DC boost converter and Ohm's Law

Question

A circuit has a resistance of 1 Ohm, a DC power supply rated at 9W is delivering power to the circuit. Using a DC boost-converter that takes in an input of 9W can it output different voltages? Even if the resistance is 1 ohm?

If the resistance is 1 Ohm and the power is 9W, from ohms law voltage is 3V and current is 3A, but can I use a DC boost converter to increase voltage to higher voltages?

EDIT: The location of the resistor is between power supply and the load, if a converter or constant current was added, the resistor will be after the converter/constant current source and before the load.

Answer 1

The text below is only valid for an ideal resistor (e.g. one without parasitic effects, which should be good enough for the considerations below)

For the answer below I interpreted the question as to ask for information on this schematic:

^{simulate this circuit – Schematic created using CircuitLab}

Since you did not mention any voltages, I'll answer a little bit more generic:

It is not possible is to have a resistor that does not adhere to Ohms law (R=U/I). A simple resistor will adhere to this law/equation at all times, therefore: If you want to build a boost converter that outputs e.g. 5V into a 1 Ohm load, it will have to deliver 25W of power. (P = U^2 / R). This is certainly possible.

But: The boost converter would need to be supplied with enough energy to do that. If your boost converter has an efficiency of e.g. 85%, you would need to supply the converter with ~30W. In this case, your power supply would not be sufficient.

A DC/DC converter can not produce energy/power out of nothing. It is a tool to change the voltage of some supply to another voltage with acceptable losses, nothing more.

If you only have a 9W power supply you can never supply a resistor with a voltage higher than 3V (U = sqrt(P*R), same eq. as above, and that only if you have an efficiency of 100%), you simply do not have the power.

What will happen if you try depends on your power supply.

Some of the possibilities:

• The power supply might switch itself off, since your exceeded its specification
• The power supply might go into current limitation (and basically become a constant current source)
• The power supply might oscillate
• The power supply might get to hot and destroy itself (hopefully not, but many cheap and or badly designed ones do)

Regardless of that, you can not have a resistor e.g. dissipate 20W of power on a 9W supply.

Edit: Further explanation regarding "constant current mode"

Regarding your second comment, and my point about the constant current source. This was just meant as an explanation what a real-world power supply might do if you try to consume more power than the power supply can deliver. Behaving like a constant current source is one thing that can happen in that case:

Some power sources (e.g. most lab supplies) are built in a way that they have a set voltage and a set current. Whichever one is the one that is the limiting point at the moment, will be the one used. Say we set the power supply to 1V and 1A, and connect a variable resistor. When you turn the variable resistor to its maximum resistance e.g. 10k the power supply will be in constant voltage mode (voltage at 1V and current at I=U/R=100uA). If you turn the resistance down, e.g. 0.5 Ohm, the power supply will go into constant current mode (current at 1A, voltage at U=IR=500mV). But the power delivered will never exceed 1W (P=UI)

Answer 2

This is a very interesting question about the fundamentals of energy conservation, power transfer/dissipation and efficiency, in the context of boost converters.

Short answer, quoting your questions:

A circuit has a resistance of 1 Ohm, a DC power supply rated at 9W is delivering power to the circuit. Using a DC boost-converter that takes in an input of 9W can it output different voltages? Even if the resistance is 1 ohm?

Yes. Of course, it will depend on the specific boost converter design. However, in general, a boost converter will raise the input voltage up to an output voltage independently of its series path resistance. That's why they are also called step-up converters. As it is explained below in more detail, the series path resistance (of which your 1 Ohm resistor is now part of), will INCREASE the power losses (reduce the overall EFFICIENCY), having little impact on the capability of the boost converter to raise the output voltage.

If the resistance is 1 Ohm and the power is 9W, from ohms law voltage is 3V and current is 3A, but can I use a DC boost converter to increase voltage to higher voltages?

You are applying Ohm's law as if the boost converter was a short-circuit to ground! Your calculation of 3V/3A is right only if you put the 9W source directly on the 1Ohm resistor.

A boost converter will demand current in repetitive pulses, usually in two phases: 1) Demand of current following a linear or exponential curve, from 0 to a certain value of Amps, while using that current to "charge-up" an internal storage component (inductor or capacitor). 2) Without demanding current from the power source, the energy from the storage component is transferred to the output (load), while raising its voltage.

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Long answer. First, some context and background that will be used for the reasoning below:

• A boost converter is a type of SMPS (Switched Mode Power Supply) where the DC output voltage is higher than the DC input voltage. It is commonly said the the input voltage is raised up to the output voltage.

• A boost converter usually works in two or more timed stages,

  1. During the first phase, some energy is captured from the external power supply (DC voltage input) and stored.
  2. During the second phase, the stored energy is transferred to the load while raising the output voltage.
• We will consider here a basic generic topology: the single inductor flyback boost converter operating in DCM (Discontinuous-Conduction Mode), which will serve us for arriving some conclusions without losing generalization. We could equally well arrive the same conclusions using any other well known topologies, for instance those using capacitors instead of inductors.

Now, we have a single inductor which will serve us for storing some energy during the first phase of the conversion:

• Let's say that the first phase lasts for \$T_{on}\$. During that time, the boost converter will connect the input voltage directly to the inductor, so that the inductor will start "charging-up" energy. An ideal inductor (no parasitic capacitance or resistance), will increase its current linearly up to its maximum allowed current (called saturation current). Let's suppose \$T_{on}\$ is short enough so that the saturation current is not reached, then:

  \$I_{inductor} (t) = \frac{V_{input}}{L}.t \$, where the slope of the curve is \$\frac{V_{input}}{L}\$.

• At the end of \$T_{on}\$, the energy stored in the inductor is: \$E_{inductor} = \frac{1}{2}L.I_{T_{on}}^2\$
• Getting to the point, now we can see that all the resistors in the series path between the input voltage and the inductor will dissipate (loss) some power (heat), according to the Ohm's Law: \$P_{losses}(t) = R.{I(t)}^2\$
• As the current curve is approximately linear (now the effect of R in the circuit makes the current curve to actually be an exponential curve), we can integrate the energy lost in the resistor: \$E_{losses}=\int{P_{losses}(t)}.dt=\frac{1}{3}R.I_{T_{on}}^3\$

• And we reach a rough estimation of the efficiency of our converter for the first phase of the conversion:

  \$\rho=\frac{E_{inductor}}{E_{inductor}+E_{losses}}=\frac{1}{1+\frac{2}{3}\frac{R}{L}}\$

• Where in the equation above you can substitute R for the total series path resistance of your circuit: physical resistors, intrinsic resistance of the switching elements (MOSFET?) and equivalent series resistance of your inductor.
• The above is only a first order approximation (simplified) for the case of a single flyback inductor boost converter. The real, more accurate calculation will add the second phase/stage of the converter and all the devices in your circuit.

For a more in-depth introduction to the topic, you can check this article by Texas Instruments:

http://www.ti.com/lit/an/slva061/slva061.pdf

Answer 3

9 Watt power means P=V*I but normally voltage will be constant ,but how much current we can drown from the ckt will be decided by the manufacture of the component. Ex: 4.5v*2A=9W ,5v*1.8A=9W in this case power is same but current and voltage will varied,hence in power calculation current is measure roll, we are not take directly 1V*9A=9W ,if in datasheet manufacture is mentioned about current means not a problem..u can boost the voltage.