Mains Powered

 

Many circuits can be powered directly from the mains with the aid of a series capacitor (C1). The disadvantage of this approach is that usually only one half cycle of the mains wave form can be used to produce a DC voltage. An obvious solution is to use a bridge rectifier to perform full-wave rectification, which increases the amount of current that can be supplied and allows the filter capacitor to be smaller. The accompanying circuit in fact does this, but in a clever manner that uses fewer components. Here we take advantage of the fact that a Zener diode is also a normal diode that conducts current in the forward direction. During one half wave, the current flows via D1 through the load and back via D4, while during the other half wave it flows via D3 and D2. Bear in mind that with this circuit (and with the bridge rectifier version), the zero voltage reference of the DC voltage is not directly connected to the neutral line of the 230-V circuit. This means that it is usually not possible to use this sort of supply to drive a triac, which normally needs such a connection. However, circuits that employ relays can benefit from full-wave rectification.

The value of the supply voltage depends on the specifications of the Zener diodes that are used, which can be freely chosen.

C2 must be able to handle at least this voltage. The amount of current that can be delivered depends on the capacitance of

C1. With the given value of 220 nF, the current is approximately 15 mA. A final warning: this sort of circuit is directly connected to

mains voltage, which can be lethal. You must never come in contact with this circuit! It is essential to house this circuit safely in a suitable enclosure.

source : Elektor Circuit Collections

Simple mV Source

 

This design can be used to simulate millivolt (mV) sensor signals for industrial control systems. Most of the new sensors used to day include some form of ‘intelligence’ at the measurement head, that is, the point at which the sensor comes into contact with what it is to measure. At this point, the sensor signal is conditioned/digitized and fed into a microcontroller that transmits a digital representation of the sensor value to the remote control system. However, there are still a number of ‘elderly’ control systems still in the field that have the intelligence remote from the sensor head. These systems rely on field wiring to convey the measured signal back to the control system.

During commissioning of these types of plant, it is useful to simulate the sensor signal to ensure amongst other things, that the sensor signal gets back to the correct terminals on the control system as they invariably pass through various junction boxes on the way. It can also be used to ensure that the control system operates correctly in response to the sensor signal. The design shown here has been used by the author to ‘bench test’ a control system prior to being installed. Please note that the design is only suitable for simple simulation and is not accurate enough for calibration purposes. Power from a ‘plugtop’ PSU (when bench testing) or a battery is fed to three current sources (diodes). Of these, I1 generates a 1.00 mA current signal, which when switched across the 100-Ω pot creates a 100-mV signal. Likewise, I2 generates a 0.25-mA signal which generates 25 mV across the pot. Current source I3 develops 3.0 mA and is used to illuminate the LED to give a power indication. The selected current source is switched via S2 to the 10-turn pot. Switch S1 is used to cleverly swap the polarity of the output signal. If the Type MTA206PA DPDT switch from Knitter is used, you get a centre-off position which actually shorts out the output signals (S1 pins 2 and 5) together, ensuring a zero output signal.

The current sources, despite being pretty expensive, are not very accurate — they have 10% tolerance! (hence the unsuitability for calibration use). If the output is too high, the tolerance can be ‘trimmed’ by fitting a bleed resistor (R1, R2) as shown in the diagram. The current sources are manufactured by Vishay/Siliconix and stocked by Farnell. The circuit draws a current of about 4.25 mA.

Measuring Inductors

Often you find yourself in the position of needing to wind your own coil for a project, or maybe you come across an unmarked coil in the junkbox. How can you best find out its inductance? An oscilloscope is all you need. Construct a resonant circuit using the coil and a capacitor and connect it to a square wave generator (often part of the oscilloscope itself) Adjust the generator until you find the resonant frequency f. When C is known (1000 pF) the inductance L may be calculated from:

L = 1 / (4π2⋅f2⋅C)

If you are also interested how good the coil is i.e. what is its quality factor or Q, you can use the oscilloscope again. If the level of the damped oscillation drops to 0.37 (= 1/e) of the maximum after about 30 periods, then the Q factor of the coil is about 30

The Q factor should be measured at the intended operating frequency of the coil and with its intended capacitor. The coupling capacitor should by comparison be a much smaller value.

source : Elektor Circuit Collections 

Two Position Dimmer

This super-simple dimmer consists of only two components, and it can easily be built into a mains switch. If you do this, don’t forget to first switch off the associated branch circuit in the fuse box, since the mains voltage is always dangerous! The circuit does not need much explanation. When S1 is closed, the lamp works at full strength, and the position of S2 does not matter. When S1 is open and S2 is closed, the capacitor causes a voltage drop, so the lamp is dimmed. The power dissipation of the capacitor is practically zero, so the circuit does not generate any heat. The resistor prevents sparking when S2 is closed while S1 is already closed. The value of the capacitor can be matched to the power of the lamp to be dimmed; it should be between 2 and 6 µF. Be sure to use a class X2 capacitor. Also, don’t forget that thiscircuit works only with resistive (non-inductive) loads. Unpredictable things can happen with an inductive load!

source : Elektor Circuit Collections 2000-2014

±5-V Voltage Converter

A symmetrical ±5 V power supply is often needed for small, battery-operated operational amplifier projects and analogue circuits. An IC that can easily be used for this purpose is the National Semiconductor LM 2685. It contains a switched capacitor voltage doubler followed by a 5-V regulator. A voltage inverter integrated into the same IC, which also uses the switched-capacitor technique, runs from this output voltage. The external circuitry is limited to two pump capacitors and three electrolytic storage capacitors.

The IC can work with an input voltage between +2.85 V and +6.5 V, which makes it well suited for battery-operated equipment. The input voltage is first applied to a voltage doubler operating at 130 kHz. The external capacitor for this is  connected to pins 13 and 14. The output voltage of this doubler is filtered by capacitor C3, which is connected to pin 12. If the input voltage lies between +5.4 and +6.5 V, the voltage doubler switches off and passes the input voltage directly through to the following +5-V low-dropout regulator, which can deliver up to 50 mA. C4 is used as the output filter capacitor.

All that is necessary to generate the –5-V output voltage is to invert the +5-V voltage. This is done by a clocked power-MOS circuit that first charges capacitor C2, which is connected between pins 8 and 9, and then reverses its polarity. This chopped voltage must be filtered by C5 at the output. The unregulated –5 V output can supply up to 15 mA. The LM 2865 voltage converter IC also has a chip-enable input (CE) and two control inputs, SDP (shut down positive) and SDN (shut down negative). If CE is set Low, the entire IC is switched off (shut down), and its current consumption drops to typically 6 µA. The CE input can thus be used to switch the connected circuit on or off, without having to disconnect the battery. The SDP and SDN inputs can be used to switch the VPSW and VNSW outputs, respectively. These two pins are connected to the voltage outputs via two low-resistance CMOS switches. This allows the negative output to be separately switched off, whereby the voltage inverter is also switched off. Switching off with SDP not only opens the output switch but also stops the oscillator. There is thus no longer any input voltage for the –5 V inverter, so the –5 V output also drops out. The SDP and SDN inputs are set Low (< 0.8 V) for normal operation and High (>2.4 V) for switching off the associated voltage(s).

source : Elektor Circuit Collections 2000-2014