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Voltage feedback

Imagine you connect your 'Lesson 7' power supply to a device that switches a lamp on and off. This device requires 4.75 ... 5.25V supply voltage to operate correctly. When the lamp is off, it draws just 5mA. When the lamp is on, the current flow rises to, say, 1A. You connect the device to the power supply with wires that have a total resistance of 1Ω. You use a volt meter to adjust P2 to an output voltage of 5V. After a while, the lamp switches on. The voltage drop over the wires will now be 1V, leaving just 4V for the device. This is not enough and the device shuts down, turning off the lamp. The device now receives 5V again and the lamp switches on. The supply voltage drops to 4V, and the lamp goes off, and so on, and so on...

The remedy for this is eighter using very thick wires or voltage feedback. When you want to permanently connect the device to a power supply, the first option is probably the best, especially when the wires are short. In this chapter however, we want to build a lab power supply. You may already have noticed R14. This is a 0.5Ω current sense resistor which will be discussed later in this chapter. This means that even if we use thick wires, the voltage drop will always be at least 0.5V per ampere.

In the schematic above the feedback terminals are FB+ and FBGND. R19 and R16 devide the actual output voltage by 2 and R13 feeds it back to the non-inverting pin of opamp U2. The inverting pin of U2 is connected to P2 via R17. U3 is a cheap 15V voltage stabalizer. To keep it as stable as possible, it is fed by separate transformer winding (or a separate transformer), bridge rectifier (G1) and smoothing electrolytic (C3). So Vp=0.5∙VOUT and Vn ranges from 0 ... 15V. Lesson 9 tought us that if Vp > Vn, U2's output is 15V. T5 will be saturated and the base voltage of the driver darlingtom T3/T6 will be 0V. The load at the output terminals will discharge C2 until Vp < Vn. At that moment, U2's output will become 0V, T5 opens, and a current starts to flow in R4 and the bases of T3 and T6. This current will saturate the darlington. A large current will charge C2 and feed the load. This will continue until Vp > Vn. At that point everything starts all over again. Again we see that opamp U2 tries to keep Vp equal to Vn, just like an opamp amplifier. In matter of fact, U2 is used as an amplifier: it amplifies the voltages over P2 (=Vn) 2 times (because R19 and R16 devide the output voltage by 2). Since Vn ranges from 0 ... 15V, VOUT will range from 0 ... 30V.

Without R21 and R22, we should not turn on the power supply until FB+ and FBGND were both connected! If FB+ and FBGND were disconnected, U2 might have thought that the output voltage were too low, and a voltage of 40V or more might have appeared across the output terminals!

When T5 closes, there may still be up to 30V across C2. Since The base voltage of T3 is 0V, there will be -30V between the base of T3 and the emitter of T6. However, T6 cannot withstand voltages less than -7V. For T3, VBE must be greater than -60V. R5 and R20 devide the -30V, so that T6 will survive.

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