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Rev A.

Symbol used is a Schottky Power Diode, which are Avalanche type clamps but like Zeners use the same notation for current, \$i_Z\$

Although the energy stored in current in the Inductor is released into the Zener E= ½ LI² , I(t) is a ramp and not constant.

But what is the Power Dissipation?

Yet we know when the transistor is switched off, Iz=IL and for now, if we let Vce(sat)=0, the voltage across the inductor = 5V , thus the final power input while the ramp reaches a peak current , Ip is P = ½ VIp ( triangle is ½ the area of the VI product square)

The Zener will discharge the current faster but with a fast step voltage and some triangular ramp decay time. {edit} The discharge time depends on the voltage ratio. THe inductor Ramp times will be equal when the voltage across the inductor is equal and opposite polarity or Vz=2Vcc V=LdI/dt and both L and dI=Ip are constants at time Tu=dt of current ramp up for the ramp-down time, Td. The discharge time duration Td = Vcc/Vz * T \$Td/Tu=\dfrac{Vz-Vcc}{Vcc}\$

We already concluded Vdt = LIp = some fixed value.

Although the semiconductor doping of Avalanche diodes is lighter, has a higher breakdown voltage, is faster and has a different conduction mechanism, the notation for such diodes often still uses \$i_Z\$.

Rev A.

Symbol used is a Schottky Power Diode, which are Avalanche type clamps but like Zeners use the same notation for current, \$i_Z\$

Although the energy stored in current in the Inductor is released into the Zener E= ½ LI² , I(t) is a ramp and not constant.

But what is the Power Dissipation?

Yet we know when the transistor is switched off, Iz=IL and for now, if we let Vce(sat)=0, the voltage across the inductor = 5V , thus the final power input while the ramp reaches a peak current , Ip is P = ½ VIp ( triangle is ½ the area of the VI product square)

The Zener will discharge the current faster but with a fast step voltage and some triangular ramp decay time. {edit} The discharge time depends on the voltage ratio. THe inductor Ramp times will be equal when the voltage across the inductor is equal and opposite polarity or Vz=2Vcc V=LdI/dt and both L and dI=Ip are constants at time Tu=dt of current ramp up for the ramp-down time, Td. The discharge time duration Td = Vcc/Vz * T \$Td/Tu=\dfrac{Vz-Vcc}{Vcc}\$

We already concluded Vdt = LIp = some fixed value.

Although the semiconductor doping of Avalanche diodes is lighter, has a higher breakdown voltage, is faster and has a different conduction mechanism, the notation for such diodes often still uses \$i_Z\$.

Simulation proof

Other info

Schottky Power Diodes are also Zeners in the 20 to 40V range at high currents and may be chosen to handle this for some rated Watt and Joule values.

Although the energy stored in current in the Inductor is released into the Zener E= ½ LI² , I(t) is a ramp and not constant.

But what is the Power Dissipation?

Yet we know when the transistor is switched off, Iz=IL and for now, if we let Vce(sat)=0, the voltage across the inductor = 5V , thus the final power input while the ramp reaches a peak current , Ip is P = ½ VIp ( triangle is ½ the area of the VI product square)

The Zener will discharge the current faster but with a fast step voltage and some triangular ramp decay time. {edit} The discharge time depends on the voltage ratio.

V=LdI/dt and both L and dI=Ip are constants at time T=dt of current ramp up for the ramp-up time, T.

The discharge time duration Td = Vcc/Vz * T since we already concluded Vdt = LIp = some fixed value.

Although the semiconductor doping of Avalanche diodes is lighter, has a higher breakdown voltage, is faster and has a different conduction mechanism, the notation for such diodes often still uses \$i_Z\$.

Rev A.

Symbol used is a Schottky Power Diode, which are Avalanche type clamps but like Zeners use the same notation for current, \$i_Z\$

Although the energy stored in current in the Inductor is released into the Zener E= ½ LI² , I(t) is a ramp and not constant.

But what is the Power Dissipation?

Yet we know when the transistor is switched off, Iz=IL and for now, if we let Vce(sat)=0, the voltage across the inductor = 5V , thus the final power input while the ramp reaches a peak current , Ip is P = ½ VIp ( triangle is ½ the area of the VI product square)

The Zener will discharge the current faster but with a fast step voltage and some triangular ramp decay time. {edit} The discharge time depends on the voltage ratio. THe inductor Ramp times will be equal when the voltage across the inductor is equal and opposite polarity or Vz=2Vcc V=LdI/dt and both L and dI=Ip are constants at time Tu=dt of current ramp up for the ramp-down time, Td. The discharge time duration Td = Vcc/Vz * T \$Td/Tu=\dfrac{Vz-Vcc}{Vcc}\$

We already concluded Vdt = LIp = some fixed value.

Although the semiconductor doping of Avalanche diodes is lighter, has a higher breakdown voltage, is faster and has a different conduction mechanism, the notation for such diodes often still uses \$i_Z\$.

Schottky Power Diodes are also Zeners in the 20 to 40V range at high currents and may be chosen to handle this for some rated Watt and Joule values.

Although the energy stored in current in the Inductor is released into the Zener E= ½ LI² , I(t) is a ramp and not constant.

But what is the Power Dissipation?

Yet we know when the transistor is switched off, Iz=IL and for now, if we let Vce(sat)=0, the voltage across the inductor = 5V , thus the final power input while the ramp reaches a peak current , Ip is P = ½ VIp ( triangle is ½ the area of the VI product square)

The Zener will discharge the current faster but with a fast step voltage and some triangular ramp decay time.

V=LdI/dt and both L and dI=Ip are constants at time T=dt of current ramp up for the ramp-up time, T.

The discharge time duration Td = Vcc/Vz * T since we already concluded Vdt = LIp = some fixed value.

Although the semiconductor doping of Avalanche diodes is lighter, has a higher breakdown voltage, is faster and has a different conduction mechanism, the notation for such diodes often still uses \$i_Z\$.

Schottky Power Diodes are also Zeners in the 20 to 40V range at high currents and may be chosen to handle this for some rated Watt and Joule values.

Although the energy stored in current in the Inductor is released into the Zener E= ½ LI² , I(t) is a ramp and not constant.

But what is the Power Dissipation?

Yet we know when the transistor is switched off, Iz=IL and for now, if we let Vce(sat)=0, the voltage across the inductor = 5V , thus the final power input while the ramp reaches a peak current , Ip is P = ½ VIp ( triangle is ½ the area of the VI product square)

The Zener will discharge the current faster but with a fast step voltage and some triangular ramp decay time. {edit} The discharge time depends on the voltage ratio.

V=LdI/dt and both L and dI=Ip are constants at time T=dt of current ramp up for the ramp-up time, T.

The discharge time duration Td = Vcc/Vz * T since we already concluded Vdt = LIp = some fixed value.

Although the semiconductor doping of Avalanche diodes is lighter, has a higher breakdown voltage, is faster and has a different conduction mechanism, the notation for such diodes often still uses \$i_Z\$.