Function of Capacitor in Flyback converter circuit?

Thread Starter

abdulwahab.hajar

Joined Jun 14, 2016
93

Hello everyone

I want to make a flyback converter, it will have to output 3.3V DC. Where the input is initially an power line 230V AC signal.

I have found this circuit which is essentially a flyback converter, controlled by a small controller which seems reasonable, this circuit first takes as an input a DC signal therefore, the incoming AC signal would be rectified to match that of a steady DC signal. Then passed on to the aforementioned circuit which is attached below.

I understand some aspects of the working of the circuit, however I would like to grasp it fully:
When the current passes through the transistor,a positive voltage is on the dot on N1 indicating the polarity of the coil on the left, an opposite polarity occurs across the second coil in the right, however this would lead to a current which can't pass through the diode since it would be reverse biased.
Therefore, ultimately current stops passing in the right section of the circuit.
and somehow the coil on the left stores energy??? How is that when it's a DC input wouldn't it act as a short circuit???

Then when the switch is open (Current no longer passes through the transistor), the energy in the left coil is somehow transferred to the coil on the right ( I guess by the back induced emf in the left coil, which will produce a current in the right coil???), and this current produced can pass through the diode thus charging the capacitor and supplying an output voltage...

When the switch is closed again (current passes through the transistor) the charged energy stored in the capacitor simply keeps driving the output.. Is this correct? and the cycles continue

1) One thing I don't understand is; what is the function of the capacitor Cin? I need to understand its function to determine its value...
2) I know that Cout is used to store the energy from the transformer, to later be transferred to the output, however quantitavely speaking how large a capacitor would I need?

Thank you all I know it's a big read. I appreciate the help!
 

ericgibbs

Joined Jan 29, 2010
8,753
I want to make a flyback converter, it will have to output 3.3V DC. Where the input is initially an power line 230V AC signal
hi a,
On the circuit you have posted, the circuit expects a DC voltage input.
The Cin cap acts a current source when the transistor conducts and is drawing current from the DC supply.
The cap helps to maintain the voltage level close to the circuit and also reduces the pull down of the supply voltage.

E
 

Thread Starter

abdulwahab.hajar

Joined Jun 14, 2016
93
hi a,
On the circuit you have posted, the circuit expects a DC voltage input.
The Cin cap acts a current source when the transistor conducts and is drawing current from the DC supply.
The cap helps to maintain the voltage level close to the circuit and also reduces the pull down of the supply voltage.

E
But why would it act as a current source, when there is already the DC supply which is a current source, isn't that right?
 

ericgibbs

Joined Jan 29, 2010
8,753
hi,
If the DC supply has a very low output impedance and was able to supply the peak current thru wires of very low resistance then we probably could avoid using a cap.
But consider if the DC supply could only supply say 100mA and the Boost circuit transistor tried to draw 200mA when switched On hard, the DC voltage will drop below its regulated level.
Also if the DC supply was say 2 or 3 mtrs away from your circuit, every time the transistor switched On, the current could cause a voltage drop due to the resistance of the connecting cable.

A Cap close to the transistor would supply that peak current and the DC supply would be less effected, I would use 100uF and 100nF in parallel.

E
 

ebp

Joined Feb 8, 2018
2,332
It is very important to realize that the thing that looks like a transformer in a flyback convert is not a transformer in the conventional sense but rather an inductor with two windings. One winding (usually called the primary, just for convenience though that is more accurately applied to a true transformer) is used to store energy in the inductor when the switch is ON, and the other winding (usually called the secondary, again just for convenience) when the switch is off.

Assuming everything is starting from a state with no current flowing anywhere, no charge on the output capacitor and no energy stored in the inductor:

When the switch turns on, the current in the inductor primary begins to rise linearly according to the expression
δi/δt = V/L
and energy is stored according to the expression
E = (L·i^2)/2 (energy in Joules)

As you note, the phasing of the windings and the diode at the output prevent current from flowing in the secondary when it is flowing in the primary.

When the switch turns off, the energy stored in the inductor has to go somewhere. Remember that the energy is stored as a magnetic field. Now the phasing and the output diode allow current to flow in the "secondary." The current drops linearly with time, just as it rose linearly with time when the switch was on. If the turns ratio were 1:1, the initial current in the secondary would be exactly equal to the current in the primary at the instant the switch turned off. Some of the energy will go to the load, but some of the energy will be stored in the capacitor.

After sufficient cycles, sufficient energy will be stored in the capacitor to bring the voltage up to the regulation point. As soon as this has happened, the control circuity will reduce the duty cycle so that the energy delivered to the secondary is exactly the amount required by the load. When the switch is ON, the capacitor supplies all of the energy to the load, so its voltage drops a little. Normally the capacitance is high enough so the voltage change in a single full cycle as the cap is first charged then discharged is very small.

Many flyback converters are designed so that all of the energy stored by the primary winding is delivered to the capacitor and load each time the switch turns off. This is called "discontinuous current" operation. Some designs use higher inductance so not all of the energy is transferred. This is called "continuous current" operation, even though the current in the secondary winding still goes to zero each time the switch turns on. Each mode has its merits and disadvantages.
 

Thread Starter

abdulwahab.hajar

Joined Jun 14, 2016
93
It is very important to realize that the thing that looks like a transformer in a flyback convert is not a transformer in the conventional sense but rather an inductor with two windings. One winding (usually called the primary, just for convenience though that is more accurately applied to a true transformer) is used to store energy in the inductor when the switch is ON, and the other winding (usually called the secondary, again just for convenience) when the switch is off.

Assuming everything is starting from a state with no current flowing anywhere, no charge on the output capacitor and no energy stored in the inductor:

When the switch turns on, the current in the inductor primary begins to rise linearly according to the expression
δi/δt = V/L
and energy is stored according to the expression
E = (L·i^2)/2 (energy in Joules)

As you note, the phasing of the windings and the diode at the output prevent current from flowing in the secondary when it is flowing in the primary.

When the switch turns off, the energy stored in the inductor has to go somewhere. Remember that the energy is stored as a magnetic field. Now the phasing and the output diode allow current to flow in the "secondary." The current drops linearly with time, just as it rose linearly with time when the switch was on. If the turns ratio were 1:1, the initial current in the secondary would be exactly equal to the current in the primary at the instant the switch turned off. Some of the energy will go to the load, but some of the energy will be stored in the capacitor.

After sufficient cycles, sufficient energy will be stored in the capacitor to bring the voltage up to the regulation point. As soon as this has happened, the control circuity will reduce the duty cycle so that the energy delivered to the secondary is exactly the amount required by the load. When the switch is ON, the capacitor supplies all of the energy to the load, so its voltage drops a little. Normally the capacitance is high enough so the voltage change in a single full cycle as the cap is first charged then discharged is very small.

Many flyback converters are designed so that all of the energy stored by the primary winding is delivered to the capacitor and load each time the switch turns off. This is called "discontinuous current" operation. Some designs use higher inductance so not all of the energy is transferred. This is called "continuous current" operation, even though the current in the secondary winding still goes to zero each time the switch turns on. Each mode has its merits and disadvantages.
Much appreciated sir!
 

Thread Starter

abdulwahab.hajar

Joined Jun 14, 2016
93
T
hi,
If the DC supply has a very low output impedance and was able to supply the peak current thru wires of very low resistance then we probably could avoid using a cap.
But consider if the DC supply could only supply say 100mA and the Boost circuit transistor tried to draw 200mA when switched On hard, the DC voltage will drop below its regulated level.
Also if the DC supply was say 2 or 3 mtrs away from your circuit, every time the transistor switched On, the current could cause a voltage drop due to the resistance of the connecting cable.

A Cap close to the transistor would supply that peak current and the DC supply would be less effected, I would use 100uF and 100nF in parallel.

E
Thank you!
 

ebp

Joined Feb 8, 2018
2,332
I should have noted:
In buck and boost topologies and their derivatives, at some point in the cycle current is flowing from the input to the output "directly" through the inductor. In a flyback converter all energy from the input source is first stored during part of the cycle then delivered to the output in another part of the cycle. At no time is there a mechanism for current to flow from input to output. The version of the flyback converter that uses only a single winding on an inductor is often called a "buck-boost" converter because the output voltage can be higher or lower than the input. It is neither a buck nor a boost because of that temporal separation of when and how current flows. The single-winding version is properly called either a flyback converter or an inverting converter.

Contrary to what you will find somewhere on AAC (tutorials or something like that - I don't remember), the term "flyback" comes from the use of a dual-function circuit in magnetically-deflected cathode ray tubes, as in television receivers. Energy was stored in inductance as the electron beam was deflected "slowly" across the screen, then transferred to the high voltage power supply for the CRT during the "retrace" time when the beam was deflected back to the "starting" side of the tube. Horizontal retrace was commonly called "flyback" - the beam flew back at high speed to the other side of the screen. Magnetically deflected CRTs are becoming are rarity (as are electrostatically deflected, as for oscilloscopes).
 
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