Appunti di Power Electronics

CH1 + Constant voltage CV

V_CH1 = 10V

I_CH1 = 20mA

I_CH = I_CH1 + I_CH2

Increasing I_CH2 = 50μA, we measured I_CH = 0.18A

Why we have some voltage difference? V_CH1 = 10V, V_CH2 = 9.98V

CH2 + Constant current CC

I_CH2 = 1mA

V_CH2 = 9.98V (tiny difference)

CH2 used as a voltmeter

ΔV

ΔV (due to resistance across wires)

This phenomenon is significant for critical values of current

V_R = V_CH1 - R_w·I

• LAB ACTIVITY
• 2 DC/DC converters
• linear regulator
• switching regulator

HEAV: ALL: CH1 in

ΔV_CH1 = 10V

Therm 28.2.2024

1. 90% 100W
2. 95% 95W out

P_loss = P_IN - P_OUT and γ = P_OUT / P_IN (efficiency)

Half dissipated power!

R_TH: P_dis, important term to consider!

P_loss = (P_IN - P_OUT) ⋅ P_OUT

P_OUT = P_OUT (1/η - 1)

P_OUT = P_OUT (1-1/η)

depends on the cooling system:

ΔT = R_Th ⋅ P_dis

P_OUT = P_DIS (η/(1-η))

efficiency of the circuit

All currents and voltage will be periodic. So we have

ν(t) = ν(t+T₀) since T₀ = switching period

i(t) = i(t+T₀) and f_s = 1/T₀ switching frequency

ν(t) = ν(t+T₀) = a₀ + ∑₁ⁿ a_i ⋅ sin(2πω/ T₀ ⋅ t) + ∑₁ⁿ b_i ⋅ cos(2πω/ T₀ ⋅ t)

ν̂ = harmonic component of the signal (AC)

average value of ν(t): a₀ = 1/T_A ∫₀^T_A ν(t) dt = V

DC component of the signal

ν(t) = V + ν̂(t)

Δi_L represents the peak-to-peak variation

ripple i_L = Δi_L/I_L = 50%

i_MAX = 3A

i_MIN = 1A

I_L = 2A

Δi_L = 1A

ν̂_L = î_L - I_L

figure of merit that shows how close we are to DC components

I_load > I_RA → I_in ≡ I_out

V_out constant

V_in →

So we have

η = ^{V_out I_out}/_{Vin Iin}

50% efficiency

Plost of power: thermal dissipation

η ∝ 1/V_in

Switching regulator

η = ^{V_out I_out}/_{Vin Iin} ≈ constant

in order tokeep the sameefficiency

As V_in increases, I_in decreases and

vice-versa; the goal is to keep the efficiency

as constant as possible.

Switching regulator can regulate the output voltage introducing a

small ripple waveform superimposed over a DC component.

Having a higher efficiency, the switching regulator can operate at lower

temperatures (at a given power) compared with the linear regulator.

The presence of switching mechanisms can introduce important noise

contributions. The linear regulator can be preferred for the absence of noise.

Also, it provides more stable output voltage, avoiding ripple.

R

V = V(L)

t

V_g

2^nd order filter with

η = 100% (no dissipation!)

√(ripple

R + C_f) ≈

noise

approximation)

η = 100%

(theoretically!)

STEP-DOWN or BUCK

CONVERTER

DC/AC

Conversion performed by SPDT η = 100%, no power dissipation

ΔV/V = 1/16 * 1/LC * T_d * D

ripple = 1/16 * 1/LC * f_sw

Ripple equation obtained by small ripple approximation. If the ripple is high, the circuit needs to be analyzed in a different way.

D << 1

cannot be changed due to target output voltage

It's possible to act on these parameters.

↓ lower when L or C (or both) are increased.

↓ lower quadratically when the frequency switching is increased.

∅ from 2/3 and 2 for the same ripple.

The limitation of increasing f_sw is the semiconductor device.

Boost Converter (Step-up)

V_L = V_G - D + (V_G-V) * D'

| = V_G - V' D' = ∅

Balance Equation

V = V_G/D' = V_G/1-D

The output V > V_G (since 0 < D < 1)

For what concerns the current:

I_L = I_G

If η = 100% (true under our assumptions)

η = V * I / V_G * I_G = 1

since V/V_G = 1/D'

then I/I_G = D'

I = D' * I_G = D' * I_L

I_L = I/D' = V/RD'

About i_c, remember that depends by the converter.

i_c = i_l2 - I

i_c = -I_L + Δi_L - I = -0.23 A at the beginning of switching period

i_c = I_L + Δi_L - I = 0.056 A at the end of switching period

The two triangles are similar. So, we can say

0.23t_A =

ΔQ = ¹/₂ 0.23 ∙ 2.91_Ct_A = 0.39 nC

Δv = ^ΔQ/_2C = ^0.34/_2∙10^-5 = 17 mV

And finally

ripple = ^{17 ∙ 10-3}/₄ = 0.42%

Everything treated until now is correct considering an ideal SPDT. But now we will understand how the things will change considering real cases.

n-channel MOSFET

V_GS > V_TH ON

NORMALLY OFF TRANSISTORbetter behavior in case of error

Bidirectional switch

the charge of battery grows when the light is applied; when there is no light the battery will discharge. A new diode must be applied.

What happens when the current, in 2, turns negative?

V_L = (V_g - V) D - D₂ = 0

I_C = I_O => I = I (charge balance equation)

I_C = 1/j   1/2   (D₁ + D₂) I_peak   V_L L

2) i_L? Use the schematic!

i_{L peak} = V_g / L. DT_T = 5A

D_{2 T} = i_{L peak} | (-L / V) = 5   9.6 / 12 = 4μ_s

^{i_{L peak}} 5

3) i_C?

i_C = -i_F - I

i_F = {0 nos on ( t < DT_T )

i_L mos off ( DT_T < t < T_P )

I = -1A

i_C = -i_F - I

time length → 4 / 5 → t_Q / 4 = 4 / 5

height → 5

ΔQ = 1/2   t_a i_{L peak}

= 1/2   3.2 × 10^-6   4 = 6.4 nC

ΔV = ΔQ / 2C = 6.4 / 2 · 10 = 0.32 V

V_Q = 0.32 / 12 = 2.6 %

1) For each inductor, extract the V_L equation

2) For each capacitor, extract the I_C equation

3) If ig fix, then extract the I_g equation

Which are the dependent and independent variables?

Independent variables: V_g, D, V_CX, I_X, R_on, R_L, V_on, R₀

V_CX = V_C

Copper - losses boost converter

I_g = I_L

V_L = ¹/_D [(V_g - R_LI_L) + (V_g - R_LI_L - V_C)D]

I_C = ¹/_D [^-(V_C)/_R D + (I_L - ^V_C/_R)D]

V_L = V_g - R_LI_L - V_CD

I_C = ID - ^V_C/_R

Voltage controlled generator

We're in periodic waveform operation. So, V_L = 0 and I_C = 0