2 Choppers

Electrical Power Engineering 3

Power Electronics dc-dc converters (Choppers) Converts dc → dc

Converts input dc voltage (eg battery) to output dc voltage. Usually, but not always, Vin > Vout

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dc-dc Converters

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Average voltage:

Vo ( average ) × (ton + toff ) = Vi × ton

The average voltage is such that the yellow area = the green area.

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Vo ( average ) × (ton + toff ) = Vi × ton Therefore:

Vo ( average) =

ton × Vi ton + toff

= D × Vi Where D = switch duty ratio

D is the proportion of the cycle that the switch is on.

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Vo = D.Vi Average output controlled by D

D controls the output voltage

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Switch • MOSFET – easy to switch off – low powers ( < 100 amp) – high frequency ( up to 1 MHz)

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Switch • Thyristor – difficult to switch off – high power – low frequency ( < 1 kHz)

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Switch • GTO – easy to switch off – high power – low frequency ( < 50 kHz)

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Switch • IGBT – easy to switch off – medium power – medium frequency ( < 100 kHz)

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Chopper with Inductive Load

di v=L dt

Switch opened: – very high voltages – may destroy switch

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Chopper with Inductive Load Include “freewheeling diode”

If the load is inductive, a “freewheeling diode” is required to provide a path for the current when the switch is opened, otherwise very high voltage spikes will occur which may destroy the switch, due to:

v = L.

di dt

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Continuous/Discontinuous Operation Discontinuous:

Here, the current falls to zero during the switch “off” time.

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Continuous/Discontinuous Operation Continuous:

Here, the inductance is large enough to maintain the current > 0 during the switch “off” time.

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Load with Back EMF • Most loads can be approximated as:

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DC Motor Load

Vc = E = dc motor back EMF RL = open circuit Rs = armature resistance

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Battery Charger

Vc = E = battery voltage RL = load on battery (if connected) Rs = smoothing inductor resistance ≈ 0

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DC Power Supply

Vc = voltage on smoothing capacitor RL = load on power supply Rs = smoothing inductor resistance ≈ 0

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Continuous Mode with Back EMF

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Discontinuous Mode with Back EMF

Note that when the current has fallen to zero, there is no volt drop across Rs and Ls, therefore v2 = VC. Thus the average voltage is slightly higher than the previous case, due to the extra (shaded) area. The equation Vo = D.Vi does not apply in this case.

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If Rs = 0

vL = L.

di = v2 − VC dt

S on: v2 = Vi S off: v2 = 0

If Rs is negligible, then the current varies linearly with time, making the analysis much easier (no exponentials!). The slope can be calculated for each segment from:

di dt Therefore: vL = L.

v2 − VC = L.

di dt

Where (v2-VC) = voltage across the inductor. If Rs is ignored:

iaverage =

VC RL

v2 ( average) = VC

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Average voltage across an inductor over one complete cycle in steady state = 0

[Average current through a capacitor over one complete cycle in steady state = 0]

For any inductor operating in steady state conditions:

vL = L.

di dt

Therefore, integrating over one cycle:

∫ vL .dt = ∫ L.di = L.[i ] = L.(I 2 − I1 ) I2 I1

Where

I1 = current at start of the cycle I2 = current at end of the cycle

But under steady state conditions, I1 = I2 Therefore, over one complete cycle, under steady state conditions,

vL .dt = 0 Therefore:

∫

Average voltage across an inductor over a complete cycle, under steady state conditions, is zero.

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vl L .

di dt

vL i

In this case the average voltage across the inductor = 0, and the current is in steady state (current at the end of the cycle = current at the start of the cycle).

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vl L .

di dt

i

vL

In the figure above the average voltage is clearly not zero, resulting in the current steadily rising. Thus the circuit is not in steady state.

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Step-Up Chopper

• S closed: v2 = 0

• Average vL = 0

• S open:

• Average v2 = E

v2 = Vdc

In the step-up chopper, when switch S is “on” current iL builds up in the inductor, round loop E, L, S and back to E. No current flows through the diode during this period. As S is “on”, v2 = Vdc for this period. When S is switched “off”, the only path for the inductor current iL is through the diode to Vdc. (It is assumed that capacitor C will hold Vdc approximately constant). As the diode is conducting, v2 = Vdc for this period, giving the waveform for v2 shown on the next slide.

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v2 = average of E

E = Vdc ×

toff ton + toff

Vdc 1 = E 1− D For the Step-Up Chopper, if VL(average) = 0, then E is the average of v2.

E = Vdc . = Vdc .

toff ton + toff

(1 − D ).T T

Where T is the switching period. Hence:

Vdc 1 = E 1 −duty D ratio) must be between 0 and 1, Vdc > E As D (switch ie. This is a step-up chopper.

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Vdc 1 = E 1− D

Step-Up Chopper

The output voltage is always greater than the input voltage. However, the output current is less than the input current, so we aren’t getting energy from nowhere – the laws of physics are not broke. This is similar to a step-up transformer.

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Chopper DC Motor Drive With Regeneration

The circuit above combines the step-up and step-down choppers, such that iL can be either positive or negative. If iL is positive, power (=Vdc.iL) is positive, ie. power is from the motor to Vdc. This is regeneration and, if used with a dc motor, can be used for regenerative braking, where the motor is slowed down by feeding the kinetic energy of rotation back to the supply, recovering this energy. An application of this is in battery powered electric vehicles, where braking is achieved by feeding the vehicle kinetic energy back to the battery, thus recharging the battery and extending the period between charging.

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• Normal (motor) operation: – S2 is permanently off – S1 is switched on and off – iL negative, therefore power (=Vdc .I) is negative

During normal motor operation S1 is switched on and off (step-down chopper mode) with D1 providing a path for the inductor current when S1 is switched off. During this mode S2 is permanently off.

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• Regenerative braking – – – – –

S1 is permanently off S2 is switched on and off iL positive, therefore power ( = Vdc .I) is positive Power flow from dc machine to supply Regenerative braking

For regenerative braking, S1 is permanently off, and S2 is switched on and off (step-up chopper mode). When S2 is on, current iL builds up from source E, through R, L, S2 and back to E. When S2 is switched off, the only path for the current through the inductor to flow is through diode D2 back to the supply, ie. feeding energy back to the supply. For this circuit to work, the supply has to be capable of receiving energy (eg there must not be a diode blocking reverse current).

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