# "High" Voltage regulation

• posted

In my never ending quest for a cheap and simple high voltage regulation I have come up with a simple dual capacitor mode. This is similar to a Buck circuit without the inductor.

The concept is simple: One capacitor is charging to the pre-destined voltage while the other is discharging into the load. These are simple RC circuits and the time constants are readily known. The load is not connected directly to the load because it will potentially see high transient voltages from the supply. Also the capacitors are charged at a rate close to what it is being discharged by the load. This is to reduce switching frequency and rise/fall times.

The switching is done with make before break on the inner two so that the load is never disconnected from a capacitor. Alternatively one could simply use one side with a capacitor in parallel with the load to supply the current while the other capacitor is charging. Either case will work but I have specific reasons for choosing the first case.

My main concern is driving the mosfets so the gate voltage is within spec. In fact I'm not quite sure how to do this.

My requirements is a programmable constant voltage supply from around

0 voltages to 1kV driving a load of around 100kohms to 1Mohms. The regulation should be less than 1% or better.

From my calculations I easily get less than 0.1%. i.e., t = -R*C*ln(Vc/ V) and for R = 10^5, C = 10^(-5), Vc/V = 1 +- a/100 where a is the percent regulation gives about 1ms for 0.1%. Hence discharging the capacitor into the load will drop approximately 0.1% of its voltage in

1ms.
• posted

You may as well use a single series fet+resistor, pwm controlled, and one output cap. Or just use a linear regulator. All are about equally efficient.

John

• posted

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If you don't disconnect the load there will be a voltage spike and also a voltage dividing effect. i.e., the load is connected to the voltage source directly and depending on the ratio of the two resistors the load will generally see a higher voltage than you might want. Hence the reason why the load is disconnected. If the voltage supply resistance is too small then it requires faster rise/fall times to quickly stop charging the capacitor as it will end up charging up much faster.

The same problem still exists with your idea though. The fet's gate will need to be held at a voltage that is within the source/drain of usually around +-20V max. But if the fet is sorta "floating" I am unsure how to accomplish such a task. At least in a way that keeps with the simplicity of the design.

I've thought about using a simple linear regulator idea but unfortunately similar problems as well as other problems exist. Mainly in this case the regulation seems much poorer and it is less efficient. Although I'm not too interested in efficiency as I am regulation. Trying to find cost effective HV bjt's is a bit of a problem as compared to fets.

• posted

Take a look at one of the PD patents i posted on my site, maybe it will either be useful as-is or give you some ideas:

• posted

your circuit avoids neither of those problems.

a linear series regulator is no less efficient than a series resistor for dropping voltage, but it handles load fluctuations better.

you should do an energy audit on your design.

• posted

That circuit is about as efficient as a linear regulator.

[....]

I have done a design that did that. How much ripple can you stand?

If you make the basic flyback switcher and drive a transformer with it, getting 300V on the secondary isn't hard. A voltage tripler with the two diodes in the flyback direction and a choke in series with the diode in the forwards direction gives 1000V.

The reason to stay down at 300V is that I found a transformer that could do that.

• posted

One of my products has a power switcher (variable output, 25 to 200 volts) using a series switching fet and a BIG power resistor feeding a BIG output cap. A microprocessor/ADC checks the output voltage once every millisecond and turns the fet on or off. Works great, and keeps the room warm. It's about as efficient as a linear reg, but simpler, and the heat's mostly in the resistor, not in the fet.

Sort of a delta-sigma regulator.

John

• posted

No so. A linear regulator must drop the full voltage difference. The capacitor resistors are only there to reduce the charging rate on the capacitor so that the fet switching times are within spec. Ideally no capacitor resistance would be needed and the capacitors would charge up to the programmed voltage instantaneously and the switches would cut the capacitor off at the programmed voltage. The capacitor resistors are not analogous to the bjt in a linear regulator. In they behave very differently in what they are doing. The bjt acts as a current controlled resistance while the capacitor resistors are only reducing the charging times and effect the duty cycle. It's a different story for large loads but in this case for loads larger than

100kohms they are not even close to being the same.

Last line... 1%. Ideally 0.1% or even 0.01%.

I only need to reduce the voltage rather than increase it. I have a main power supply that will give the 1000V and I need to reduce this subcircuits that will use some fraction of the mains. The circuit I'm using is simply a variation of a buck circuit taking into account the low loads I'm driving. It replaces the inductor with a resistor and prevents the voltage dividing effect when the load is connected directly to the power supply which the inductor would take care of naturally. In fact I might eventually end up going with a buck circuit if I can find decent low cost inductors to get the regulation I need. I just found that the circuit above, at least ideally, seems to give me great regulation that I don't think I can get with the standard buck.

The charging phase of one capacitor is

Vc(t) =3D (V - Vc0)*(1 - e^(-t/Rc/C)) + Vc0 t_charge =3D -Rc*C*ln((V-Vc)/(V - Vc0))

The discharging phase is

Vc(t) =3D Vc0*e^(-t/R/C) t_discharge =3D - R*C*ln(Vc/Vc0)

t_charge and t_discharge basically determine the duty cycle and frequency but also regulate the rise/fall times of the fet. One can reduce Rc but this reduces the efficiency of the fets and potentially decreases the regulation. Alternatively it could be balanced out by using a higher capacitance. t_charge must obviously be smaller than t_discharge. For ideal switches t_charge can simply be chosen to be 0 and hence it should be clear that it would be much more better than a linear regulator.

Given

R =3D 10^5, Rc =3D 10^3, C =3D 10^(-5), V =3D 1000, Vc =3D 100, Vc0 =3D 0.1= % within Vc then

t_charge ~=3D 1us t_discharge ~=3D 1ms

That is, it takes 1us to charge the capacitor and 1ms to discharge it. Taking the rise and fall time to be 1/100th of this time requires a rise/fall time of ~ 10ns. Increasing the capacitance to 100uF reduces this requirement by a factor of 10 as does increasing Rc by a factor of 10.

• posted

r

Similar to what I'm suggesting doing. I assume though your load is very large compared to mine. At most I would be dissipating

1000^2/10^5 =3D 10W. I'm mainly doubling up so that when the fet is off for one capacitor it is on for the other sot hat the load is never disconnected from a capacitor as I need to supply it continuously. As I mentioned one could probably do this just by adding a diode and capacitor after the first capacitor so that it acts as a temporary storage while the other capacitor is being charged.

How do you drive the gate of the fet with the uP?

• posted

Optoisolator, with a small stock DC/DC converter to supply floating power for the gate driver. The whole rig just precharges a capacitor bank before we kick in the BIG power supply!

John

• posted

the only diference is the power to control the circuit

their purpose does not change their performance.

If you have 1000v on one end and 10v on the other end for evert milliwatt you pull out of the 10V end you're burning 990v in the resistor.

the magnetic fields created by the infinite currents would destroy your device, and it still wouldn't be any more efficient. instead of heating up it would be producing radio waves.

if you put 1mA in at 1000V and only get 1mA out at 10V you still have only 1% efficiency.

it's clearly no more efficient, see above.

if you want improved efficiency use a buck regulator if you want reduced ripple use a larger filter capacitor. or a higher frequency,

• posted

You seem to be running some kind of pSpice, so why not use it properly to audit the local component losses or efficiency?

There was another thread you might have been interested in, but it seems to have vanished from the 'record'. Started by J.Slaughter on Jan18 of this year under the heading 'tranformerless transformers', a search for TPS60503 seems to get some results.

Perhaps another reader can post a reference; I have only partial contents of this thread archived locally.

RL

• posted

On Mon, 04 May 2009 11:00:27 -0700, James Rollins wrote:

The energy lost in charging a capacitor through a resistor doesn't depend upon the resistor's value. It doesn't matter whether the resistor is

1kOhm or 1Ohm or 1millioOhm or 1microOhm.

If you let current flow from one capacitor to another through a purely resistive path (zero inductance), the energy/power loss is proportional to the difference in voltages, and independent of the resistance.

If you want better efficiency than a linear circuit and don't want to use an inductor, use a switched capacitor topology, charging capacitors in series and discharging in parallel (for step-down, conversely for step-up), e.g.:

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• posted

So you are saying that no power is dissipated by R1 and R2? You are extremely mistaking. It is true that if you simply have an RC circiut that the the average power dissipation in Rc =3D R1 =3D R2 is 0 because the instantanous power decreases exponentially with time and hence the average power approaches zero.

Essentially you are saying because we are charging a capacitor that electrons flowing through Rc do not heat it up and no power dissipation.

Do you not agree that the power dissipation in the resistor Rc is V_Rc^2/Rc? and that V_Rc =3D V - Vcap and Vcap =3D (V - V0)*(1 - e^(-t/Rc/ C)) + V0?

You can easily compute the average power and it is not 0. Sure it does approach zero but there is another circuit involved. In this case the case the power is extremely small which is totally different than a linear regulator and that is exactly the point. It is not zero as you are claiming though but it on the order of 10mW for my application.

But for a linear regulator the power dissipation is the regulator must be (aV)^2/R where a =3D V/VL and VL is the load voltage. It is only efficient when dropping a small portion of the voltage but becomes more inefficient as we require a lower voltage on the load. Linear regulators also need a few volts of headroom further reducing efficiency.

If you even just glance at the circuit I posted you can see it is very similar to a buck circuit with the the inductor replaced by a resistor. It has the similar effect of reducing the charging on the capacitor as both resist current flow. It does not prevent current changes nor does it temporarily store energy. The inductor in a buck circuit prevents the current supplied to the load and capacitor from changing instantaneously while for the resistor it can change instantaneously but this is prevented by preventing the load from ever being connected to the power supply. The inductor also stores energy but this is technically not needed as the capacitor can do this job too.

This circuit replaces the inductor with a resistor and a symmetric charging/discharging phase to simulate the inductors ability to resist instantaneous current changes. It doesn't do this perfectly and there is a practical limit but it does do this. For small loads such as around 10^5ohms one can easily get around 1% regulation which high efficiency for the a wide voltage range. As the load is increased or the voltage reduction decreased the efficiency approaches that of a linear regulator.

There are a few other notable differences between this circuit and a true inductor based buck circuit. The main difference is that the control circuitry is more complex and 4 switches are needed instead of

2 along with 2 capacitors instead of 1.

The circuit principle is very simple: Charge capacitor to programmed voltage. Use capacitor as power source for load. Repeat.

To do this in practice effectively one must have two capacitors so that while one is attached to the load the other is being charged. This way the load always has power. This is not necessary for all applications but is for mine. Also one must prevent the capacitor from charging past the programmed voltage. This is difficult if you have ever tried to charge a capacitor as it charges extremely fast. In fact too fast for any practical switch to handle. Hence by adding a series resistance in the charging phase this reduces the charging rate to acceptable levels and allows for mosfets to disconnect it from the power supply relatively quickly.

• posted

No; did you even read what I wrote? I'm saying that the energy lost is the charge transferred multiplied by the voltage drop (E=QV); the resistor's

*value* doesn't have any effect. If you use a lower R, the current simply increases. Halving the resistance means double the charging current and half the charging time, so twice the I^2.R power dissipation for half the time, so the same energy loss.

This holds for charge transferred between V1 and C1 via R1, between V2 and C2 via R2, and between C1 and C2 via the unspecified parasitic resistance.

Your circuit is no more efficient than a linear regulator. Current in is equal to current out. Any voltage drop results in a proportional drop in efficiency.

OTOH, a buck converter behaves like a transformer: power in equals power out (plus losses, but none of these are inherent; you can reduce such losses almost without limit). Any voltage drop results in a corresponding drop in input current.

Any circuit that involves transferring current through a purely resistive path can't do any better than a linear regulator. If you want transformer-like (power-conserving) behaviour, you need an inductor or a switched series-parallel topology.

With a buck converter, the energy difference corresponding to the voltage drop across the inductor isn't energy lost; it's energy which is stored in the inductor's magnetic field. When you open the switch, the energy is recovered, as you're now passing current in the same direction, but

*up* a voltage differential rather than down it.

With a series-parallel topology, switching from series to parallel halves the voltage while conserving energy. E=(1/2).C.V^2. If each capacitor's value is C, the capacitance of the pair is C/2 in series but 2.C in parallel, so parallel has 4x the capacitance but half the voltage, so the same energy.

You still have the resistive losses when charging and discharging, but in each case the capacitor voltage is much closer to the input or output voltage, so the energy loss is much lower.

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