# Error in my efficiency calculation for FWB rectifier?

A full-wave bridge rectifier with an RL load is connected to a 120V source. If the load resistance is 10.8 $\Omega$ and L is very large, find:

(d) RMS value of load current

(e) Average current in each diode

(f) RMS current in each diode

(g) Power supplied to the load

(h) Ripple factors of the load voltage and current

(i) Rectifier efficiency

So,

(a) $$V_{o(avg)}=\frac{2V_m}{\pi} =\frac{2\sqrt{2}V_s}{\pi} =108.04 \text{ V}$$

(b) $$\frac{V_{o(avg)}}{R}=\frac{2V_m}{\pi R}=10 \text{ A}$$

(c) $$\text{Should just be when V_{m} is peak and R is minimum, hence, }10 \text{ A}$$

(d) $$\text{Very large L as given in the question}$$ $$\text{therefore the RMS current can be assumed to be equal to I_{o(avg)}, hence, }10 \text{ A}$$

(e) $$I_{D(avg)}=\frac{I_{o(avg)}}{2}=5 \text{ A}$$

(f) $$I_{D(RMS)}=\frac{I_{o(avg)}}{\sqrt{2}}=3.54 \text{ A}$$

(g) $$P_o=I_{o(avg)} \cdot V_{o(avg)}=1080.4 \text{ W}$$

(h) $$RF_V=\sqrt{\frac{V^2_{RMS}}{V^2_{o(avg)}}-1} = 0.619$$

$$RF_I=\sqrt{\frac{I^2_{RMS}}{I^2_{o(avg)}}-1} = 0$$

(i) $$\text{Efficiency}=\frac{V_{o(avg)}\cdot I_{o(avg)}}{V_{RMS}\cdot I_{RMS}} = \frac{108.048\cdot 10}{\frac{120\sqrt{2}}{2}\cdot 10} = 1.273$$

Assuming, $V_{RMS} = \frac{V_s \sqrt{2}}{2}$

Obviously the efficiency shouldn't be greater than one... What have I done wrong?

Following Felthry's comment, I notice the above formula for $V_{RMS}$ only stands for half-wave rectification, hence, $V_{RMS} = 120 \text{ V}$ and the efficiency ~90%.

• Right off the bat I can tell you that when something says "a 120V source" without specifying, what is meant is that V_RMS=120V. – Hearth Apr 30 '17 at 22:07
• @Felthry Ahhh... Note that I need the output RMS voltage, though, not the input RMS voltage. Is it still 120V? I think Vrms = Vs*sqrt(2)/2 only applies to half-wave rectifiers. If so, that was a simple mistake... – jmdlok Apr 30 '17 at 22:57
• An ideal full-wave voltage rectifier does not change the RMS voltage. Even a non-ideal one would change it by only a miniscule amount, in this case. – Hearth Apr 30 '17 at 22:59
• @Felthry Alright, thanks for your help. I'll amend question. – jmdlok Apr 30 '17 at 23:00
• I'll make this into an actual answer so that you can mark the question as answered, then. – Hearth May 1 '17 at 0:01