Diagram Plus guide

It is all too often necessary to augment the power supply of an existing electronic circuit because exactly the voltage that you need is missing. The circuit presented here may provide a solution in a number of cases, since it can be used to convert a single-ended supply voltage into a balanced set of supply voltages. That’s not so remarkable by itself, but the special feature of this circuit is that this is accomplished without using difficult to obtain, exotic ICs. All of the components used in the circuit are ones that every electronics hobbyist is likely to have in a drawer somewhere.

The heart of the circuit is formed by an ‘old reliable’ 555 timer, which is wired here as a free-running oscillator with a frequency of approximately 160 kHz. The oscillator is followed by two voltage-doubling rectifiers, consisting of C1, D1, D2, C3 and C7, D3, D4, C5. They are followed in turn by two voltage regulators to stabilise the positive and negative voltages generated in this manner. The duty cycle of the 555 is set to approximately 50 percent using R1 and R2. The square-wave signal at the output of the timer IC has a DC offset, which is eliminated by C4 and R3.

The amplitude of the output signal from the 555 is approximately equal to the supply voltage less 1.5 V, so with a 12-V input voltage, there will be a square-wave signal on pin 3 with an amplitude of approximately 10.5 Vpp. With respect to ground (across R3), this is this +5 V / –5 V. Although this yields a symmetric voltage, its positive and negative amplitudes are somewhat too small and it is not stabilized. In order to split the square-wave signal into sufficiently large positive and negative amplitudes, C1/D2 are added for the positive voltage, causing the positive half to be doubled in amplitude.

For the negative half, the same effect is achieved using C7/D3. Following this, the two signals are smoothed by D1/C3 and D4/C5, respectively. Both voltages are now high enough to be input to normal 5-V voltage regulators, yielding symmetric +5V and –5V supply voltages at the output. The input voltage does not have to be regulated, although it must lie between +11V and +18V. The maximum output current is ±50mA with an input voltage of 12V. This circuit is an excellent choice for generating auxiliary voltages, such as supply voltages for low-power opamps. Naturally, the fact that the converter can be powered from the in-vehicle voltage of a car is a rather attractive feature.

By configuring a comparator and a transistor to control the oscillator in a charge pump circuit, you enable the pump to generate a regulated output of in principle any desired value. Charge pump ICs can either invert or double an input voltage (for example, 3 V to –3 V or 3 V to 6 V). The charge pump itself does not regulate the output voltage and one running off 3 V is not normally capable of generating intermediate output voltage levels like 5 V. However, by adding a comparator and a reference device, you can create arbitrary output levels like 5 V and regulate them as well. Charge pump IC1 (a MAX660) has an internal oscillator whose 45 kHz operation transfers charge from C1 to C2, causing the regulated output to rise.

When the feedback voltage (pin 3 of IC2) exceeds 1.18 V, the output of comparator IC2 (a MAX921) goes high, turning off the oscillator via T1. The comparator hysteresis (easily added on IC2) is zero here simply because no hysteresis is required in the control loop. The oscillator when enabled generates two cycles, which is sufficient to drive VOUT slightly above the desired level. Next, the feedback turns the oscillator off again. The resulting output ripple will depend mainly on the input voltage and the output load current. Output ripple may be reduced at the expense of circuit efficiency by adding a small resistor (say, 1 ?) in series with C1. You’ll find that ripple also depends on the value and ESR associated with C1 - smaller values of C1 transfer less charge to C2, producing smaller jumps in V OUT.

Most small internal-combustion engines commonly used in the model-building world use glow plugs for starting. Unfortunately, glow plugs have an operating voltage of 1.5 V, while fuel pumps, starter motors, chargers and the like generally run on 12 V. This means that a separate battery is always needed to power the glow plug. The standard solution is to use an additional 2-V lead storage battery, with a power diode in series to reduce the voltage by approximately 0.5 V. However, this has the annoying consequence that more than 30 percent of the energy is dissipated in the diode. Naturally, this is far from being efficient. The converter presented here allows glow plugs to be powered from the 12-V storage battery that is usually used for fuelling, charging, starting and so on.

A car battery can also be used as a power source. Furthermore, this circuit is considerably more efficient than the approach of using a 2-V battery with a series power diode. The heart of the DC/DC converter is IC1, a MAX 1627. The converter works according to the well-known step-down principle, using a coil and an electrolytic capacitor. Here the switching stage is not integrated into the IC, so we are free to select a FET according to the desired current level. In this case, we have selected a 2SJ349 (T1), but any other type of logic-level FET with a low value of RDSon would also be satisfactory. Of course, the FET must be able to handle the required high currents. Diode D1 is a fast Schottky diode, which must be rated to handle the charging currents for C2 and C3. This diode must also be a fairly hefty type. The internal resistances of coil L1 and capacitors C2 and C3 must be as low as possible. This ensures efficient conversion and prevents the components from becoming too warm. The resistor network R2/R3 causes 87 percent of the output voltage to be applied to the FB pin of IC1.

This means that an output voltage of 1.5 V will cause a voltage of approximately 1.3 V to be present at the FB pin. The IC always tries to drive the switching stage such that it ‘sees’ a voltage of 1.3 V on the FB input. If desired, a different output voltage can be provided by modifying the values of R2 and R3. When assembling the circuit, ensure that C5 and C1 are placed as close as possible to IC1, and use sufficiently heavy wiring between the 12-V input and the 1-5-V output, since large currents flow in this part of the circuit. A glow plug can easily draw around 5 A, and the charging current flowing through the coil and into C2 and C3 is a lot higher than this!