Polarity Inverters

What are polarity inverters and what are used for.

Picture showing images of polarity interter schematics and the title of the post.

Sometimes we design and build a circuit that needs a dual power supply. But, in certain cases, we really need just a positive voltage to power a circuit and the negative is only used for some special polarization that doesn’t really need the same amount of power used for the positive. Consider, for example, a circuit with a depletion channel MOSFET that requires a negative voltage just for the polarization of its gate.

Example of Depletion Mosfet charateristics, where we can see that the gate voltage can be negative.

In such cases, it is economically better to use a different approach than having a full fledged dual power supply. This approach is called “polarity inversion”, resulting in a device that is able to convert the positive voltage of a power supply into a low current negative voltage.

A polarity inverter is, therefore, a circuit that is capable of taking a positive voltage with respect to the ground and generate a negative voltage also with respect to the ground, so that we can have both a positive and a negative voltage available at the same time to power another circuit, without using a dual power supply.

In principle, the inverter is based on the following circuit.

Schematic describing the working principle of the polarity inverter.

There are two capacitors and two diodes, and a switch that connects the positive of the first capacitor alternatively to the positive voltage source and to the ground.

When the switch is set toward the positive voltage, capacitor C1 starts charging through the first diode, which closes the circuit toward the ground. Given enough time, the voltage at the capacitor increases up to the input voltage minus the voltage drop on the diode.

For example, if the input voltage is 9V, the capacitor will charge to about 8.4V.

This is represented in the following diagram by the first pulse on Vin and the corresponding voltage on C1.

Wavw diagrams showing voltages in the various parts of the previous circuit.

Now, once the capacitor is charged, we move the switch toward ground. Doing so, we open the circuit that connects capacitor C1 to the input voltage and, instead, we connect the same end of the capacitor toward ground.

This way, the voltage at the capacitor C1 is now providing a forward polarization to the second diode, the one on the right, and therefore we have a closed circuit that goes from capacitor C1, to capacitor C2 and through the second diode.

If we choose the two capacitors with the same capacitance, half of the charges on capacitor C1 will transfer to capacitor C2 and, as a result, both capacitors C1 and C2 will end up with half of the original charge and, therefore, with half of the original voltage that was on C1.

This is represented by the second part of the above diagram, where now the input voltage is zero, but capacitors C1 and C2 are at half the original voltage.

On the next cycle, we move the switch back toward the power supply, so capacitor C1 is again charged to the input voltage. In this case, however, the second diode is inversely polarized, so capacitor C2 is isolated and cannot either charge nor discharge, thus it keeps the previous value of voltage.

Moving the switch back to the ground, C1 gives now some more charge to C2 and, therefore, its voltage drops a bit while C2 voltage, instead, increases more.

And you can now see that if I keep switching back and forth, adding more cycles to the diagram, both C1 and C2 keep retaining more and more charges, and their voltage keep increasing so that, after a number of cycles, C2 has reached about the same voltage as the input.

Now, note how capacitor C2 is connected to the ground on its positive side, and the other end is offering its negative voltage to the the output of the circuit that is thus negative with respect to the ground.

If you look at the last of the four diagrams, in fact, you can see how the output voltage becomes more and more negative with respect to the ground, with a tendency to reach the 8.4 V we mentioned before.

So, if we keep moving the switch back and forth quickly, after reach that state we can sustain it, even if we remove a little amount of charge from C2 at each cycle, due to a load that we could put across its leads.

This circuit is called a charge pump, because is able to pump charges into the second capacitor, even if it is not directly connected to the input voltage.

Note that if we start applying a strong load to the output, C2 won’t be able to recharge fast enough and its voltage will start dropping. And that is why we cannot use this polarity inverter for loads comparable to those that we can put directly on the original power supply.

But, how do we move a switch fast enough to obtain this functionality?

The trick is to replace the mechanical switch with a a solid state one, and control it with a square wave oscillator, the so called astable multivibrator.

One way to do that is to use a 555 timer, like in the following schematic.

Practical realization of a polarity inverter using a 555 timer.

The circuit on the right half side is exactly the same as the one in the previous schematic. However, on the left half side, the mechanical switch has been replaced with a 555 timer setup as an astable multivibrator, with a duty cycle close to 0.5.

Pin 3 of the 555, which is the output pin, will move alternatively from the voltage of the power supply to the ground, thus working as if it was the switch of the previous schematic.

The oscillation frequency is provided by R1, R2 and C4, which I calculated in this example to provide a frequency of about 30 kHz with a duty cycle very close to 0.5.

If you would like to know more about the 555 timer, I suggest you to watch the video I made about one year ago where I describe what it is and how it works. Here is the link to the video.

In order to be able to support relatively higher currents with the polarity inverter, we need to be able to recharge the capacitors at a faster pace, which translates in a higher current. One way do so is by using the output of the 555 to pilot a couple of transistors with a high value of beta, the coefficient that express the amplification in current of the transistors. With a higher available current, the capacitors will charge faster and, therefore, it will be possible to handle a higher load current.

Here is an example circuit that can provide higher currents:

Practical schematic of the previous polarity inverter adapted to provide a higher current at its output.

This circuit is basically identical to the previous one but, instead of applying the output voltage of the 555 directly to the charge pump, made of C1, C2, D1, and D2, the 555 controls the two transistors 8050 and 8550, respectively an NPN and a PNP.

With these transistors, we can still connect the positive lead of C2 to the positive of the power supply and to the ground alternatively, and we can force the charges in and out of the two capacitors to move at a faster pace.

The two resistors R3 and R4 are necessary to limit the amount of current through the base of the transistors. Too much current in there would have two unwanted side effects:

First, the transistors could burn because of too much current.

Second, even if the transistors did not burn, they would still go deep into saturation, which would make them spend more time moving between the on and off states and causing the circuit not to work as expected.

In addition to that, since the voltage at the output of the 555 does not change instantaneously between 0 and Vin, there would be a period, during the transition, where both transistors would be on at the same time. As a result, the input voltage would be short circuited for a little while during each cycle, which is a condition definitively to avoid.

To fix the problem, I added those two Zener diodes to the circuit. The Zener diodes create a gap between 4.7V and 5.1V that will prevent the transistors from being both on at the same time, thus fixing the short circuit problem.

Here is how it works.

During the transition from 0 to 9V on pin 3 of the 555, transistor 8550 will be on in the interval between 0 and 4.7V.

During the interval between 4.7V and 5.1V both transistors will be off and, finally, during the transition between 5.1V and 9V, transistor 8050 will be on.

Viceversa, during the transition from 9V to 0, the opposite sequence will happen: first, transistor 8050 will be on, then both transistors will be off, then transistor 8550 will be on, alone.

And that is why the two zener diodes make sure that the two transistors will never be on at the same time, thus protecting them and the power supply.

The final effect will still be the same: the positive lead of C2 will be alternatively connected to the positive and to the ground, making the charge pump to work, and creating the negative output.

To conclude, polarity inverters have their usefulness in certain situations, but are not good enough to replace a full fledged dual power supply.

So, when do we use one or the other?

We will use the polarity inverter in those cases where only a little load is required on that specific pole, whereas the majority of the load would depend on the single power supply.

Whenever we need considerable and comparable amount of power on both the positive and the negative poles, we will need to use a dual power supply.

And finally, if you would like to see the polarity inverter in action, you may want to watch this video, which I posted back in December 2020.

Theremin v.2 Power Supply Design

theremin-v2-power-supply

For the new version of the Theremin, I have chosen to use a dual 12V power supply. This will have more flexibility because it will allow me to use more sophisticated units, possibly using op-amps.

The circuit is very basic: it uses a dual 14V transformer (not shown in the schematic) capable of providing 1.5A at its output.

A dual transformer is made up as in the following picture.

center-tapped-transformer

Is has a primary winding that is connected to the AC power supply outlet, and a secondary winding with a center tapped wire that is usually put to ground on the low voltage circuit side.

Voltage between either end wire of the secondary and the center tapped wire is usually the same (with the exception of specifically made transformers), which we call V.

The voltage measured between the two end wires of the winding is instead two times V or 2V.

Sometimes, instead of having a single secondary winding, we have two, carrying the exact same voltage. In this case, we can connect together the two closest wires and consider that as the center tapped wire. Then everything works as the first kind of transformer.

transformer-trans64

The AC current of the transformer is converted in to a DC current through the usage of a bridge rectifier and the capacitors C1 and C2.

The bridge rectifier converts the sine wave coming from the transformer into a fully rectified wave.

Full-wave_rectified_sine

Then, the capacitor that follows (in this case C1 and C2) starts charging over the ascending sides of the wave and discharging, partially, over the descending sides of the wave, basically filling the wave in between crests and making it look like more a straight horizontal line with some disturbance in it that we call ripple (the red line in the following picture).

ripple

In general, depending on the use of the power supply, we define a maximum value of the ripple that the circuit can handle.

In our case, we need to make sure that the voltage at the input of the regulators never goes below 14.5V, according to the data sheet, otherwise the regulator will not function properly.

The peak voltage provided by the transformer is its RMS value multiplied by the square root of 2, or:

peak_voltage

The minimum voltage we can have at the input of the regulator is:

regfulator_input_voltage

This is the max value of ripple that we can sustain.

To calculate the capacitor necessary to obtain this ripple, we use the following formula:

capacitance calculation

where f is the frequency of the alternate current which, in the USA, is 60Hz, and Ix is the maximum current that the power supply needs to provide.

So, we would need a capacitance value, for C1 and C2, of 2358uF.

However, the Theremin circuit will really not draw 1.5A from the power supply, so we can stay a little conservative, and use the closest value below the calculated one, which is 2200uF.

At this point we can safely say that the voltage on the output of the regulators will be exactly 12V (positive or negative, depending on the output side).

To further help the regulator, and preventing the current through it to go too close to the 1.5A threshold, where the regulator would not work anymore because the ripple becomes too high, we add to the output of each regulator another electrolytic capacitor, this one with a value at least equal to the capacitance value that we did not put at the input side. Since at the input side we put a capacitance of 2200uF instead of 2358uF, we will need a capacitor of at least 158uF.

However, to stay totally safe, I decided to use a capacitor at least 5 times higher, so I used the value of 1000uF for C3 and C4.

And finally, I added an extra capacitor (C5 and C6) to shunt toward ground any RF frequency that would travel back from the Theremin oscillators toward the power supply. A 0.1uF value is what is suggested by the data sheet of the regulator, so I used just that.

Why did I use this capacitor if there was already a 1000uF in there?

The reason hides in the way the electrolytic capacitors behave. In short, the electrolytic capacitors do not work well at high frequencies, so we need to add the extra 0.1uF capacitor, which is not an electrolytic one, to work in that range of frequencies. And since the range of frequencies is much higher than the one of the 110Vac outlet, a very small capacitance is enough to do the job.

 

DC Electronic Load V.3

Back in August 2018, I presented a DC electronic load on my YouTube channel (V.2). For that, I used an old 2N3055 transistor in a Darlington configuration with 2 more transistors to be able to get enough gain to use it.

100W_el_load_v2

Although declared useful for 100W, I was never able to make it work at those powers due to the limited dissipation capabilities of the power transistor and the heat sink. The max power dissipation I could have from that device was about 20W.

Today, at the anniversary of that presentation, I have created a new version of of the DC electronic load. This new version is based on a MOSFET that can work alone as a load, adjusting the current only through an appropriate voltage on its gate, avoiding the need of having a Darlington circuit with multiple transistors.

The schematic of this new version of the DC electronic load is based on a single MOSFET capable of driving the necessary current, up to 5A and a voltage divider connected to the battery, providing the appropriate voltage to the gate of the MOSFET.

electronic_load-v.3

In order to make it work correctly, the trim-pot RV1 needs to be tuned to obtain a voltage of 1.5V on pin 3 of the potentiometer that regulates the amount of current flowing through the MOSFET, which provides a better use of the multi-turn potentiometer that regulates the actual value of the current.

A combination of digital voltmeter and am-meter, like in the previous version of the DC load, takes care of providing information about the power supply under test.

The device is powered through a 9V battery and it is connected in such a way that the voltage is measured through the yellow wire of the digital voltmeter, wile the current is measured putting the am-meter in series with the MOSFET, with the thick red wire on the source, and the thick black wire toward the negative connector, through a 5A fuse that is used, mostly, to protect the am-meter itself against currents too high of those it can handle.

I made a new case for this new version of the DC load. The main difference is the location of the heat-sink, which is now located on the back panel rather than the top of the device. The new heat-sink is also attached to the back panel through 4 separators, which allow for a better air flow and cooling of the unit when it is used for long period of times.

Here is an OpenSCAD view of the box and the corresponding code to create it.

v3_box_view

$fa=0.5;
$fs=0.5;

//main section
rotate([180, 0, 0]) translate([0, 10, -2])
{
// front panel
difference()
{
cube([150, 80, 2]);
translate([27.5, 40, -1]) cube([45.8, 27.7, 4]);
translate([52, 20, -1]) cylinder(d=6.2, h=4);
translate([120, 60 , -1]) cylinder(d=9, h=4);
translate([108, 20 , -1]) cylinder(d=9, h=4);
translate([132, 20 , -1]) cylinder(d=9, h=4);
translate([3,3,0.5]) linear_extrude(height=2) text(“eleneasy.com – DC load – 25W max.”, size = 6);
translate([36,18,0.5]) linear_extrude(height=2) text(“off”, size=5);
translate([61,18,0.5]) linear_extrude(height=2) text(“on”, size=5);
translate([109,44,0.5]) linear_extrude(height=2) text(“current”, size=5);
}
translate([102.25, 15, 0]) cube([2, 10, 2]);
translate([126.25, 15, 0]) cube([2, 10, 2]);
translate([111.75, 15, 0]) cube([2, 10, 2]);
translate([135.75, 15, 0]) cube([2, 10, 2]);
translate([12, 20, -36]) cube([27, 2, 35]);
translate([39, 4, -36]) cube([2, 18, 35]);

// left panel
translate([0, 0, -60]) cube([2, 80, 60]);

// right panel
translate([148, 0, -60]) cube([2, 80, 60]);

// bottom panel
translate([0, 0, -60]) cube([150, 2, 60]);

// top panel
translate([0, 78, -60])
{
cube([150, 2, 60]);
}

// screws supports
translate([2, 2, -58]) difference()
{
cube([10, 10, 58]);
translate([5, 5, -1]) cylinder(d=2, h=16);
}
translate([138, 2, -58]) difference()
{
cube([10, 10, 58]);
translate([5, 5, -1]) cylinder(d=2, h=16);
}
translate([2, 68, -58]) difference()
{
cube([10, 10, 58]);
translate([5, 5, -1]) cylinder(d=2, h=16);
}
translate([138, 68, -58]) difference()
{
cube([10, 10, 58]);
translate([5, 5, -1]) cylinder(d=2, h=16);
}
}

// back cover
translate([0, 10, 0]) difference()
{
cube([146, 76, 2]);
translate([5, 5, -1]) cylinder(d=4, h=4);
translate([5, 71, -1]) cylinder(d=4, h=4);
translate([141, 5, -1]) cylinder(d=4, h=4);
translate([141, 71, -1]) cylinder(d=4, h=4);
translate([73, 38, -1]) cylinder(d=40, h=4);
translate([126, 60, -1]) cylinder(d=12.5, h=4);
translate([130.5,51,0.5]) rotate([0, 0, 180]) linear_extrude(height=2) text(“5A”, size=5);
translate([(146-55)/2, (76-50)/2, -1]) cylinder(d=4, h=4);
translate([146-(146-55)/2, (76-50)/2, -1]) cylinder(d=4, h=4);
translate([146-(146-55)/2, 76-(76-50)/2, -1]) cylinder(d=4, h=4);
translate([(146-55)/2, 76-(76-50)/2, -1]) cylinder(d=4, h=4);
}

Assembling the circuit is pretty straightforward, and it is done partially in the air and partially  on a perforated board.

We just need to make sure we provide the cables with the right thickness for the current we need to support.

In my case, I used stranded cables with an 18 gauge. These cables are necessary between the thick am-meter cables, the MOSFET source and drain, and the external connectors.

Every other connection can be made with 22 gauge cables.

And finally, the heat-sink should have a resistance of 0.82 Centigrade degrees per watt or less, to prevent the MOSFET from becoming too hot. Note that this will not save the MOSFET in case you draw a current too high. The product between the current and the voltage as provided by the measurements display must never exceed 25W, and the current should never exceed 5A, or the MOSFET will burn.

The tuning is done by measuring the voltage between the terminal 3 of the potentiometer and the ground, with the circuit on, but not connected to any external power supply. The trim-pot has to be adjusted such that the measured voltage equals 1.5V, which is just below the minimum voltage necessary to make the MOSFET conduct current. This way, when turning on the apparatus with the potentiometer all the way to the counter-clockwise position, there will be no current. Then, moving the potentiometer in the clockwise direction, current will start flowing.

Testing of the unit is done attaching it to a power supply that provides different test voltages while we adjust the current with the multi-turn potentiometer on the DC load unit. Just make sure not to exceed 25W of power at any given time. Doing so could damage the MOSFET itself.

I plan to use this DC load in all my future projects that require a power supply of 25W or less, to test the power supply itself. Besides checking that the power supply works fine, you could also check that the ripple of the output voltage does not exceeds your requirements. That can be done connecting the power supply output to an oscilloscope while the DC load draws the current.

And finally, here are a couple of picture of the finished device.

20190816_112540.jpg

20190816_112614

Happy experiments!