A sample text widget

Etiam pulvinar consectetur dolor sed malesuada. Ut convallis euismod dolor nec pretium. Nunc ut tristique massa.

Nam sodales mi vitae dolor ullamcorper et vulputate enim accumsan. Morbi orci magna, tincidunt vitae molestie nec, molestie at mi. Nulla nulla lorem, suscipit in posuere in, interdum non magna.

Voltage Regulation and Power Loss

Real PS & Load

This is one of those posts that is this hard of me to categorize and actually even to give a title because there is so much to say about so little.  However, it is very important for several reasons. First, we will be talking about energy conservation. Second, it is heading directly toward some concepts necessary to understand “the real op-amps”. Third, it will explain the need for the 10 ohm resistor on the computer power supply I suggested for use as a “lab supply”, in a previous post.

In the diagram, we have a real power supply that consists of a 10 V ideal source and a 1Ω resistor. A question often asked is:  “What value load resistor will receive the most power from this power supply?”

Calculations for different size load resistors.

In the spread sheet, I did the calculations for various sizes of load resistor.  This is standard Ohm’s law and wattage calculations.  As you can see in the table, the most power is transferred when the load resistor is equal to the power supply internal resistance.  At resistances above that value, the current decreases and reduces power, At resistances below that value the voltage across the load resistor decreases and reduces the power available to the load.   However, one very important point.  At the point of maximum power transfer, 50% of the power is wasted in the internal resistance in the power-supply. That wasted power will generate heat that must be removed from the power source.  Often this is done using a fan causing more power to be wasted.

The obvious answer is to reduce the internal resistance in the power-supply, but that also has a cost.  That means larger diameter wires and larger components in the power supply and that costs money and materials as well as space.  Audio equipment specifies the impedance the speakers, usually 8Ω for home units and 4Ω for automotive sound systems.  The reason is to get maximum power transfer to the speakers.  Obviously, audio signals are AC signals, but the same concept applies.  As we start working with the op-amps we will have to concern ourselves on both sides of this. The op-amp often will be the load on a source signal and it will be the power source to the next device.

In our simple circuit the voltage across the load varied by a great amount depending upon the amount of current it wanted to draw from the load.  This is similar measuring water pressure at the end of a long section of pipe.  If no water is flowing the pressure will be high, but once the toilet is flushed while you are in the shower the pressure drops.  (Hopefully in that case both the hot water and cold water pressure drops by the same amount.)

Linear Regulating Power Supply

Often this is done by installing a variable resistor in series with the power supply to act as a regulator.  The regulator resistor is actually an electronic circuit and is automatically adjusted to maintain a constant voltage on the load. To use this circuit means the actual power source must produce more voltage than is required by the load.  It also means even more power is wasted than before, because the regulator is designed to produce a voltage drop and therefore more wasted watts.    This circuit works similar to adjusting the pressure on the end of a hose by adjusting a valve up-stream.  In that case also, power is lost across the valve and heat is generated.

A conceptual diagram of a switching regulator.

A more modern way of regulating the voltage is the switching regulating power supply.  Again, the voltage of the source is higher than the voltage needed by the load. This power supply has a storage unit built in it to hold power while a switch rapidly switches on and off. The storage unit is a capacitor and a device we will learn in the not too distant future.  The switch is constantly toggled at the same frequency, but the on-time vs. the off-time is changed depending upon how much current is being drawn by the load.  This on-time vs. off-time ratio is called the duty cycle.

What makes this more superior than the older “linear regulator” show above?  When the switch is on (closed) and conducting there is very very little voltage drop across it and we will say the ideal amount of zero.   Zero volts X any amps = zero power loss.  When the switch is off (open) no current can flow and again there is no power loss.  The power supply can be designed to provide high amperage and good regulation while consuming wasting very little power.

There is one small problem with this and one I had to cope with while using the old computer power supply for a lab supply.   The switch shown in the conceptual diagram is actually an electronic device.  This device has what is called “leakage current” through it.  This can be pictured as a very high ohm value resistor in parallel with the switch.  If no current is being drawn out of the storage unit the storage unit will quickly develop the same voltage as the power source.  This is why it was necessary to install a fixed load on the power supply and I used the 10Ω resistor across the 5V portion of the power supply.

My version of the power supply is the low cost, no frills and almost all junk parts version.  A nicer looking one is shown on this web site: Converting a ATX power-supply to a lab supply.  Should you decide to build the nice looking one, be careful about making sure to follow the instructions about leaving the power supply off a long time before opening it up.  Also, make sure you do not get any metal chips inside the power supply when you drill the holes.

The summing amplifier circuit.

Answer to the question about the summing amplifier.   The equation is solved exactly the same way as the inverting amplifier equation. Kirchhoff’s current law is used to sum the currents and the current through each resistor is determined by assuming the inverting input (-) is zero volts, the same as the non-inverting input (+) because the amplifier is “happy”.  Doing the same substitutions as in the inverting amplifier, except there are more of them this time. we can get the output equation of:

Vout = -Rf ( V1/R1 + V2/R2 + V3/R3).

This was a very long and very detailed post.  Probably somewhere in it there is an oversight or something said backwards and I have missed it.  If I have confused you please feel free to e-mail me and I will fix the problem.

The next post on actual op-amps will probably involve a video so we can be looking at the same data at the same time.  It will also have an actual op-amp data sheet available for your download.

Thank you for your time.    Gary

If you find this and other posts educational and enjoyable please consider subscribing using one of the methods shown on the home page.  Please share the site with others that may enjoy this.


Print Friendly

Leave a Reply

You can use these HTML tags

<a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>