Bridge type full wave rectifier, Lecture-XI.

Analog systems and applications, lecture — XI.

Bridge type full wave rectifier.

This article belongs to a series of lectures on analog electronics, the paper goes by the name “Analog Systems and Applications” for the physics honors degree class. All lectures of this series will be found here. This comprises of the 11th lecture of the series. The lecture was delivered on 12th February 2018.

In our last lecture we discussed the center tapped full wave rectifier and derived its most important parameters. Check that lecture for a conceptual continuity with the present lecture. Prior to that we discussed in much detail the case of the half wave rectifier and its parameters. Those two lectures can be accessed here.

Today we will discuss the other type of full wave rectifier, viz. bridge type rectifier. Lets draw a suitable diagram for the same and discuss its operational ideas.

The bridge type full wave rectifier

(i) It consists of 4 diodes D₁, D₂, D₃ and D₄.

(ii) Diagonally opposite corners ‘A‘ and ‘C‘ are connected across the secondary ‘S‘ of the power transformer. Diagonally opposite corners ‘B‘ and ‘D‘ are connected to the load resistance ‘R_L‘.

(iii) During positive or first half cycle D₁, and D₃ conduct as A is positive wrt C. During negative half cycle or second half of signal D₂, and D₄ conduct as C is positive wrt A.

(iv) Bridge rectifier does not require center-tapping at the secondary of the transformer.

Let the voltage across the secondary of the transformer be given by V = V₀ sin ωt. Here V₀ is the maximum transformer voltage or amplitude of voltage signal. Also ω/2π is the frequency of the AC signal.

(i) During the half cycle when A is positive relative to C; D₁, and D₃ are “forward biased”, D₂, and D₄ are “reverse biased”. i_L1 flows through R_L, in the direction ABEFDCSA. i_L2 = 0.

(ii) Similarly during next-half cycle C is positive relative to A; D₂, and D₄ are “forward biased”, D₁, and D₃ are “reverse biased”. i_L2 flows through R_L, in the direction CBEFDASC. i_L1 = 0.

Thus current always flows in the same direction; along EF, through R_L.

Lets write down the current.

$i_{L_{1}} =I_0 \sin \omega t=\frac{V_0}{2R_f+R_s+R_L} \,\sin \omega t,\,\,i_{L_{2}}=0$ during $0\le \omega t \le \pi$ .

AND

$i_{L_{2}} =-I_0 \sin \omega t=\frac{V_0}{2R_f+R_s+R_L} \,\sin (\omega t +\pi),\,\,i_{L_{1}}=0$ during $\pi\le \omega t \le 2\pi$ .

So maximum current through load is $\boxed{I_0=\frac{V_0}{2R_f+R_s+R_L} }$ .

The average or dc and rms value of current, rectifier efficiency and ripple factor can be given by comparing with the same of a center-tapped full wave rectifier.

$\boxed{I_{dc}=\frac{2I_0}{\pi}=\frac{2V_0}{\pi(2R_f+R_s+R_L)} }$ .

$\boxed{V_{dc}=\frac{2V_0}{\pi \big[1+\frac{2R_f+R_s}{R_L} \big]}}$

$\boxed{I_{rms}=\frac{V_0}{\sqrt 2(2R_f+R_s+R_L)} }$ .

Rectifier efficiency (in %) is given by $\boxed{ \%\,of\,\eta = \frac{81.06}{1+\frac{(2R_f+R_s)}{R_L}}}$ .

Ripple factor is $\boxed{r=0.48}$ .

PIV across each diode is now V₀ {instead of 2V₀}.

Advantage.

(i) Small transformer can be used.

(ii) Center-tap on secondary of transformer is not required.

(iii) PIV per diode is V₀, instead of 2V₀ for center-tapped full wave rectifier. This is helpful when higher voltage is required.

(iv) TUF (transformer utilization factor) increases to 0.811, current flowing in secondary is purely alternating. This helps for large power delivery.

(v) Low cost.

(vi) High reliability.

Disadvantage.

(i) 2 extra diodes are needed. This increases voltage drop and reduces output voltage. This creates problem when low voltage output is required.

(ii) This has poor voltage regulation compared to center tapped full wave rectifier.

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April 30, 2020

[…] (lecture-8 and 9); center tapped full wave rectifier (lecture-10); bridge type full wave rectifier (lecture-11); current and voltage sources and Norton’s and Thevenin;’s and Norton’s […]

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Bridge type full wave rectifier, Lecture-XI.