Let us return to the camera flash circuit, redrawn in Fig. 7.25, which was discussed in the introduction of this chapter. The Xenon flash has the following specifications: Voltage required for successful flash: { minimum 50 V maximum 70 V Equivalent resistance: 80 Ω For this particular

Question

Let us return to the camera flash circuit, redrawn in Fig. 7.25, which was discussed in the introduction of this chapter. The Xenon flash has the following specifications:

Voltage required for successful flash: [latex]\{_{maximum\,70\,V}^{ minimum \,50 \,V}[/latex]

Equivalent resistance: 80 Ω

For this particular application, a time constant of 1 ms is required during flash time. In addition, to minimize the physical size of the circuit, the resistor [latex]R_{1}[/latex] must dissipate no more than 100-mW peak power. We want to determine values for [latex]V_{S}, C_{F}[/latex], and [latex]R_{1}[/latex]. Furthermore, we wish to determine the recharge time, the flash bulb’s voltage, current, power, and total energy dissipated during the flash.

Accepted Answer

We begin by selecting the source voltage, [latex]V_{S}[/latex]. Since the capacitor is applied directly to the Xenon bulb during flash, and since at least 50 V is required to flash, we should set [latex]V_{S}[/latex] higher than 50 V. We will somewhat arbitrarily split the difference in the bulb’s required voltage range and select 60 V for [latex]V_{S}[/latex].
Now we consider the time constant during the flash time. From Fig. 7.25, during flash, the time constant is simply

[latex] au_{F} = R_{B} C_{F}[/latex] 7.22

Given [latex]t_{F}[/latex] = 1 ms and [latex]R_{B}[/latex] = 80 Ω, we find [latex]C_{F}[/latex] = 12.5 μF.
Next, we turn to the value of [latex]R_{1}[/latex]. At the beginning of the charge time, the capacitor voltage is zero and both the current and power in [latex]R_{1}[/latex] are at their maximum values. Setting the power to the maximum allowed value of 100 mW, we can write

[latex]P_{Rmax} = \frac{V^{2}_{S}}{R_{1}}= \frac{3600}{R_{1}}[/latex] = 0.1 7.23

and find that [latex]R_{1}[/latex] = 36 kΩ. The recharge time is the time required for the capacitor to charge from zero up to at least 50 V. At that point the flash can be successfully discharged. We will define t = 0 as the point at which the switch moves from the bulb back to [latex]R_{1}[/latex]. At t = 0, the capacitor voltage is zero, and at t = [latex]\infty[/latex], the capacitor voltage is 60 V; the time constant is simply [latex]R_{1} C_{F}[/latex]. The resulting equation for the capacitor voltage during recharge is

[latex]v_{CF}(t) = K_{1} + K_{2} e^{-t/ au} = 60 - 60e^{-t/R_{1} C_{F}}[/latex] V 7.24

At t= [latex]t_{charge} , v_{CF}[/latex](t) = 50 V. Substituting this and the values of [latex]R_{1}[/latex] and [latex]C_{F}[/latex] into Eq. (7.24) yields a charge time of [latex]t_{charge}[/latex] = 806 ms—just less than a second. As a point of interest, let us reconsider our choice for [latex]V_{S}[/latex]. What happens if [latex]V_{S}[/latex] is decreased to only 51 V? First, from Eq. (7.23), [latex]R_{1}[/latex] changes to 26.01 kΩ. Second, from Eq. (7.24), the charge time increases only slightly to 1.28 s. Therefore, it appears that selection of [latex]V_{S}[/latex] will not have much effect on the flash unit’s performance, and thus there exists some flexibility in the design.
Finally, we consider the waveforms for the flash bulb itself. The bulb and capacitor voltage are the same during flash and are given by the decaying exponential function

[latex]v_{B}(t) = 60e^{-1000t}[/latex] V 7.25

where the time constant is defined in Eq. (7.22), and we have assumed that the capacitor is allowed to charge fully to [latex]V_{S}[/latex] (i.e., 60 V). Since the bulb’s equivalent resistance is 80 Ω, the bulb current must be

[latex]i_{B}(t) = \frac{60e^{-1000t}}{80} = 750e^{-1000t}[/latex] mA 7.26

As always, the power is the v-i product.

[latex]p_{B}(t) = v_{B}(t)i_{B}(t) = 45e^{-2000t}[/latex] W 7.27

Finally, the total energy consumed by the bulb during flash is only

[latex]w_{B}(t) = \int_{0}^{\infty}{p_{B}(t) dt} =\int_{0}^{\infty}{45e^{-2000t} dt} =\frac{45}{2000}e^{-2000t}\mid_{\infty}^{0} = \frac{45}{2000}[/latex] = 22.5 mJ 7.28

Question 7.12: Let us return to the camera flash circuit, redrawn in Fig. 7...

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