Regulator input circuit

Thread Starter

jimjohn

Joined Jan 28, 2008
8
Hi,
Here I am attaching a input portion of a regulator circuit.I don't know the actual use of this circuit.I think this circuit is to deliver more power and will keep the input of the regulator at a constant level.Is it right?
Can anyone please help me ?
 

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studiot

Joined Nov 9, 2007
4,998
This type of circuit is called a pre-regulator. It's purpose is to condition the raw input voltage to the main regulator probably to reduce an unecessarily high input voltage and perhaps to improve overall regulation.

The pass transistor FZT692B is an emitter follower, the output being set by the zener. the other two diodes are protection against reverse connection from faults or perhaps inductive connections.

The second transistor BC847B is used to limit the output current of the pass transistor. It only operates when the current through the two 1 ohm resistors (the output current) is enough to make the voltage across the base emitter junction enough to switch on the transistor. When this transistor operates it diverts current away from the pass transistor base, thus reducing the output.
 

hgmjr

Joined Jan 28, 2005
9,027
At around 300 to 350 milliamps, the second transistor is programmed by the 2 one ohm resistors to foldback. It's there to protect the regulator from delivering power to the load during an overload condition.

hgmjr
 

Ron H

Joined Apr 14, 2005
7,063
At around 300 to 350 milliamps, the second transistor is programmed by the 2 one ohm resistors to foldback. It's there to protect the regulator from delivering power to the load during an overload condition.

hgmjr
I guess I'm a nitpicker, but I don't see any foldback there. Foldback means that the output current will be less with a short to ground than it is at peak load (less than current limiting threshold). One way to do this is to add a resistor from the base of the current sense transistor to +V.
Without foldback, a short to GND causes very high dissipation in the pass transistor.
 

Thread Starter

jimjohn

Joined Jan 28, 2008
8
Thank u for the explanation.

But i don't know what is the use of 1uF capacitor across base and emitter of FZT692B

and why the ckt designers are using resistor or capacitor combination for a non standard value (eg:instead of 2R it is two 1 R resistors;Why they are not using a near standard values 2.2R or 1.5R)
 

rwmoekoe

Joined Mar 1, 2007
172
But i don't know what is the use of 1uF capacitor across base and emitter of FZT692B
i guess the regulator cicuit that this cicuit is feeding is a switching power supply that pulls intermittent relatively high current. the 1uF cap is to prevent short period current limiting episodes from occuring, but still provides long period over current prevention. thus, the 1uF cap is to 'average' the current sensing.

and why the ckt designers are using resistor or capacitor combination for a non standard value (eg:instead of 2R it is two 1 R resistors;Why they are not using a near standard values 2.2R or 1.5R)
the 3 - 220uF caps at the output is probably for accomodating high intermittent current draw from the client switching power supply. while the single 68nF deals with high frequency that the large caps misses (due to parasitic inductance in large caps), and tuned to the working frequency of the switching power supply.
 

hgmjr

Joined Jan 28, 2005
9,027
I guess I'm a nitpicker, but I don't see any foldback there. Foldback means that the output current will be less with a short to ground than it is at peak load (less than current limiting threshold). One way to do this is to add a resistor from the base of the current sense transistor to GND.
Without foldback, a short to GND causes very high dissipation in the pass transistor.
Upon closer examination, I see your point Ronh. Indeed the term "foldback" does not apply to this supply since the supply continues to deliver the 300 to 350 milliamps into the short-circuited output while the short-circuited output condition is present.

For every 3 volts applied to the power supply's input, there will be an estimated 1W of power dissipated by the pass-transistor. I would hope that the pass-transistor has been supplied with sufficient heat-sinking to permit it to withstand the heat to which it will be subjected in the event of a sustained output short-circuit condition.

hgmjr
 

rwmoekoe

Joined Mar 1, 2007
172
Hello sir,
It provides a continuous supply.
i understand. what i mean by being intermittent is the natural way a switching power supply is working.
it is being intermittent that's why it is called the 'switching' mode, being either on or of at a time, never half on, two third on or anything like that.
imagine the client switching power supply is delivering half of it's input voltage at 500mA to it's load.
at the input of this switcher (output of our pre-regulator), the current draw would be about 500mA peak too, only it goes half the time if the duty cycle is 50%.
now without the 1uF cap, the pre-regulator would limit the output current to about 300mA when the switcher is drawing 500mA, while in average actually it is only about 250mA. with the 1uF cap in place, the current sensing is averaged over a period of time as per the period of the switcher.
 

Ron H

Joined Apr 14, 2005
7,063
I'm wondering why someone put a linear regulator in front of a part that can handle 40V on the input, and has current-limiting and thermal protection.
 
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