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1
Inclusion of Switching Loss in the
Averaged Equivalent Circuit Model
The methods of Chapter 3 can be extended to include switching loss in
the converter equivalent circuit model
• Include switching transitions in the converter waveforms
• Model effects of diode reverse recovery, etc.
To obtain tractable results, the waveforms during the switching
transitions must usually be approximated
Things that can substantially change the results:
• Ringing caused by parasitic tank circuits
• Snubber circuits
• These are modeled in ECEN 5817, Resonant and Soft-
Switching Phenomena in Power Electronics
2
The Modeling Approach
Extension of Chapter 3 Methods
Sketch the converter waveforms
– Including the switching transitions (idealizing assumptions
are made to lead to tractable results)
– In particular, sketch inductor voltage, capacitor current, and
input current waveforms
The usual steady-state relationships:
vL = 0, iC = 0, ig = Ig
Use the resulting equations to construct an equivalent
circuit model, as usual
3
Buck Converter Example
• Ideal MOSFET, p–n diode with reverse recovery
• Neglect semiconductor device capacitances, MOSFET
switching times, etc.
• Neglect conduction losses
• Neglect ripple in inductor current and capacitor voltage
4
Assumed
waveforms
Diode recovered charge Qr,
reverse recovery time tr
These waveforms assume
that the diode voltage
changes at the end of the
reverse recovery transient
• a “snappy” diode
• Voltage of soft-recovery
diodes changes sooner
• Leads to a pessimistic
estimate of induced
switching loss
5
Inductor volt-second balance
and capacitor charge balance
As usual: vL = 0 = DVg – V
Also as usual: iC = 0 = IL – V/R
6
Average input current
ig = Ig = (area under curve)/Ts
= (DTsIL + trIL + Qr)/Ts
= DIL + trIL /Ts + Qr /Ts
7
Construction of Equivalent Circuit Model
From inductor volt-second balance: vL = 0 = DVg – V
From capacitor charge balance: iC = 0 = IL – V/R
8
Input port of model
ig = Ig = DIL + trIL /Ts + Qr /Ts
9
Combine for complete model
The two independent current sources consume power
Vg (trIL /Ts + Qr /Ts)
equal to the switching loss induced by diode reverse recovery
10
Solution of model
Output:
V = DVg
Efficiency: = Pout / Pin
Pout = VIL Pin = Vg (DIL + trIL /Ts + Qr /Ts)
Combine and simplify:
= 1 / [1 + fs (tr /D + Qr R /D2Vg )]
11
Predicted Efficiency vs Duty Cycle
Switching frequency 100 kHz
Input voltage 24 V
Load resistance 15
Recovered charge 0.75 µCoul
Reverse recovery time 75 nsec
(no attempt is made here to
model how the reverse
recovery process varies with
inductor current)
• Substantial degradation of
efficiency
• Poor efficiency at low duty
cycle
Buck converter with diode reverse recovery
0.00%
10.00%
20.00%
30.00%
40.00%
50.00%
60.00%
70.00%
80.00%
90.00%
100.00%
0 0.2 0.4 0.6 0.8 1
Duty cycle
Efficiency
12
Boost Converter Example
Model same effects as in previous buck converter example:
• Ideal MOSFET, p–n diode with reverse recovery
• Neglect semiconductor device capacitances, MOSFET
switching times, etc.
• Neglect conduction losses
• Neglect ripple in inductor current and capacitor voltage
13
Boost
converter
Transistor and diode
waveforms have same
shapes as in buck
example, but depend
on different quantities
14
Inductor volt-second balance
and average input current
As usual: vL = 0 = Vg – D V
Also as usual: ig = IL
15
Capacitor
charge balance
iC = id – V/R = 0
= – V/R + IL(D Ts – tr)/Ts – Qr /Ts
Collect terms: V/R = IL(D Ts – tr)/Ts – Qr /Ts
16
Construct model
The two independent current sources consume power
V (trIL /Ts + Qr /Ts)
equal to the switching loss induced by diode reverse recovery
The result is:
17
Predicted V/Vg vs duty cycle
Boost converter with diode reverse recovery
0
1
2
3
4
5
6
7
8
0 0.2 0.4 0.6 0.8 1
Duty cycle
V/Vg
Switching frequency 100 kHz
Input voltage 24 V
Load resistance 60
Recovered charge 5 µCoul
Reverse recovery time 100 nsec
Inductor resistance RL = 0.3
(inductor resistance also inserted
into averaged model here)
With RL only
With RL and diode reverse recovery
18
Summary
The averaged modeling approach can be extended to
include effects of switching loss
Transistor and diode waveforms are constructed,
including the switching transitions. The effects of the
switching transitions on the inductor, capacitor, and
input current waveforms can then be determined
Inductor volt-second balance and capacitor charge
balance are applied
Converter input current is averaged
Equivalent circuit corresponding to the the averaged
equations is constructed

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Chapter 04a

  • 1. 1 Inclusion of Switching Loss in the Averaged Equivalent Circuit Model The methods of Chapter 3 can be extended to include switching loss in the converter equivalent circuit model • Include switching transitions in the converter waveforms • Model effects of diode reverse recovery, etc. To obtain tractable results, the waveforms during the switching transitions must usually be approximated Things that can substantially change the results: • Ringing caused by parasitic tank circuits • Snubber circuits • These are modeled in ECEN 5817, Resonant and Soft- Switching Phenomena in Power Electronics
  • 2. 2 The Modeling Approach Extension of Chapter 3 Methods Sketch the converter waveforms – Including the switching transitions (idealizing assumptions are made to lead to tractable results) – In particular, sketch inductor voltage, capacitor current, and input current waveforms The usual steady-state relationships: vL = 0, iC = 0, ig = Ig Use the resulting equations to construct an equivalent circuit model, as usual
  • 3. 3 Buck Converter Example • Ideal MOSFET, p–n diode with reverse recovery • Neglect semiconductor device capacitances, MOSFET switching times, etc. • Neglect conduction losses • Neglect ripple in inductor current and capacitor voltage
  • 4. 4 Assumed waveforms Diode recovered charge Qr, reverse recovery time tr These waveforms assume that the diode voltage changes at the end of the reverse recovery transient • a “snappy” diode • Voltage of soft-recovery diodes changes sooner • Leads to a pessimistic estimate of induced switching loss
  • 5. 5 Inductor volt-second balance and capacitor charge balance As usual: vL = 0 = DVg – V Also as usual: iC = 0 = IL – V/R
  • 6. 6 Average input current ig = Ig = (area under curve)/Ts = (DTsIL + trIL + Qr)/Ts = DIL + trIL /Ts + Qr /Ts
  • 7. 7 Construction of Equivalent Circuit Model From inductor volt-second balance: vL = 0 = DVg – V From capacitor charge balance: iC = 0 = IL – V/R
  • 8. 8 Input port of model ig = Ig = DIL + trIL /Ts + Qr /Ts
  • 9. 9 Combine for complete model The two independent current sources consume power Vg (trIL /Ts + Qr /Ts) equal to the switching loss induced by diode reverse recovery
  • 10. 10 Solution of model Output: V = DVg Efficiency: = Pout / Pin Pout = VIL Pin = Vg (DIL + trIL /Ts + Qr /Ts) Combine and simplify: = 1 / [1 + fs (tr /D + Qr R /D2Vg )]
  • 11. 11 Predicted Efficiency vs Duty Cycle Switching frequency 100 kHz Input voltage 24 V Load resistance 15 Recovered charge 0.75 µCoul Reverse recovery time 75 nsec (no attempt is made here to model how the reverse recovery process varies with inductor current) • Substantial degradation of efficiency • Poor efficiency at low duty cycle Buck converter with diode reverse recovery 0.00% 10.00% 20.00% 30.00% 40.00% 50.00% 60.00% 70.00% 80.00% 90.00% 100.00% 0 0.2 0.4 0.6 0.8 1 Duty cycle Efficiency
  • 12. 12 Boost Converter Example Model same effects as in previous buck converter example: • Ideal MOSFET, p–n diode with reverse recovery • Neglect semiconductor device capacitances, MOSFET switching times, etc. • Neglect conduction losses • Neglect ripple in inductor current and capacitor voltage
  • 13. 13 Boost converter Transistor and diode waveforms have same shapes as in buck example, but depend on different quantities
  • 14. 14 Inductor volt-second balance and average input current As usual: vL = 0 = Vg – D V Also as usual: ig = IL
  • 15. 15 Capacitor charge balance iC = id – V/R = 0 = – V/R + IL(D Ts – tr)/Ts – Qr /Ts Collect terms: V/R = IL(D Ts – tr)/Ts – Qr /Ts
  • 16. 16 Construct model The two independent current sources consume power V (trIL /Ts + Qr /Ts) equal to the switching loss induced by diode reverse recovery The result is:
  • 17. 17 Predicted V/Vg vs duty cycle Boost converter with diode reverse recovery 0 1 2 3 4 5 6 7 8 0 0.2 0.4 0.6 0.8 1 Duty cycle V/Vg Switching frequency 100 kHz Input voltage 24 V Load resistance 60 Recovered charge 5 µCoul Reverse recovery time 100 nsec Inductor resistance RL = 0.3 (inductor resistance also inserted into averaged model here) With RL only With RL and diode reverse recovery
  • 18. 18 Summary The averaged modeling approach can be extended to include effects of switching loss Transistor and diode waveforms are constructed, including the switching transitions. The effects of the switching transitions on the inductor, capacitor, and input current waveforms can then be determined Inductor volt-second balance and capacitor charge balance are applied Converter input current is averaged Equivalent circuit corresponding to the the averaged equations is constructed