding to discharge the large capacitor.
Very hazardous!

--- End quote ---

 :o
I hadn't noticed that!, thanks for pointing it out.
They saved the cost of a discharge resistor. But the more complex DPDT switch instead of a DPST would offset that, don't you think?

--- End quote ---
There's also the potential for it to go wrong, as hinted above. The DPDT switch would have to be rated to give full isolation from the mains.

I don't think it's about cost, but discharging the capacitor quickly, without having to use a low value, high power resistor which would get very hot. A safer solution would be to use a suitable resistor and another set of properly isolated switch contacts or a small relay.
ArthurDent:

--- Quote from: Hero999 on November 21, 2018, 11:54:28 am ---
--- Quote from: ArthurDent on November 21, 2018, 04:26:01 am ---"So how does a floating regulator create/have a low impedance to ground if it never references to it?"

There is a difference between impedance and resistance. A supply's output could have an extremely high resistance to ground at D.C. but with a capacitor from the output to ground could basically be a short circuit at R.F.. A lot of supplies I have say the output can be floated +/- XXX volts.

--- End quote ---
That's true, but with a DC power supply we always need a very low DC resistance. The the linear regulator is too slow, then the output impedance will rise with increasing frequency, which is why a decoupling capacitor is used.


--- Quote from: 001 on November 21, 2018, 08:02:16 am ---
--- Quote from: spec on November 21, 2018, 07:54:59 am ---

The term floating is really a misnomer, because all voltage regulators are connected one way or another to 0V, they must be to function. Having said this, it is possible to have the regulator actually isolated from the 0V line by using an opto isolator for example.

Getting back to the 200V supply, you would connect a 100R resistor between the output and sense terminals and a 15k9 resistor from the sense terminal to 0V.

--- End quote ---

--- End quote ---

--- End quote ---

That's not what I said at all. I said "high resistance to ground", not impedance. I also said "to ground", not to 0 volts. The output impedance between the output terminals of a power supply should be low but we generally want the resistance to ground of a bench supply to be high or infinite. This gives us the ability to wire one supply in series with another without shorts and this is what is done with dual supplies where they can be switched for either series or parallel operation. Supplies floating with respect to ground allow us to do this or a single output supply to be wired either + or - with respect to ground.

If you look at the schematic 001 posted in post #20 to show this is what he meant. You will see this in the lower right hand corner of the schematic as shown below. The circuit has high resistance but low impedance. With a capacitor to ground the 'impedance' decreases with increasing frequency.
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