Hi, I dont understand why is the 3 diodes (In the red box) is connected this way. Can anyone explain to me hows this 3 diodes work? Also I have a issues on connecting the transformer. On the circuit diagram, the GND was connected to the middle of the coil but how should i connect it to a real transformer?

 

They are forward conducting and provide a fixed +ve voltage reference due to the constant volt drop across each.
The transformer is correct, it is full wave centre tapped common.
Max.

 

Hi,

i bought a transformer

http://sg.rs-online.com/web/p/pcb-t...5522677633D4E4F4E45267573743D3530342D36383426

How should i connect the GND? If I used 12V and 0V for the 2 diode, then i got no where to connect the GND. (only using 1 output for now).

Also is there any site i can read to understand more about the 3 diodes? Kinda lost. Knowing that voltage will drop by 2.1V. So if my input to the circuit is 12V, I will get a output of around 10V?

Lastly, for the smoothing Capacitor, I know that to calculate the value of the capacitor is given by V ripple = Iload / 2 f C. For the Vripper can i assume it as Vavg? So i can calculate using Vavg = Vpeak * 0.637.

Thank You

 

As you have a dual secondary transformer, you can hook it up in various ways, if you want 12vac out then the two secondaries can be connected in parallel but have to be phased correctly, also you would need a full wave bridge instead of the to diodes shown previously.
Smoothing capacitor is dependant on required %ripple and maximum load.
Look up full wave DC supply.
Also the 12v sec will give you ~16vdc output after smoothing.
The voltage across the diodes will be approx 2.1v at all times.
Max.

 

the set up is a "long tailed pair" or differential amp,

The diodes are for setting the constant voltage across the 270R resistor, to make a constant current of approx 5.2mA flowing in the 2N3904 C/E, to turn on the BD436,

the voltage at the other 2N3904 transistor has to get to 2.1V before it turns on , which is set by the preset resistor, when this happens the first 2N3904 turn off thus turning off the output supply, then the cycle repeats, and so you have your regulated output.

To connect the transformer windings in series with a center tap, follow this diagram http://www.hammondmfg.com/par_serD.gif

the connection at terminals 6,7 is your ground terminal, 5 and 8 go to the diodes.

 

Lastly, for the smoothing Capacitor, I know that to calculate the value of the capacitor is given by V ripple = Iload / 2 f C. For the Vripple can i assume it as Vavg? So i can calculate using Vavg = Vpeak * 0.637.

1.414 times the RMS AC voltage of the transformer (minus the voltage loss of a diode or two, depending on which configuration you use) will be your peak filter capacitor voltage under no load. The formula I use for ripple is 1.414 C Er F = I
That will give you the loss of voltage between peaks of the power line.
Vavg has no usefulness in this kind of work.
What matters is the lowest voltage between pulses of power and how much voltage is used up in the regulator. If you have enough voltage to start with, the output will be smooth.

 

Why not just use a LM317 and be done with it? It will be much more stable, and provide better regulation than the posted circuit. Transformer, rectifier, filter remain the same.

 

Why not just use a LM317

Lots of people look up these regulator circuits because they are not aware of the improvements available. The best use for the old circuits is education. The best way to make a regulator is to buy the modern chip.

 

I would like to note that in some parts of the world, fancy-pants integrated circuits are not available but discreet components are obtainable.

 

Yep. I have one of those running right now.

http://forum.allaboutcircuits.com/threads/help-timing-light-circuit-improvement.105447/

 

Here is why I would use a LM317 costing a few cents. The only thing that the posted circuit does better is that it has a lower drop-out voltage.

 

Thank for the reply,

I was thinking, if i set my Vripple to be 0.8mV. With a output current of 100mA. After using the formula, the cap i need is at least 1250uF?

Also for the long tailed pair, if there a different voltage between the 2x 2N3904, the different will be the gain to amplifies my output? So i will get something like 2.1V (constant from the 3 diodes) * (the voltage different between the 2 transistor).

 

Why would you want the first filter capacitor to limit the ripple voltage to 0.0008 volts?
Because you don't know that the regulator circuit removes most of the ripple voltage?

 

...Because you don't know that the regulator circuit removes most of the ripple voltage?

All in all, it is not a very good regulator...

 

All in all, it is not a very good regulator...

Oh. That explains it. For 1/10th of an amp at a 120 Hz rectifying frequency and 0.0008 Vripple (p-p) you will need 736,570 uf.