J
James Rollins
Guest
On May 5, 10:23 am, Nobody <nob...@nowhere.com> wrote:
extremely mistaking. It is true that if you simply have an RC circiut
that the the average power dissipation in Rc = R1 = 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 = V - Vcap and Vcap = (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 = 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.
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 = R1 = 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 = V - Vcap and Vcap = (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 = 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.