The Power Supply
THE CAPACITOR-FED POWER SUPPLY
This type of power supply is VERY DANGEROUS and is not allowed in many countries.
I have NEVER built one or used one as I value my life.
I am including it because it is a CONSTANT CURRENT POWER SUPPLY and there are many different types of CONSTANT CURRENT power supplies for illuminating components such as strings of LEDs and these power supplies are perfectly safe.
This discussion covers how to work with a CONSTANT CURRENT POWER SUPPLY or CONSTANT CURRENT CIRCUIT.
But first we have to discuss the MAINS
Fig 1. The "Mains" WAVEFORM
|The "MAINS" is dangerous because the voltage
is much higher than stated on any appliance.
It is actually 180v or 345v for the 240v mains.
The ACTIVE lead rises 340v then falls 340v below "ground" 50 times per second and this will produce a CURRENT FLOW through your body that will kill you very quickly.
Obviously you don't want to touch the mains and that's why everything connected to the MAINS is covered with plastic.
Plastic is an insulator and that's why you can handle hair-dryers and electric drills etc.
Mains transformers are also safe because they have a layer of plastic between the primary and secondary winding.
This layer of plastic is called INSULATION.
The primary winding is connected to the mains and the secondary winding is connected to your project.
The insulation between the two windings means you can NEVER touch the "MAINS."
Ask yourself this simple question: "Would you poke a knife or fork into a toaster?" No. Because you may touch the heating element and you may get a shock.
But you will pick up a hair-dryer because the case is made of plastic.
The plastic is an insulator and you trust plastic to isolate the "MAINS."
The same with a mains transformer. The plastic provides insulation (called ISOLATION).
The energy from the mains is converted to magnetic flux and is passed to the metal core of the transformer. This flux passes through the turns of the secondary winding to produce energy and that is how the energy is passed from the mains to your project. It is through magnetism.
We are not going into great detail about the transformer.
But we are going to cover some of the basic facts.
In the diagram above, the winding connected to the
"mains" is called the PRIMARY and the low-voltage winging is called the
A BASIC POWER SUPPLY
The circuit shows a Power Supply using a transformer.
The insulation is between the primary and secondary where the two lines have been placed. The two lines indicate a magnetic circuit exists between the two windings.
Fig 2. The TRANSFORMER POWER SUPPLY
We can remove the transformer and replace it with
You can see one of the mains wires is connected directly to the Power Supply.
This is ok when the Neutral line is connected to 0v of the supply.
But what happens if the power supply is connected via an extension-lead and the wires are reversed!!
The active will be connected to the 0v and this is just like poking a nail into the active hole of the power-outlet on the wall.
If you are touching the frame of a metal toaster at the same time, you will get a shock.
Fig 3. The Capacitor in our discussion
The capacitor in the diagram above can be
replaced with a door and two bar magnets.
When the mains moves the magnet up, the magnet on the other side of the door moves up too.
The two magnets do not touch each other but are influenced by magnetism.
The same with a capacitor.
When the left plate of the capacitor is charged by the mains, the left plate is filled by the "electricity" from your power supply.
Fig 5. The Magnet
Place a sheet of aluminium foil on each side of
the door and connect one side to the active of the mains.
The foil is exactly like a capacitor.
Now connect the 0v of the power supply to your leg and touch the aluminium foil.
You will get a SHOCK.
The capacitor works like this.
When the active line of the mains is rising, it is putting a charge on the plate on the left side of the door.
This effect is influencing charges on the other plate by a term called ELECTROSTATIC ATTRACTION.
At the moment the charges on the plate you are touching consist of "pairs" of positive and negative charges with an equal number of each. This results in a neutral or zero volt condition.
But when the charge on the left plate increases, it pushes the opposite charges off the plate you are touching and these charges go through your body. These charges are called ELECTRICITY.
As the active line rises and falls 50 or 60 times per second, the charges pass though your body at the same rate and even though the number of charges is very small, you will feel a "tingle".
A capacitor contains two larger sheets of foil and the spacing is between the sheets is much smaller.
That's why a capacitor will give you a bigger "tingle."
Fig 6. Stand at door
A 100n capacitor will give you a 7mA
A 220n capacitor will KILL YOU !!
That's right. A current as low as 16mA will prove FATAL.
The "trip detectors" in your house will detect leakage as low as 15mA and "cut off" the power in 1 or 2 cycles.
These detectors are called RCD - Residual Current Device (detector) or GFI (Ground Fault Interrupter) save your life because they operate very quickly.
Without a detector you will be either thrown off the wires or "STUCK TO THEM LIKE GLUE."
Fig 7. The 100n capacitor
calculations are made with multiples of 100n capacitors.
This makes the calculations easy.
A 100n capacitor will pass or "produce" 7mA when connected to a bridge or 3.5mA if only a single diode is used (assuming 240v AC supply - all values are halved for 110v).
Now we come to the first basic
fact of a capacitor-fed power supply.
WHAT IS THE OUTPUT VOLTAGE ?
The output voltage of ALL capacitor-fed power supplies is 180v or 340v - depending on the supply voltage.
The output voltage of a capacitor-fed power supply only drops to 12v or 35v when a load is connected.
When the load is removed, the supply rises to 180v or 340v. That's another reason why they are so DANGEROUS.
The simplest CAPACITOR-FED power supply requires
a diode and a red LED.
These two items are called the LOAD.
The capacitor passes current in ONE DIRECTION when the mains is rising and then it passes current in the OPPOSITE DIRECTION when the mains is falling.
This is also described as the CAPACITOR CHARGING when the mains is rising and the CAPACITOR DISCHARGING when the main is falling.
Figure 1 above shows the waveform of the mains and you can see it rises 340v above ground and 340v below ground.
In fig 8, the two LOADS are on the OUTPUT of the cap-fed power supply.
When the mains is rising, the output of the power supply rises and when it is 1.7v, the red LED turns ON and the voltage does not rise any more.
This means the voltage-difference between the top of the red LED and the mains will be stored in the capacitor and the capacitor will be charged to 338v.
When the voltage of the mains falls, the output of the power supply will be NEGATIVE and when it is 0.7v NEGATIVE, the diode prevents the voltage falling. The capacitor is discharged and starts to charge in the opposite direction to 340v. It really has -340v across its terminals.
The red LED is the LOAD in one direction and the diode is the LOAD in the other direction.
The basic half-wave capacitor-fed power supply is shown in the diagram.
Each 100n of capacitance will deliver 7mA RMS (10mA peak on each half-cycle).
In the half-wave supply, the capacitor delivers 3.5mA RMS because the current is lost in the lower diode when it discharges the capacitor.
Don't use the peak value in any calculations because this only applies for a very short portion of the cycle (or half-cycle).
HALF-WAVE CAP-FED POWER SUPPLY with ZENER
Here's a clever use of a zener diode.
A zener diode effectively BREAKS DOWN in BOTH DIRECTIONS.
The 12v zener breaks down when the voltage on the "top" of the diode (the cathode) is 12v and it "breaks down" at 0.7v when the voltage is in the opposite direction.
The load will see a maximum of 12v and the zener will discharge the capacitor, ready for the next "positive cycle."
So, how does it work?
When a load is not connected, all the current from the capacitor will pass through the zener.
If the capacitor is 470n, this current will be 16mA.
When a load is connected, SOME of the current will be TAKEN from the zener and will flow through the LOAD.
If the resistance of the load is reduced, MORE current will flow though the LOAD and when it reaches 16mA, ALL the current from the capacitor will flow though the LOAD and NOTHING will flow though the zener.
If you reduce the LOAD . . . . this will happen:
The voltage across the LOAD will REDUCE to 11v, 10v, 9v, . . . BUT the current will remain at 16mA. Eventually the voltage will reduce to 1v @16mA
WHAT IS THE RATING OF THE ZENER DIODE?
The "rating" of the zener diode is the milliwatt (called the WATTAGE) dissipation. This is the amount of heat it will "get rid of" without getting too hot. The heat is mainly passed through the leads to the PC board.
Most small zener diodes are 400milliwatt (mW) and the next size is 1 watt.
To work out the dissipation of the 12v zener above, the value of the capacitor is used to work out the current and the 12v is used for the voltage part of the equation.
The equation is V x I where V is volts and I is amps.
The circuit above is working in the milliamp range, where 1,000mA = 1 amp.
For each 100n, the circuit will deliver 3.5mA
Suppose the capacitor is 470n = 16mA
The wattage dissipation for the zener will be 12x16 = 200mW.
A 400mW zener will not get too hot.
There is a small amount of heat generated when the zener is discharging the capacitor, so we allow 210mW.
HALF-WAVE CAP-FED POWER SUPPLY with ZENER circuit-2
If you want to add an electrolytic across the LOAD, you will need to add a diode to prevent the electrolytic discharging on the second-half of the cycle.
A diode connected to a capacitor
. . .
This circuit DOES NOT WORK. The capacitor will charge during the first cycle and produce an output.
But when the MAINS is "falling" the capacitor will not charge in the opposite direction because the diode will act like an OPEN CIRCUIT and not put any load on the capacitor.
The capacitor must be charged in one direction and then discharged in the opposite direction.
In other words it must deliver a current to the "POWER SUPPLY" (also called the LOAD) in BOTH DIRECTIONS.
In fig 9. the capacitor will remain charged from the first cycle and both sides of the capacitor will rise and fall at 50 or 60 times per second.
The capacitor must be discharged during the second part of the cycle so it can be charged again.
It is ONLY the charging and discharging process that causes current to flow.
The discharging process can also be called CHARGING IN THE OPPOSITE DIRECTION.
The capacitor must have a LOAD in BOTH DIRECTIONS.
Fig 9. A diode connected to a capacitor WILL NOT WORK
Although this circuit will work, the red
LED will illuminate in one direction as the
mains voltage rises but when the main voltage
FALLS, the REVERSE CURRENT through the LED will
destroy the crystal.
The red LED is
providing a LOAD in both directions but the
REVERSE direction will destroy the crystal. You
need a diode to provide the LOAD in the reverse
Fig 10. A LED connected to a capacitor WILL NOT WORK
The other type of
CAPACITOR-FED POWER SUPPLY uses a
bridge. A bridge is a set of 4 diodes. You need to refer to
other sections of this course to see how the 4 diodes "steer" to
the incoming AC waveform to produce a series of "bumps" that are
all above the 0v line.
This is called PULSATING DC or "DC with RIPPLE."
Fig 11. Cap-fed POWER SUPPLY with BRIDGE
The current produced by a capacitor-fed power supply is quite small and we use the simple rule:
100n delivers 3.5mA for a single diode, and
7mA for a BRIDGE supply.
The single diode supply actually delivers 7mA for half-a-cycle to the LED and 0mA for the other half-cycle. In the second half-cycle the capacitor is delivering 7mA to the diode and this energy is wasted or "lost." That's how we get an average of 3.5mA.
The capacitor can be increased to 220n or 470n for a higher current.
THE CAPACITOR - X2
The type of capacitor used in a cap-fed power supply must be a SPECIAL TYPE.
It must be rated at 400v AC and must be constructed with materials and insulation that will NOT BLOW UP.
These special types of capacitors have the identification X2 (X2 does not mean "times two" or "use two capacitors" !! It is a special identification that the capacitor is rated for MAINS OPERATION).
Any capacitor will work, but some will short-circuit or blow-up for no apparent reason.
Because the capacitor is charging and discharging 100 or 120 times per second, there is a certain amount of stress on the foil and insulation and that's why it must be strongly constructed.
Although there is theoretically no energy lost in the capacitor, it will heat up a small amount due to losses.
The charging and discharging is classified as RIPPLE CURRENT and this current ALWAYS causes a small amount of heating. In our case, this ripple current is the MAXIMUM for the capacitor. Normally ripple current is only a fraction of the maximum as the capacitor (such as an electrolytic will be charged to a voltage and experience a slight increase and decrease in voltage and this will cause a small current to flow. But in our case the voltage is changing FULLY and also in the reverse direction and creating maximum current-flow.
The capacitor-fed power supply is a CURRENT arrangement.
The output current depends on the value of the capacitor.
The output voltage will be 340v if NO LOAD is present.
NOW WE START
The capacitor-fed power supply is a CONSTANT CURRENT supply and they way you explain how the circuit works is entirely different to a normal power supply.
BUT FIRST WE ADD SAFETY RESISTORS
Here is a simple cap-fed power supply driving a red LED.
When the circuit is turned ON, we do not know if the mains voltage is zero, a small positive value or full 340v.
If it is 340v and the capacitor is not charged, a VERY HIGH current will flow to charge the capacitor and this will damage the LED.
To limit (reduce) this current we add a 470R resistor in series with the active line (It can be in either line).
Fig 12. Adding a SAFETY RESISTOR
Now here's the amazing part:
We can add more LEDs to the circuit and they will ALL glow:
Fig 13. Cap-fed POWER SUPPLY with 6 LEDs
|You cannot add hundreds of LEDs because as you add another LED, the voltage across the combination increases by 1.7v and when the total becomes 340v, NONE of the LEDs will illuminate. That's because there is zero volt difference between the mains voltage and the LED voltage.|
|A single-diode design is not a good design because the LEDs are only illuminated for each half-cycle. LEDs turn ON and OFF very quickly and they will appear to "flicker." A better circuit uses a BRIDGE.|
The bridge will deliver 2 pulses of energy during
each cycle and this will result in 100 or 120 "flickers" each
More LEDs can be added and they will ALL illuminate.
Fig 14. A BRIDGE Cap-fed POWER SUPPLY
To eliminate the "flicker," the output needs an electrolytic. This will
store the energy during a peak and deliver it when the mains voltage is
low. See the waveform on the circuit. The voltage
remains high enough to keep the LED constantly illuminated.
Fig 15. A BRIDGE Cap-fed POWER SUPPLY with electro
You need at least 50 LEDs in each string to prevent them being damaged via a surge through the 1k resistor - if the circuit is turned on at the peak of the waveform. As you add more LEDs to each string, the current will drop a very small amount until eventually, when you have 90 LEDs in each string, the current will be zero.
For 50 LEDs in each string, the total characteristic voltage will be 180v so that the peak voltage will be 330v - 180v = 150v. Each LED will see less than 7mA peak during the half-cycle they are illuminated. The 1k resistor will drop 7v - since the RMS current is 7mA (7mA x 1,000 ohms = 7v). No rectifier diodes are needed. The LEDs are the "rectifiers." Very clever. You must have LEDs in both directions to charge and discharge the capacitor. The resistor is provided to take a heavy surge current through one of the strings of LEDs if the circuit is switched on when the mains is at a peak.
This clever design uses 4 diodes in a bridge to produce a fixed voltage power supply capable of supplying 35mA.
All diodes (every type of diode) are zener diodes. They all break down at a particular voltage. The fact is, a power diode breaks down at 100v or 400v and its zener characteristic is not useful.
But if we put 2 zener diodes in a bridge with two ordinary power diodes, the bridge will break-down at the voltage of the zener. This is what we have done. If we use 18v zeners, the output will be 17v4.
When the incoming voltage is positive at the top, the left zener provides 18v limit (and the other zener produces a drop of 0.6v) This allows the right zener to pass current just like a normal diode. The output is 17v4. The same with the other half-cycle.
The current is limited by the value of the X2 capacitors and this is 7mA for each 100n when in full-wave (as per this circuit). We have 1u capacitance. Theoretically the circuit will supply 70mA but we found it will only deliver 35mA before the output drops. The capacitors should comply with X1 or X2 class. The 10R is a safety-fuse resistor.
The problem with this power supply is the "live" nature of the negative rail. When the power supply is connected as shown, the negative rail is 0.7v above neutral. If the mains is reversed, the negative rail is 340v (peak) above neutral and this will kill you as the current will flow through the diode and be lethal. You need to touch the negative rail (or the positive rail) and any earthed device such as a toaster to get killed. The only solution is the project being powered must be totally enclosed in a box with no outputs.
BEST CAP-FED SUPPLY
This circuit is the best cap-fed power supply. It uses 4 diodes to produce the best current from the 220n capacitor and an electrolytic to smooth out any flickering.
Here's a circuit using two 16v zeners.
If a16v zener is placed AFTER the bridge, you only need ONE 16v zener !!!!
The 0.47u will deliver 33mA.
The 16v zeners will limit the voltage to 16.7v
The output of the bridge will be 16.7v -0.7v -0.7v = 15.3v
The voltage drop across R2 will be 3.3v
This will provide 12v for the 12v zener.
38 LED LAMP
Here is a 38 LED lamp using a capacitor-fed power supply to illuminate 38 white LEDs.
The total voltage across the LEDs is 38 x 3.6 = 138v. This means the incoming voltage is 340v - 138 = 202v. The 334 capacitor will deliver about 20mA.
The fastest way to learn electronics is to work with a faulty circuit.
The following circuits have been presented on the web from people who do NOT understand electronics.
A FAULTY CIRCUIT:
It has 2
A FAULTY CIRCUIT:
A FAULTY CIRCUIT:
Here's another faulty circuit.
How can you work out what is happening ?
Consider the top terminal is connected to the active and the lower terminal is connected to neutral.
The active rises to 345v and falls to minus 345v while the neutral wire does not move.
When the active rises, it charges to the 220n and delivers a current through the top 1N4004 diode.
You can consider the neutral wire to be at the same potential as the earth connection on the power supply and the second zener diode and lower 1N4004 diode are not needed.
If you like to think the active and neutral "spread apart" as the active rises, you will be able to see the neutral wire drops slightly until the 12v zener conducts and drops 0.7v across its leads.
The active now drops and the first 12v starts to conduct when the voltage is -0.7v. This allows the 220n to discharge and then charge in the opposite direction to -345v. When the active line starts to rise from its voltage of -345v, the capacitor starts to discharge and then start to charge in the positive direction.
As you can see, the second zener and lower diode can be removed from the circuit. They have no effect on the operation of the circuit.
A FAULTY CIRCUIT:
Here's another circuit that does not work.
Each 100n will deliver 3.5mA in half-wave for 250v input. This becomes 1.75mA for 110v or 3.85mA for the circuit above.
But the 10k resistor will drop a voltage when 3.85mA is flowing.
The voltage-drop will be 0.00385 x 10,000 = 38v.
This means the 220n does not see 110v, but 72v and the current will be less than 3.85mA.
Jose Pino says the circuit will deliver 100mA !!!!!
It will deliver 3mA or less !!!!!
A FAULTY CIRCUIT:
What is the first transistor doing ??? It is simply putting a 270 ohm load across the supply. It is assisting the 1k by providing a base current for the second transistor and current to keep the zener in conduction, but simply connecting the 270R to the base of the second transistor will do the same thing, or decreasing the 1k will provide additional current.
This circuit WILL NOT WORK.
Where is the diode to discharge the 2u2 capacitor ????
A FAULTY CIRCUIT:
The three 1u capacitors will pass 3 x 70mA = 210mA.
The voltage drop across R2 will be 0.21 x 100 = 21v. The wattage will be V x I = 21 x 0.21 = 4.4watts
The voltage drop across R4 will be 0.21 x 10 = 2.1v.
The characteristic voltage-drop across a 1 watt LED is 3.6v
The voltage needed by the 1 watt LED and R4 is 2.1 + 3.6 = 5.7v
The 4.7v zener will not allow the LED to illuminate.
The circuit is just another mess by Professor D.Mohankumar. It is untested and DOES NOT WORK !!
A FAULTY CIRCUIT:
The bridge is around the wrong way !!!
A FAULTY CIRCUIT:
The capacitors will pass about 180mA.
The output voltage will be 340v. The 180v electrolytics will BLOW UP !
The 100R resistor will dissipate 3.2 watts.
The impedance of 2.2u and 0.47u is about 1,300 ohms. The 470R across the capacitors will pass another 60mA, making a total for the supply 240mA. The 470R will dissipate about 1.7 watts.
Very few people know anything about designing a capacitor-fed power supply.
You can see all these mistakes on the web.
The circuits above have not been tested and most of them will not work or BLOW UP !
CAPACITOR-FED POWER SUPPLIES are very dangerous.
All the components are "live." This means they are all at a potential (voltage) that can be as high as 340v.Touching any of them is the same as touching the active wire of the mains.
Almost every circuit will pass more than 15mA and this amount of current will throw you across the room or give you a nasty shock if you touch a component with one hand and touch something that is earthed with another part of your body.
When designing and testing this type of power supply I always use an isolation transformer.
You can make your own isolation transformer by
connecting two identical transformers "back-to-back."