Diode rectifiers (uncontrolled rectifiers) POWER ELECTRONICS Diode rectifiers (uncontrolled rectifiers) Mustafa A. Fadel Assistant Lecture Electrical Engineering Department College of Engineering, Mustansiriyah University

Block diagram of an uncontrolled diode rectifier circuit Rectifier is a circuit that converts an AC signal to a DC signal AC supply Transformer Rectifier Block diagram of an uncontrolled diode rectifier circuit Note: In a diode rectifiers, the power flows only from the AC source to the DC side. Applications of Uncontrolled Rectifiers • DC power supply for consumer electronic products such as radios, TVs, DVD players, mobile phone chargers, computers, laptops and so on (low power) • DC motor drives (high power)

Uncontrolled rectifier Depending on the type of input source, rectifiers are classified into two main groups: Uncontrolled rectifier Single phase Half-wave Full-wave Three phase Important equations 𝑦 𝑎𝑣 = 1 𝑇 0 𝑇 𝑦(𝑡) 𝑑𝑡 𝑦 𝑟𝑚𝑠 = 1 𝑇 0 𝑇 𝑦(𝑡) 2 𝑑𝑡

Single-phase half-wave rectifier (R load) Circuit diagram Waveforms

η= 𝑃 𝑑𝑐 𝑃 𝑎𝑐 = 𝑉 𝑑𝑐 𝐼 𝑑𝑐 𝑉 𝑟𝑚𝑠 𝐼 𝑟𝑚𝑠 𝑉 𝑑𝑐 = 𝑉 𝑚 𝜋 The average value of output voltage 𝐼 𝑑𝑐 = 𝑉 𝑑𝑐 𝑅 = 𝑉 𝑚 𝜋 𝑅 The average value of load current 𝑉 𝑟𝑚𝑠 = 𝑉 𝑚 2 The rms value of output voltage 𝐼 𝑟𝑚𝑠 = 𝑉 𝑟𝑚𝑠 𝑅 The rms value of load current Peak inverse voltage across the diode 𝑃𝐼𝑉= 𝑉 𝑚 η= 𝑃 𝑑𝑐 𝑃 𝑎𝑐 = 𝑉 𝑑𝑐 𝐼 𝑑𝑐 𝑉 𝑟𝑚𝑠 𝐼 𝑟𝑚𝑠 The efficiency of rectification

The effective (rms) value of the ac component of output voltage 𝑉 𝑎𝑐 = 𝑉 𝑟𝑚𝑠 2 − 𝑉 𝑑𝑐 2 The form factor (a measure for the shape of output voltage) FF= 𝑉 𝑟𝑚𝑠 𝑉 𝑑𝑐 The ripple factor (a measure for the ripple content) RF= 𝑉 𝑎𝑐 𝑉 𝑑𝑐 = 𝑉 𝑟𝑚𝑠 2 − 𝑉 𝑑𝑐 2 𝑉 𝑑𝑐 = 𝐹𝐹 2 −1

𝐻𝐹=𝑇𝐻𝐷= 𝐼 ℎ 𝐼 𝑠1 = 𝐼 𝑠 2 − 𝐼 𝑠1 2 𝐼 𝑠1 = 𝐼 𝑠 𝐼 𝑠1 2 −1 The harmonic factor or total harmonic distortion (a measure for distortion of a waveform] of the input current 𝐻𝐹=𝑇𝐻𝐷= 𝐼 ℎ 𝐼 𝑠1 = 𝐼 𝑠 2 − 𝐼 𝑠1 2 𝐼 𝑠1 = 𝐼 𝑠 𝐼 𝑠1 2 −1 Where Is1 is the rms value of the fundamental component of the input current. And Is is the rms value of the input current.

If Φ is the angle between the fundamental component of the input current and the voltage, the displacement factor 𝐷𝐹= cos ∅ The input power factor 𝑃𝐹= 𝑉 𝑠 𝐼 𝑠1 cos ∅ 𝑉 𝑠 𝐼 𝑠 = 𝐼 𝑠1 cos ∅ 𝐼 𝑠

Single-phase half-wave rectifier (RL load) without freewheeling diode Due to inductive load, the conduction period of the diode D1 will extend beyond 180 degree until the current becomes zero. Circuit diagram Waveforms

𝑉 𝑑𝑐 = 𝑉 𝑚 2𝜋 1− cos (𝜋+𝜃) 𝜃= 𝑡𝑎𝑛 −1 𝜔 𝐿 𝑅 , 𝜔=2 𝜋 𝑓 𝐼 𝑑𝑐 = 𝑉 𝑑𝑐 𝑅 The average value of the output voltage 𝑉 𝑑𝑐 = 𝑉 𝑚 2𝜋 1− cos (𝜋+𝜃) Where the angle θ can be calculated as: 𝜃= 𝑡𝑎𝑛 −1 𝜔 𝐿 𝑅 , 𝜔=2 𝜋 𝑓 The average value of the load current 𝐼 𝑑𝑐 = 𝑉 𝑑𝑐 𝑅 The average value of the output voltage (and hence the current) can be increased by making θ=0, which is possible by adding a freewheeling diode Dm across the load.

Single-phase half-wave rectifier (RL load) with freewheeling diode Circuit diagram Waveforms The effect of the freewheeling diode Dm is to prevent the negative voltage appearing across the load, and as a result, the magnetic stored energy is increased. At t=π/ω, the current from D1 is transferred to Dm and this process is called commutation of diodes. Depending on the load time constant, the load current may be discontinuous.

Single-phase center-tap full-wave rectifier (R load) Circuit diagram Waveforms

𝑉 𝑑𝑐 = 2𝑉 𝑚 𝜋 𝐼 𝑑𝑐 = 𝑉 𝑑𝑐 𝑅 = 2 𝑉 𝑚 𝜋𝑅 𝑉 𝑟𝑚𝑠 = 𝑉 𝑚 2 𝐼 𝑟𝑚𝑠 = 𝑉 𝑟𝑚𝑠 𝑅 𝑉 𝑑𝑐 = 2𝑉 𝑚 𝜋 The average value of output voltage 𝐼 𝑑𝑐 = 𝑉 𝑑𝑐 𝑅 = 2 𝑉 𝑚 𝜋𝑅 The average value of load current 𝑉 𝑟𝑚𝑠 = 𝑉 𝑚 2 The rms value of output voltage 𝐼 𝑟𝑚𝑠 = 𝑉 𝑟𝑚𝑠 𝑅 The rms value of load current Peak inverse voltage across each diode 𝑃𝐼𝑉= 2𝑉 𝑚

Single-phase bridge full-wave rectifier (R load) Circuit diagram

𝑉 𝑑𝑐 = 2𝑉 𝑚 𝜋 𝐼 𝑑𝑐 = 𝑉 𝑑𝑐 𝑅 = 2 𝑉 𝑚 𝜋𝑅 𝑉 𝑟𝑚𝑠 = 𝑉 𝑚 2 𝐼 𝑟𝑚𝑠 = 𝑉 𝑟𝑚𝑠 𝑅 The average value of output voltage 𝑉 𝑑𝑐 = 2𝑉 𝑚 𝜋 The average value of load current 𝐼 𝑑𝑐 = 𝑉 𝑑𝑐 𝑅 = 2 𝑉 𝑚 𝜋𝑅 The rms value of output voltage 𝑉 𝑟𝑚𝑠 = 𝑉 𝑚 2 The rms value of load current 𝐼 𝑟𝑚𝑠 = 𝑉 𝑟𝑚𝑠 𝑅 Peak inverse voltage across each diode 𝑃𝐼𝑉= 𝑉 𝑚 Waveforms

Single-phase bridge full-wave rectifier (RL load) With a resistive load, the load current is identical in shape to the load voltage. In practical applications, most loads are inductive. The load current shape and magnitude depend on both the load resistance R and inductance L. Circuit diagram Waveforms

Single-phase bridge rectifier with very large inductive load

E#1 If a single-phase bridge rectifier supplies a very high inductive load such as a dc motor. the turns ratio of the transformer is unity. Determine a) the HF of the input current, and b) the input PF of the rectifier. Note the output (load) current is constant and ripple free due to the highly inductive load.

Using Fourier series, the input current is can be analyzed as 𝑖 𝑠 (𝑡)= 𝐼 𝑑𝑐 + 𝑛=1,2,3,… ∞ ( 𝑎 𝑛 cos 𝑛𝜔𝑡 + 𝑏 𝑛 𝑠𝑖𝑛 𝑛𝜔𝑡 ) 𝐼 𝑑𝑐 = 1 2𝜋 0 2𝜋 𝑖 𝑠 (𝑡) 𝑑 𝜔𝑡 =0 𝑎 𝑛 = 1 𝜋 0 2𝜋 𝑖 𝑠 (𝑡) cos 𝑛𝜔𝑡 𝑑 𝜔𝑡 =0 𝑏 𝑛 = 1 𝜋 0 2𝜋 𝑖 𝑠 (𝑡) sin 𝑛𝜔𝑡 𝑑 𝜔𝑡 = 4 𝐼 𝑎 𝑛𝜋 ∴ 𝑖 𝑠 𝑡 = 4 𝐼 𝑎 𝜋 𝑠𝑖𝑛 𝜔𝑡 1 + 𝑠𝑖𝑛 3𝜔𝑡 3 + 𝑠𝑖𝑛 5𝜔𝑡 5 +… Therefore, the rms value of the input current is ∴ 𝐼 𝑠 = 4 𝐼 𝑎 𝜋 2 1+ 1 3 2 + 1 5 2 +… = 𝐼 𝑎

Therefore, the rms value of the fundamental component of the input current is ∴ 𝐼 𝑠1 = 4 𝐼 𝑎 𝜋 2 =0.9 𝐼 𝑎 Therefore, the harmonic factor is 𝐻𝐹=𝑇𝐻𝐷= 𝐼 𝑠 𝐼 𝑠1 2 −1 = 1 0.9 2 −1 =0.4843 𝑜𝑟 48.43% Since Φ=0, the displacement factor is 𝐷𝐹= cos ∅ =1 Therefore, the input power factor 𝑃𝐹= 𝐼 𝑠1 cos ∅ 𝐼 𝑠 =0.9 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

Multiphase rectifier

Three phase bridge rectifier Circuit diagram

Waveforms

𝑉 𝑑𝑐 = 3 3 𝑉 𝑚 𝜋 =1.654 𝑉 𝑚 𝑉 𝑟𝑚𝑠 = 3 2 + 9 3 4𝜋 1 2 𝑉 𝑚 =1.6554 𝑉 𝑚 The average value of output voltage 𝑉 𝑑𝑐 = 3 3 𝑉 𝑚 𝜋 =1.654 𝑉 𝑚 The rms value of output voltage 𝑉 𝑟𝑚𝑠 = 3 2 + 9 3 4𝜋 1 2 𝑉 𝑚 =1.6554 𝑉 𝑚 Peak inverse voltage across each diode 𝑃𝐼𝑉= 3 𝑉 𝑚

Currents through diodes E#2 A three-phase bridge rectifier supplies a high inductive load such that the average load current Idc= 60 A and the ripple content is negligible. Determine the ratings of the diodes if the line-to-neutral voltage of the supply is 120V, 60Hz. As shown, the average value of the diode current 𝐼 𝑑(𝑎𝑣) = 𝐼 𝑑𝑐 3 = 60 3 =20 A The rms value of the diode current 𝐼 𝑑(𝑟𝑚𝑠) = 1 2𝜋 𝜋/3 𝜋 𝐼 𝑑𝑐 2 𝑑𝜃 = 1 2𝜋 × 𝐼 𝑑𝑐 2 × 2𝜋 3 = 𝐼 𝑑𝑐 3 =46.19A The peak inverse voltage across the diode PIV= 3 𝑉 𝑚 = 3 × 2 ×120=294𝑉 Currents through diodes

E#3 For the following current waveform, determine the average, rms, peak value of the current. 𝐼 𝑎𝑣 = 𝐼 1𝑎𝑣 + 𝐼 2𝑎𝑣 + 𝐼 3𝑎𝑣 𝐼 1𝑎𝑣 =150× 𝑡 3 𝑇 =150× 400 5000 =12𝐴 𝐼 2𝑎𝑣 = 2 𝐼 𝑚 𝜋 × 𝑡 2 − 𝑡 1 𝑇 = 2×150 𝜋 × 100 5000 =1.91𝐴 𝐼 3𝑎𝑣 =100× 𝑡 5 − 𝑡 4 𝑇 =100× 200 5000 =4𝐴 𝐼 𝑎𝑣 =12+1.91+4=17.91𝐴

𝐼 𝑟𝑚𝑠 = 𝐼 1𝑟𝑚𝑠 2 + 𝐼 2𝑟𝑚𝑠 2 + 𝐼 3𝑟𝑚𝑠 2 𝐼 1𝑟𝑚𝑠 =150× 𝑡 3 𝑇 =150× 400 5000 =42.43𝐴 𝐼 2𝑟𝑚𝑠 = 𝐼 𝑚 2 × 𝑡 2 − 𝑡 1 𝑇 × = 150 2 × 100 5000 =15𝐴 𝐼 3𝑟𝑚𝑠 =100× 𝑡 5 − 𝑡 4 𝑇 =100× 200 5000 =20𝐴 𝐼 𝑟𝑚𝑠 = 42.43 2 + 15 2 + 20 2 =49.25𝐴 𝐼 𝑝 =300𝐴