How do most voltage regulator ICs work? Are they the same as hooking up a variable resistor and a voltmeter and turning the knob until you get the desired voltage?
Electronic – How does an LM7805 voltage regulator work
resistorsvoltage-regulator
Related Solutions
I had this answer ready to send.
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So -
In a shunt regulator Iload flows fron Vin to Vout via a series resistor Rseries.
A "shunt element" connects from Vout to ground.
The shunt element dissipates power when the load does not need it.
Above circuit from here - good shunt regulator explanation.
Answer is straight forward
To get more power out reduce Rseries as required - see below
Shunt element MUST be able to dissipate required maximum power out (for when there is NO load)
Series dropper resistor MUST be able to handle maximum load current
Power in series resistor = Imax^2 x Rseries
ALL shunt regulator circuits have a designed voltage rating.
If you exceed the maximum design level the voltage output will drop below design level.
Shunt regulators are in some ways easier to design for max possible power level than most other types.
When there is NO load the shunt element is taking ALL the power output hat is possible.
When you add a load some of this power is stakn by the load instead.
When all the power is going to the load and none to the "shunt" the output voltage drops.
As long as the voltage is OK at no load then you know the maximum possible power out - it is te power being taken by the shunt element.
In practice a very small amount of power will be required at "dropout" for the shunt but this is typically 1% or less of the total power available.
So:
Given Vin, Vout, Power wanted.
Imax = Power_wanted/Vout
Rin <= Vdrop/Imax
or Rin <= (Vin_min-Vout) / Imax
or Rin <= *Vin_min - Vout) x Vout / Power_wanted
Select required Rin.
If Vout = correct then shunt regulator WILL provide desired power - as long as shunt element does not burn up.
For the original question that this is based on:
You MUST tell us what the regulator is (IC name at least).
You MUST tell us what the load is.
A circuit is a very very very good idea.
The heat produced by a voltage regulator isn't a function of the voltage, it's a function of the power.
You can calculate the power the voltage reg will dissipate by taking the voltage-drop across the regulator, times the current flowing through the regulator.
E.G: $$Power = Volts * Amps$$
In other words, if you have a 30V source, and your device is running off 3V, you have \$30V - 3V = 27V\$ across the regulator. However, if your device is only drawing ~3 mA: $$27V * 0.003A = 0.081W$$
You would only have a dissipation of 81 milli-watts, which wouldn't even get too warm to the touch.
However, if you have a 5V input, with a 3V output (giving 2V across the regulator), yet your deice is drawing 1A, you have: $$2V * 1A = 2W$$
You have a power dissipation of 2 watts!
Basically, there is more involved in evaluating heat production then just the voltages
Best Answer
Voltage regulators achieve "stiffness" via a feedback control loop, where "stiffness" means that a large change in load current causes a small change in voltage.
Both switching and linear regulators include a control loop (historically analog... some of the newer switchers use digital control loops) to adjust some parameter of the circuit so that the output voltage remains constant in the presence of load current changes and input voltage changes.
In a linear regulator the circuit parameter is the pass transistor drive circuit (which produces base current for an NPN/PNP power transistor, gate voltage for a MOSFET).
In a switching regulator the circuit parameter is the duty cycle of the switching element(s).
So there's really two areas you need to understand if you want to get into the details of how regulators work: