An electronics hobbyist who has built his own stereo amplifier wants to add a back-lit display panel to his creation for that professional look. His panel design requires seven light bulbs—two operate at 12 V/15 mA and five at 9 V/5mA. Luckily, his stereo design already has a quality 12-V dc

Question

An electronics hobbyist who has built his own stereo amplifier wants to add a back-lit display panel to his creation for that professional look. His panel design requires seven light bulbs—two operate at 12 V/15 mA and five at 9 V/5 mA. Luckily, his stereo design already has a quality 12-V dc supply; however, there is no 9-V supply. Rather than building a new dc power supply, let us use the inexpensive circuit shown in Fig. 2.49a to design a 12-V to 9-V converter with the restriction that the variation in [latex]V_{2}[/latex] be no more than ±5% In particular, we must determine the necessary values of [latex]R_{1}[/latex] and [latex]R_{2}.[/latex]

Accepted Answer

First, lamps [latex]L_{1}[/latex] and [latex]L_{2}[/latex] have no effect on [latex]V_{2}[/latex]. Second, when lamps [latex]L_{3}–L_{7}[/latex] are on, they each have an equivalent resistance of

[latex]R_{eq}\,=\,\frac{V_{2}}{I}\,=\,\frac{9}{0.005}[/latex] = 1.8 kΩ

As long as [latex]V_{2}[/latex] remains fairly constant, the lamp resistance will also be fairly constant. Thus, the requisite model circuit for our design is shown in Fig. 2.49b. The voltage [latex]V_{2}[/latex] will be at its maximum value of 9 + 5% = 9.45 V when [latex]L_{3}–L_{7}[/latex] are all off. In this case [latex]R_{1}[/latex] and [latex]R_{2}[/latex] are in series, and [latex]V_{2}[/latex] can be expressed by simple voltage division as

[latex]V_{2}[/latex] = 9.45 = 12 [latex]\left[\frac{R_{2}}{R_{1}+ R_{2}}\right] [/latex]

Rearranging the equation yields

[latex]\frac{R_{1}}{R_{2}}[/latex] = 0.27

A second expression involving [latex]R_{1}[/latex] and [latex]R_{2}[/latex] can be developed by considering the case when [latex]L_{3}–L_{7}[/latex] are all on, which causes [latex]V_{2}[/latex] to reach its minimum value of 9–5%, or 8.55 V. Now, the effective resistance of the lamps is five 1.8-kΩ resistors in parallel, or 360 Ω The corresponding expression for [latex]V_{2}[/latex] is

[latex]V_{2}[/latex] = 8.55 = 12[latex]\left[\frac{R_{2}//360}{R_{1}+ (R_{2}//360)}\right] [/latex]

which can be rewritten in the form

[latex]\frac{\frac{360R_{1}}{R_{2}}+\,360\,+\,R_{1}}{360}\,=\,\frac{12}{8.55}[/latex] = 1.4

Substituting the value determined for [latex]R_{1}/R_{2}[/latex] into the preceding equation yields

[latex]R_{1}[/latex] = 360[1.4 - 1 - 0.27]

or

[latex]R_{1}[/latex] = 48.1 Ω

and so for [latex]R_{2}[/latex]

[latex]R_{2}[/latex] = 178.3 Ω

Question 2.34: An electronics hobbyist who has built his own stereo amplifi...

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