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 Fusion916 14th November 2012 09:31 PM

Such as this schematic:

http://www.sentex.ca/~mec1995/circ/ampboost.gif

I understand that at low currents the BJT is off and the current is supplied by the voltage regular, then as the current increases a I*R vbe develops on Q1, until the point of Vbe(on) then Q1 starts to conduct.

My question is, how is the feedback maintained to provide constant output voltage? I don't quite understand the feedback path. Vbe should adjust with the load current, but is the current path through the load straight from Vdc to vout to Q1 or is R1 somehow in the path?

 Fusion916 14th November 2012 09:32 PM

Also, how does the performance change when R2 is removed? I seen this circuit both with and without R2.

 lineup 14th November 2012 10:26 PM

R2 is a base stopper for the big transistor.
It is there for to make the input to the transistor suitable.
It might work without R2, but we can say R2 is there for safety.

If you look into schematics of power amplifiers, you can see these Base stopper resistors.
They are in small value. 2.2, 4.7 or 10 Ohm.

 martin clark 14th November 2012 10:52 PM

Quote:
 how is the feedback maintained to provide constant output voltage? I don't quite understand the feedback path
It's very simple: look at the current paths.

if output voltage drops due to loading, the regulator tries to maintain the voltage output and that requires it supplies more current into the output. That current must be reflected in the input current to your '7812' input; which increases the voltage across R1 and therefore, Q1 is driven harder.

In effect the expected voltage drop across R1 due to variations in load current is divided by the hfe of the pass transistor.

Q1 remains under control of the 3-pin reg unless cut-off when current through R1 drops low enough that V across r1 <0.6v.

 jerluwoo 14th November 2012 11:43 PM

1 Attachment(s)
I'm sorry to inform you but that schematic and its intended use are incorrect. The circuit in post 1 is a short circuit protection scheme and should not be used as a "current boosted regulator" , for one it will inject all the noise and ripple from the input straight into the output, making the whole point of using a regulator moot, unless you simply want a voltage and dont care if its clean.
Attached is the correct way. Note c1 and c2 are solid tantalum types and should be included with these types of regulators.

 martin clark 15th November 2012 12:06 AM

Not true at all. Q1 remains under the control of the reg; worst case, the base is driven by the input filtered by R1/C1. The output is quiet!

PS conventional 3-pin regs don't actually need an output cap for stabilty, although it helps output load transient response. LDO regs however usually do require the output cap for stability since it forms part of the reg's compensation.

An input cap is always necessaary though regardless of type.

 jerluwoo 15th November 2012 02:19 AM

It is R1/C1 which turns Q1 into a common base amplifier, with ripple fed into the emitter and none at the base giving some gain to the ripple.

 gmphadte 15th November 2012 03:50 AM

One thing is sure... It will have poor load regulation.

 wahab 15th November 2012 04:11 AM

Load sensitivity of the schematic post 1 will be the one of the regulator
since the transistor is enclosed in a feedback loop while the second schematic
resistors being outside of the loop.

 sofaspud 15th November 2012 06:53 AM

For me, the first pic is the configuration usually associated with the 3-terminal regs. The 2nd pic I usually associate with the uA723 reg. The configurations seem to be somewhat interchangeable.
The feedback network within the 78xx remains functional. If Vout is reduced, that also reduces current through R1.
That's my fingers-crossed take on it.:)

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