What is the use of implementing a transistor on path of the Vcc?

Discussion in 'General Electronics Chat' started by ChethuGowda, Oct 4, 2016.

  1. ChethuGowda

    Thread Starter New Member

    Jul 22, 2016
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    I came across the circuit with an NPN transistor connected in series to the path of Vcc.
    What is the exact use of having this transistor on Vcc, while it is always on?

    More info: This 3.3v is the supply to a VCO circuit used for RF activity.
     
  2. Ramussons

    Active Member

    May 3, 2013
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    It's a Capacitor "booster", where the setup "presents" a equivalant capacitance that's typically hFe times C71. Used in low current circuits for power supply decoupling.
    In this case, it's as good as placing a 1000 or 5000 uF capacitor across the load at the emitter.
     
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  3. hp1729

    Well-Known Member

    Nov 23, 2015
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    I just see a few milliseconds of delay when power comers up.
     
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  4. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    Yes, but with much better high frequency performance.
     
  5. ChethuGowda

    Thread Starter New Member

    Jul 22, 2016
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    I could not find any mathematical explanation for effective capacitance of hFe x C71.
    Can you please provide some link for this?
    Because We have another h/w which draws power over USB port, but due to poor VBus voltage regulation the VBus drops down sometimes leading to the USB device disconnection. I would like to experiment this transistor + capacitor combination for that too.
     
  6. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    This circuit will not maintain the output voltage if the input drops below the required output. It will not help you with this problem.
     
  7. BR-549

    Well-Known Member

    Sep 22, 2013
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  8. ChethuGowda

    Thread Starter New Member

    Jul 22, 2016
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    Dear BR-549, Thanks for this link, but I was actually looking for some explanation with respect to the circuit, like how exactly it multiplies the capacitance to pose bigger capacitor value.
    One thing mentioned in the link was, it won't improve the regulation.
    I wanted to know if this part is only to filter/sustain the high frequency noise over the power supply, or it also compensates the momentary voltage dip.
     
  9. BR-549

    Well-Known Member

    Sep 22, 2013
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    "One thing mentioned in the link was, it won't improve the regulation."

    It does not say that. It says that it can not regulate, it does not say that it does not improve the regulation.

    Electronics takes a lot of study. A capacitor can do different functions. Sometimes it can transfer or relay a signal. And sometimes it can prevent a change in the signal. (absorb signal) This means instead of transferring the signal, it tries to smooth out or flatten the signal. It depends on how you use them, and what you are trying to do in a circuit.
    In your case.........the circuit is trying to reduce any change in voltage, whether the voltage is regulated or not. That in itself.....is a form of regulation.

    Go down a little more than half on this link and read some more about the theory of this circuit.

    http://www.startfetch.com/keantoken/content/Kmultiplier.php
     
  10. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    The circuit does not present a multiplied capacitance value to whatever is connected to the emitter. But the voltage at the emitter acts as if a much larger capacitor were there without the transistor. Tow things:

    1. The R-C corner frequency is 160 Hz, so the voltage at the emitter is a low-pass filtered version of whatever gorp is on the 3.3 V rail, reduced by 0.7 V. For example, 1.6 kHz noise is reduced by 20 dB, 16 kHz noise is reduced by 40 dB, etc. So there is more than a simple turn-on delay going on.

    2. If there is a pcb trace going to a decoupling capacitor and an active circuit, variations in the circuit current discharge the capacitor, and it is recharged through the pcb trace resistance and inductance. To properly decouple something like a current-feedback opamp or an FPGA can take a lot of capacitance due to large load current spikes. But with a transistor emitter follower between the decoupling cap and the circuit, there is only 1 mA of base (and capacitor) current for every 100 mA of emitter current (-ish). Thus, a small capacitor at the base appears - to the dynamic load conditions of the downstream circuit - as a 99 times larger capacitor at the emitter.

    ak
     
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