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Understanding Energy recovery circuit

garimella

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I was reading a book on power electronics by Daniel Hart and the topic was on energy recovery. There are two circuits under analysis. Circuit 1 is a typical configuration shown below. When the transistor is off, the power is entirely dissipated in resistor. The author claims that Circuit 2 is more efficient because when the transistors are off, the entire current is returned back to source. My question is ,
1. why should we add resistor in circuit-1. Will it not return the current back to source, when the transistor is off, just through diode?
2. In circuit 2 , why two diodes are required, will it not work with one diode?

circuit1.png
Circuit1

circuit2.png
Circuit2: Courtesey : Daniel.W Hart
 

wwfeldman

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in circuit 1, when the transistor is on, the current is = il , when the transistor is off, is = 0 and il circulates through the inductor, diode and resistor
the purpose of the disode is to provide a current path for il, the energy gets dissipated by the resistor.

in circuit 2, when both transistors are on, it looks like circuit 1 with the transistor on.
when both transistors are off, il goes through the diode to Vcc. essentially is reverses direction. is still = il
the second diode clamps the voltage at the "top" end of the inductor to within 1 diode drop of ground, forcing
the bottom end to be positive, as the voltage across the inductor is now dropping.

recall that v = L di/dt so that is di/dt is increasing, v is positive - when the transistors are on, top of inductor is +, di/dt increases
and when di/dt is decreasing, polarity of v has to flip, so bottom is (relatively) positive.
clamping "top" of inductor near ground gives the voltage across the inductor a reference point

need the second diode to provide the reference
 

_Eduardo_

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1. why should we add resistor in circuit-1. Will it not return the current back to source, when the transistor is off, just through diode?
Without the resistor, when the transistor turns off, the current through the coil is I (t) = e^(-t*rcoil/L) (ideal diode)
As rcoil/L is generally small, the current fading time can be too long.
 

garimella

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Without the second diode, still the current can flow through the top diode like circuit 1. I did not quite get the idea
 

wwfeldman

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current requires a closed path
the second diode provides the closed path during the recovery part

current from inductor follows diode to +Vcc (Is is now into Vcc instead of out)
the current then follows through ground through the second diode back to the inductor

when all the energy that was in the magnetic field of the inductor has been returned
to the source, Vcc, or has been dissipated as heat, the current stops

as a side comment, Vcc and ground are the + and - ends of a battery or power supply.
it doesn't look like it in the schematic, but there must be some sort of power supply
 

garimella

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Got it but why there is no resistor in the second circuit.
 

crutschow

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Without the second diode, still the current can flow through the top diode like circuit 1.
True.
But in the first circuit, there's no current flowing back to the voltage source so there's no energy recovery.
The current flows through the diode an back to the top of the inductor.
If you removed the connection to the source when the transistor turned off, it would still work the same.

For energy recovery you need the current to flow from ground through the inductor back to the source, which the second circuit does.

There is not reason for a resistor in the second circuit, since that would just dissipate some of the energy instead of returning it to the source.
The only purpose of the resistor in the first circuit is to speed up the dissipation of the inductive energy so the current stops sooner.
This can be important if the inductor is a solenoid or relay coil, for example.
 

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