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Avnet Boards General Regulating input voltage using a LDO (shunt regulator)
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  • input voltage
  • voltage follower
  • shunt regulator
Related

Regulating input voltage using a LDO (shunt regulator)

reconchris
reconchris over 2 years ago

I am trying to create a control scheme that allows voltage regulator to regulate its input voltage instead of its output voltage. I'm trying to make a shunting regulator. My current approach is

  1. Use input voltage and link it to the adjust pin so the regulator can control it input voltage.
  2. Take input voltage and first invert it so when Vin increases a falling voltage is seen on ADJ pin and the regulator passes more current thus bringing Vin down (and vise versa)
  3. Compensate for a negative inverted Vin signal coming from the op-amp by adding a voltage to it giving ADJ a positive and correctly correlated response to Vin.

So far this isn't working. Some help on how to make a functional switch mode shunt regulator would be appreciated. I am trying with an LDO to get the control scheme working.

image

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  • michaelkellett
    0 michaelkellett over 2 years ago

    Can you step back a little and explain what you are trying to achieve.

    What will be the power source, what output voltage and current do you need ?

    When you say your present circuit doesn't work - what does it do ?

    MK

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  • michaelkellett
    0 michaelkellett over 2 years ago in reply to michaelkellett

    A quick thought is that C2 would be better across the input to gound rather than the output. If you have any capacitance to ground on the output the open loop gain of the system will increase with frequency which usually results in instability.

    MK

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  • dougw
    0 dougw over 2 years ago

    How precise does this regulation need to be? How much current is worst case? What voltage is required? Would a zener diode work?

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  • reconchris
    0 reconchris over 2 years ago in reply to dougw

    I want it to work like a Zener but have the efficiency of a sync buck regulator. I have seen papers written about solar panels being shunted this way. However this is supposed to shunt a permanent magnet generator(Like a motorcycle alternator after it has been rectified to DC.) 

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  • michaelkellett
    0 michaelkellett over 2 years ago in reply to reconchris

    What is the power source - and can you give us an idea of the load current and voltage.

    The way this is done is to use a buck regulator to drive the load but attempt to control both the load and the source at the same time - non trivial if both source and load are non linear.

    MK

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  • dougw
    0 dougw over 2 years ago in reply to reconchris

    I guess the power burned off in the shunt regulation process is always going to be the same whether it is a zener or a more complex regulator - unless the current will be stored somewhere like a battery or a super capacitor, or the power is used to do something useful by powering some other load. It is possible to reduce the power in the zener itself by using it with a pass transistor.

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  • reconchris
    0 reconchris over 2 years ago in reply to michaelkellett

    The load will vary in resistance and the source will vary in impedance. Again a Zener has no problem doing this. I am trying to make a sync buck regulator that functionally act as a efficient Zener diode. By connecting the output of the buck to GND and having the ADJ pin regulate the input voltage. Forcing the buck to dump a proportional amount of current to GND to maintain a specific Vin.

    I understand a varying input voltage is regulated to drive a load at a constant voltage. The problem is this will fly out of regulating if your load is say an LED (20mA) and your input is a 800W rectified stator output being turned at max RPM. The stator will slam current into the front end of your regulator and without a path to GND will create an overvoltage condition to any regulator you recommend. 

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  • reconchris
    0 reconchris over 2 years ago in reply to dougw

    But it wont be just as efficient as a Zener. A Zener will have its Vzner*I = W as a loss in power. In the case of a buck regulator it (tanking switching loss out) it only absorbs power when the FET is conducting  so (Rds(on)*I^2) =W or the power loss. This is where bucks get their efficiency. The energy is stored in the inductor and prevents the Vin node from switching from 12V (say) and 0V when the FET hot loop is connected to GND. 

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  • michaelkellett
    0 michaelkellett over 2 years ago in reply to reconchris

    I've asked you twice already, this is the last time.

    Unless you give us a reasonably full explanation of the source and load it is not possible to offer any sensible advice.

    You also need to explain what it is you are trying to do - ie an overall description of the system.

    For example - my miling machine has a switch mode motor driver, sometimes it drives power from the mains supply to the motor, sometime it puts power back into the big capacitors in the controller, and sometimes, when it stops the spindle rapidly, it dumps power into a huge multi kW spiral resistor with a coolling fan at the top of the machine.

    The controller for this application is utterly different from that which you might use to manage a mW power level energy harvesting system.

    MK

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