Single-Phase Half-Wave Controlled Converter with RLE load

Single phase half wave circuit with rle load diagram

A single-phase half-wave controlled converter with RLE load is shown in figure above. The counter emf E in the load may be due to a battery or a dc motor. The minimum value of firing angle is obtained from the relation V_mSinωt = E.This is shown to occur at an angle θ₁ where θ₁ = sin^-1(E / V_m)

In case thyristor T is fired at an angle α < θ₁ , then E > V_s, SCR is reverse biased and therefore it will not turn on. Similarly, maximum value of firing angle is θ₂ = π - θ_{1 ,} show in above waveforms.During the interval load current i₀ is zero, load voltage v₀ = E and during the time i₀ is not zero, v₀ follows V_s curve. For the circuit, SCR T on, KVL gives the voltage differential equation as

_Vmsinωt = Ri₀ + L(di₀/dt) + E

The solution of this equation is made up of two components;

1. steady-state current component i_s and 
2. the transient current component i_t.

For convenience, i_s may be thought of as the sum of i_s1 and i_s2, where i_s1 is the steady state current due to ac source voltage acting alone and i_s2 is that due to dc counter emf E acting alone(according to superposition theorem).

i_s1 = (V_m / Z)sin(ωt - φ ) ; i_s2 = - (E / R) ; The transient current i_t is given by  i_t = Ae^-(R/L)t

Thus the total current i₀ is given by i₀ = i_s1+i_s2+i_t = (V_m / Z)sin(ωt - φ ) - (E / R) + Ae^-(R/L)t

At ωt =  α, t =  α / ω , i0 = 0,i.e 

This gives A = [(E / R ) - (V_m / Z)sin(α - φ)]e^{-(Rα / Lω)}

so i₀ = V_m/Z[sin(ωt-φ)-sin(α-φ)exp{-R/ωL(ωt- α)}] - E/R[1-exp{-R/ωL(ωt-α)}]

In this equation is applicable for  α ≤ ωt ≤ B. The extinction angle ẞ depends upon load emf E, firing angle  α and the load impedance angle φ.

Average Load Voltage V₀ = 1/2π[V_m(cosα-cosB) + E(2π+α-B)]

Average Load current I₀ = 1/2πR[V_m(cosα+cosφ₁)-E(π-φ₁-α)]

The RMS value of load current I_or = √[1/2πR²{(E²+V_s²)(π-2α)+V_or²sin2α-4V_mEcosα}]

Power delivered to load, P = I_or²R+I₀E watts

Supply power factor = (I_or²R+I₀E)/V_sI_or

Advantages of Single-Phase Half-Wave Rectifier with RLE Load

• Simplicity
• Low Cost
• Reduced Components
• Suitable for Resistive Loads
• Ease of Control
• Suitable for Low Power Applications

Disadvantages of Single-Phase Half-Wave Rectifier with RLE Load

• Low Efficiency
• High Ripple Factor
• Limited Applicability
• Transformer Utilization Factor
• Harmonic Content

Application of single-phase half-wave rectifier with RLE load

• Light Dimmers
• Heating Applications
• Battery Charging
• Low-Power Supplies
• Electroplating
• Power Supplies for DC Motors