Bridge rectifier more or less efficient than single diode half bridge? SOLVED

The voltage source is 12V RMS 60Hz.

  1. Which circuit is more efficient? Why?
  2. Which circuit will the lamp be brighter?
November 18, 2016

2 Answers

Answer by mikerogerswsm

Bit of a trick question, really, because incandescent lamps aren't resistors, they have baretter action and adjust their resistance to run at more or less constant temperature over a range of voltage inputs. There is no need for the capacitor, lamps do not need smooth dc, they are perfectly happy to run on full wave or half wave rectified ac. Use of a Schottky diode is overkill, any old 1N4000 series diode will do. The energy loss in the diode drops is negligible at this sort of voltage. In your circuits you will need 18V lamps. Without the capacitors you will need a 12V lamp for the full wave rectified circuit and a 6V lamp for the half wave rectified circuit. On 60Hz supply the half wave circuit may produce perceptible flicker and reduced lamp life. On 50Hz (as used in third world countries) the flicker will be more pronounced.

ACCEPTED +2 votes
by mikerogerswsm
November 21, 2016

Great points! I interpreted it more as a homework question to "drive this load" rather than a lamp, but yeah incandescent lamps are full of their own weirdness.

by mrobbins
November 21, 2016

wow thanks -- I didn't think about different lamp voltages too.

November 21, 2016

Answer by mrobbins


Your circuit #2 will be "more efficient" in terms of a smaller fraction of V1's power being "wasted" in the AC-DC conversion. The reason is that in the bridge rectifier, there are always two diode drops in series with charging the capacitor.

Can we quantify it? How much more efficient? I used your circuit and plotted P(V1), P(V2), P(BR1), P(D1), P(LAMP1), P(LAMP2), and did it from $t_1=\frac {1} {60}$ to $t_2=11 * \frac {1} {60}$. Then I summed over the time period (right click on graph, click "Show P(BR1) integral" etc. The totals I got (for 10 AC cycles) were:

$$\int_{t_1}^{t_2} P(V_1) dt = -1.785 J$$ $$\int_{t_1}^{t_2} P(V_2) dt = -1.249 J$$ $$\int_{t_1}^{t_2} P(BR_1) dt = 244.7 mJ$$ $$\int_{t_1}^{t_2} P(D_1) dt = 60.55 mJ$$ $$\int_{t_1}^{t_2} P(LAMP_1) dt = 1.513 J$$ $$\int_{t_1}^{t_2} P(LAMP_2) dt = 1.157 J$$

Note that the powers of the sources are negative because power is flowing "out" of the elements, rather than into them.

If we define the "efficiency" to be the energy delivered to the lamp divided by the energy delivered by the source, then:

$$\eta_1 = \frac {\int_{t_1}^{t_2} P(LAMP_1) dt} {-\int_{t_1}^{t_2} P(V_1) dt } = \frac {1.513 J} {1.785 J} = 0.848 = 84.8\%$$

$$\eta_2 = \frac {\int_{t_1}^{t_2} P(LAMP_2) dt} {-\int_{t_1}^{t_2} P(V_2) dt } = \frac {1.157 J} {1.249 J} = 0.926 = 92.6\%$$

So this confirms that circuit #2 is more efficient than circuit #1. Nice!


Let's assume "brighter" just means "more power delivered to the lamp." In that case we just compare the two energy amounts calculated above, and it's clearly circuit #1. How much brighter?

$$ \frac {1.513 J} {1.157 J} = 1.31$$

LAMP1 will be about 31% brighter than LAMP2.

Very interesting question and engineering tradeoff!

+1 vote
by mrobbins
November 20, 2016

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