Schematic Reference

However PAL sync pulses arriving every 20ms will allow the Q output to go low after 17.9ms, before being triggered high again 2.1ms later. Thus an NTSC signal will give a constant high output while a PAL signal will result in a train of pulses 2.1ms wide. The Q output from IC2a is fed to the inverting input of IC2b, the second monostable, which has a period of about 0.5s, as set by the 270kO resistor and 4.7µF tantalum capacitor at pins 6 & 7. With its input constantly high, resulting from an NTSC signal, IC2b is not triggered and its Q output remains low.
However, the pulse train from a PAL signal will constantly retrigger it, so its Q output will remain high. The period of IC2b also effectively makes it a low-pass filter which removes spurious switching due to any input glitches. The output signal is taken from the Q-bar (inverted) output, so that an NTSC signal gives a high output, while PAL gives low. For the particular application for which the circuit was developed, diode D1 and the resistor network shown drive the base of an NPN switching transistor and relay. A dual-colour 3-lead LED can also be fitted to indicate NTSC (red) or PAL (green). Note that with no video input, the output signal is high and will indicate NTSC.

As the mark/space ratio (duty factor) of the 555 output is a long way from being 1:1 (50%), it is used to drive a D-type flip-flop produced using a CMOS type 4013 IC. This produces perfect complementary square-wave signals (i.e. in antiphase) on its Q and Q outputs suitable for driving the output power transistors. As the output current available from the CMOS 4013 is very small, Darlington power transistors are used to arrive at the necessary output current. We have chosen MJ3001s from the now defunct Motorola (only as a semi-conductor manufacturer, of course!) which are cheap and readily available, but any equivalent power Darlington could be used.
These drive a 230 V to 2 × 9 V center-tapped transformer used ‘backwards’ to produce the 230 V output. The presence of the 230 VAC voltage is indicated by a neon light, while a VDR (voltage dependent resistor) type S10K250 or S07K250 clips off the spikes and surges that may appear at the transistor switching points. The output signal this circuit produces is approximately a square wave; only approximately, since it is somewhat distorted by passing through the transformer. Fortunately, it is suitable for the majority of electrical devices it is capable of supplying, whether they be light bulbs, small motors, or power supplies for electronic devices.
PCB layout:

COMPONENTS LIST
Resistors
R1 = 18k?
R2 = 3k3
R3 = 1k
R4,R5 = 1k?5
R6 = VDR S10K250 (or S07K250)
P1 = 100 k potentiometer
Capacitors
C1 = 330nF
C2 = 1000 µF 25V
Semiconductor
T1,T2 = MJ3001
IC1 = 555
IC2 = 4013
Miscellaneous
LA1 = neon light 230 V
F1 = fuse, 5A
TR1 = mains transformer, 2x9V 40VA (see text)
4 solder pins
Note that, even though the circuit is intended and designed for powering by a car battery, i.e. from 12 V, the transformer is specified with a 9 V primary. But at full power you need to allow for a voltage drop of around 3 V between the collector and emitter of the power transistors. This relatively high saturation voltage is in fact a ‘shortcoming’ common to all devices in Darlington configuration, which actually consists of two transistors in one case. We’re suggesting a PCB design to make it easy to construct this project; as the component overlay shows, the PCB only carries the low-power, low-voltage components.
The Darlington transistors should be fitted onto a finned anodized aluminum heat-sink using the standard insulating accessories of mica washers and shouldered washers, as their collectors are connected to the metal cans and would otherwise be short-circuited. An output power of 30 VA implies a current consumption of the order of 3 A from the 12 V battery at the ‘primary side’. So the wires connecting the collectors of the MJ3001s [1] T1 and T2 to the transformer primary, the emitters of T1 and T2 to the battery negative terminal, and the battery positive terminal to the transformer primary will need to have a minimum cross-sectional area of 2 mm2 so as to minimize voltage drop.
The transformer can be any 230 V to 2 × 9 V type, with an E/I iron core or toroidal, rated at around 40 VA. Properly constructed on the board shown here, the circuit should work at once, the only adjustment being to set the output to a frequency of 50 Hz with P1. You should keep in minds that the frequency stability of the 555 is fairly poor by today’s standards, so you shouldn’t rely on it to drive your radio-alarm correctly – but is such a device very useful or indeed desirable to have on holiday anyway? Watch out too for the fact that the output voltage of this inverter is just as dangerous as the mains from your domestic power sockets.
So you need to apply just the same safety rules! Also, the project should be enclosed in a sturdy ABS or diecast so no parts can be touched while in operation. The circuit should not be too difficult to adapt to other mains voltages or frequencies, for example 110 V, 115 V or 127 V, 60 Hz. The AC voltage requires a transformer with a different primary voltage (which here becomes the secondary), and the frequency, some adjusting of P1 and possibly minor changes to the values of timing components R1 and C1 on the 555.

Here is a simple but inexpensive inverter for using a small soldering iron (25W, 35W, etc) In the absence of mains supply. It uses eight transistors and a few resistors and capacitors. Transistors Q1 and Q2 (each BC547) form an astable multivibrator that produces 50Hz signal. The complementary outputs from the collectors of transistors Q1 and Q2 are fed to pnp Darlington driver stages formed by transistor pairs Q3-Q5 and Q4-Q6 (utilising BC558 and BD140). The outputs from the drivers are fed to transistors Q7 and Q8 (each 2N3055) connected for push-pull operation.  Use suitable heat-sinks for transistors Q5 through Q8. A 230V AC primary to 12V-0-12V, 4.5A secondary transformer (T1) is used.

The centre-tapped terminal of the secondary of the transformer is connected to the battery (12V, 7Ah), while the other two terminals of the secondary are connected to the collectors of power transistors T7 and T8, respectively. When you power the circuit using switch S1, transformer X1 produces 230V AC at its primary terminal. This voltage can be used to heat your soldering iron. Assemble the circuit on a generalpurpose PCB and house in a suitable cabinet. Connect the battery and transformer with suitable current-carrying wires. On the front panel of the box, fit power switch S1 and a 3-pin socket for connecting the soldering iron. Note that the ratings of the battery, transistors T7 and T8, and transformer may vary as these all depend on the load (soldering iron).