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Wednesday, September 24, 2014

Removing Nokia 6555b Logic Board >>












Electronics Circuit Application
pls: send your name to a comet

Tuesday, September 23, 2014

Regulated 5v Solar Power Supply Circuit >>

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Electronics Circuit Application

Practical Wifi Antenna circuit >>

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Electronics Circuit Application

Thursday, September 18, 2014

Huawei HSDPA usb modem service ..

                                                               Electronics Circuit Application

Wednesday, September 17, 2014

L200 Using Battery charger circuit



Circuit diagram with Parts list




Description.

A very simple battery charger circuit having reverse polarity indication is shown here.The circuit is based on IC L200 . L200 is a five pin variable voltage voltage regulator IC.The charging circuit can be fed by the DC voltage from a bridge rectifier or center tapped rectifier.Here the IC L200 keeps the charging voltage constant.The charging current is controlled by the parallel combination of the resistors R2 & R3.The POT P1 can be used to adjust the charging current.This circuit is designed to charge a 12 V lead acid battery.The transistor t1,diode D3 and LED are used to make a battery reverse indicator.In case the battery is connected in reverse polarity ,the reverse polarity indicator red LED D5 glows.When the charging process is going on the battery charging indicator green LED D4 glows.

Notes.

    The circuit can be assembled on a good quality PCB or common board.

    The values of R2 & R3 can be obtained from the equation,

(R2//R3) =( V5-2)/(Io).

Where V5 is  the charging voltage (voltage at pin 5) and Io is the charging current.

    The POT R8 can be used for fine adjustments of charging current.

    If battery is connected in reverse polarity the RED LED will glow.

    When the  charging is going on the GREEN LED will glow.

    The rectified input voltage to the charger can be 18V.
 

 Electronics Circuit Application

Thursday, September 11, 2014

Samsung Galaxy S i9000 disassembly >>Repair guides

If you want to buy a replacement screen, screwdrivers and other parts, you can order from us here: Samsung Galaxy S i9000 screen .Electronics Circuit Application

Tools required for disassembly: Phillips screwdriver, plastic tool such as a guitar plectrum or credit card.

    Remove your back cover, battery, sim and memory card. Seven screws need to be removed. The screw circled in white is a different size to the others so make sure it goes back in the right place.
1


Run something like a credit card around the join of the battery compartment to release it from the rest of the phone.>>>>Electronics Circuit Application

2


With the battery compartment removed your phone should look like below. A number of ribbon connectors (circled) need to be disconnected. Each attaches like a plug and socket. Just lever each one up to unplug. The power button, indicated by the arrow, is stuck to the side of the phone's housing and should be carefully lifted away, being careful not to damage it. Do not try to disconnect it from the circuit board you just need to make sure it's not stuck to the side of the phone. You'll now be able to remove the circuit board simply by lifting it away from the rest of the phone.

3


With the circuit board removed your phone should look like below. The hands free socket and earpiece, circled below, can now be lifted away from the phone. The part is held in place with a little double sided tape and will peel away with little force.
4


With the hands free socket and earpiece removed your phone should look like below. The ringer, circled below left, needs to be levered away from the phone. It's held in place with double sided tape. Just use your screwdriver to carefully lever it up. It will still be connected by two wires. Be careful not the damage these two wires. Two more ribbon connectors (circled centre and right) also need to be disconnected. Each attaches like a plug and socket. Just lever each one up to unplug.

5

Now slide something like a credit card between the front and back of the phone and carefully lever the phone apart.

6


Your phone should now look like below. The next stage is to remove the screen. To do so first heat the front of the screen around its edge with a hot air gun or hairdryer. Then push the back of the screen from the point indicated by the arrow. The screen should then raise slightly on the other side, enough for your to slide your plastic tool under its edge and separate it from the silver frame.

7

With the screen assembly removed your phone should look like below. Now simply replace your damaged screen with new and reverse the whole procedure for reassembly. If you've found this disassembly guide useful, please support us and order the tools and parts you need from our site. We ship worldwide!
Electronics Circuit Application

Starting a fluorescent on an inverter circuits


inverter circuits


The 12V drivers for fluorescent lamps are tricky, because of the compromise between a good operating efficiency and the ability to start the lamp. The demands are in contradiction. Here are 7 starting methods:

1.1 Pure flyback single-transistor inverter, makes voltage spikes in the kilovolt range in idling condition, so that the lamp will start. Disadvantage: One filament evaporates and blackens the lamp, makes it electrically unsymmetric and shortens the lifetime. High level of radiation (EMC).

1.2 Heating the filaments by suitable transformer windings. The problem is that this heating power suppresses the voltage spikes. You get only either one of these two starting aids. But an additional single big voltage spike, (for example manually shortening the secondary winding just for a few milliseconds by pushbutton) will start the lamp.

1.3 Ionization wire along the lamp. This wire acts only where the potential difference is, it ionizes around the opposite voltage filament. We can regard this measure as a cheap and easy additional trick, but it does not make a breakthrough, is not alone able to start.

1.4 Ionization wire plus additional special high-voltage winding. This is indeed a full starter. The winding can be switched off in operation, or can be supplied by a separate starting oscillator, which is switched of as a whole. Possible, promising, but unusual.

1.5 Starter circuit on the lamp side. Such one is described in the book of Nührmann, “Professionelle Schaltungstechnik” Issue 2 Page 180. He used a rather exotic tyristor tetrode BRY20 to switch the filaments in series to the whole secondary winding for effective pre-heating. When the lamp has started, the tetrode is switched off. The oscillator is a sine-mode oscillator and the idling voltage is high enough for start, but does not contain spikes. The well-known glow-start cartridge is not applicable for small high-frequency inverters, it is for grid frequency and inductive ballast only.

1.6 Referring to the schematic given here, if one would switch off the capacitor 0,68uF (in parallel to the primary winding), the oscillator would be no longer in sine-mode, but in flyback mode and produce voltage spikes which immediately start the fluorescent. It would need either a manual start, such as a pushbutton with opening contact (NC), or an external relay circuit performing the same action automatically.

1.7 The secondary winding is made to produce a rather high voltage but without spikes. A capacitor is put in series with the secondary winding. At first, the high idling frequency and voltage pass to the lamp rather directly, as it is electrically inactive. After start, the voltage collapses from several hundred volts to the operating voltage, which is ca 70V for an 8W rod. Ionization wires can be applied in addition. Disadvantage is a certain loss in efficiency, as we (very simplified) at first generate a high voltage and then use only a part of it. The advantage is a self-start action without manual pushbuttons or external special starting circuits.
inverter circuits


It works according to the starting method of 1.7 as explained above.

The transistor has a resonant circuit in the emitter line and operates in common-collector mode. This needs a feedback voltage higher than the operating voltage, and so the feedback winding has a higher number of turns than the primary winding.

I happened to have some transistors, which outperformed all others, but they are very exotic PNP complementary types of an even more exotic 2SC1306 high-frequency power amplifier. I have tested also BD249C, they work well. According to data sheets, also a 2N 4923 should work.
Look for fast switch transistors, current 2A, able to handle high frequencies. Size TO-220 or larger.

The resistor made of 2 x 1kOhm in parallel was formerly a single one of 470 Ohm, but became a bit too hot for a long-term reliable operation, so I used 2 x 1 kOhm = 500Ohm just for the thermal capacity.

The capacitor 1uF 50V could certainly be a tantalum type, and possibly also a high-quality electrolytic one. Its purpose is to generate a voltage divider for high-frequency but in the first moment of start (after taking its charge) it makes the 500 Ohm resistor dominant for oscillator start.

The capacitor 0,68uF 400V is stressed with 12 to 30 kHz and in order to avoid undue dielectric losses it is a good idea to select a generous voltage rating. But probably 160V rating would also do it here.
The secondary capacitor is the most critical part. In idling mode, it passes ca 700V (peak/peak) over to the fluorescent, in operation it is stressed with over 200V at 12 kHz. The dielectric material has to be excellent, or we need to take care, that there is a generous margin in voltage rating.
Ceramics is not suitable. FKP-types and STYROFLEX (polystyrene) are good and could possibly be rated at 400V AC. All other more common types such as MKP or “no name” should be rated at over 1000V.
I used 2 x 5,6nF 1.5 kV in parallel, resulting in 11,2nF. This value is not critical, but influences the lamp current slightly. A good choice would be 2 x 22nF 400V (or 630V) in series. If they become warm in operation, they are overloaded.

3. Preparing the lamp

With a little bit of handwork we get improved starting performance. A normal ionization wire looks like this:

inverter circuits

You find it in hand-held 12V lamps for workshops or your car. It acts only where there is a notable potential difference, and is inactive close to where it is connected. So it works only on one end of the lamp.

I use here a cross-over double ionization. Two thin wires of 0,1mm diameter are glued to the glass rod with transparent silicone. They are connected to the aluminum rings which surround the ends of the tube. The aluminum rings generate a high electric flied strength relative to the filaments close by, as they are on opposite potential. The two wires along the tube ionize all along the tube. So we act all over the lamp with this arrangement. This is very effective and helps starting according to method 1.7 without any further means.

inverter circuits


The aluminum rings can be soldered on a small spot by applying a thick drop of acid-free flux, scraping off the oxide layer inside the drop with a sharp screwdriver, then soldering by using a rubbing action of the tip. The ionization wires should be around 0,1mm in diameter, not to take too much light, and you can win such wires by stripping a thin flexible wire.

4. Transformer

Transformer, insulation and windings

Two parallel flat ferrite rods of length 50mm, cross-section each 3,5 x 10mm
Overall ferrite cross section is 70mm2. Volume is 3500mm3
The insulation material is paper of 0,04mm thickness, boiled in a mixture of paraffin /stearin wax. Its long-term insulation property is roughly 5 kV/mm.

    Four layers of wax paper = 0,16mm insulation around the core (800V)
    Copper wire 0,8mm diameter 24 turns = primary
    Two layers of wax paper = 0,08mm (400V)
    Copper wire 0,4mm diameter 35 turns = feedback
    Six layers of wax paper = 0,24mm (1200V)
    Copper wire 0,2mm diameter 130 turns, secondary. Tap connection
    Layer insulation 2 turns of paper = 0,08mm (400V)
    Copper wire 0,2mm diameter 130 turns, second tap connection
    Layer insulation 2 turns of paper = 0,08mm (400V)
    Copper wire 0,2mm diameter 130 turns
    Layer insulation 2 turns of paper = 0,08mm (400V)
    Copper wire 0,2mm diameter 130 turns, end of secondary, overall 520 turns
    Four layers of wax paper = 0,16mm (800V against mechanical fixations)
    Binding with sewing thread

The whole thing is slowly dipped into nearly boiling hot wax, ca 90C. It is important not to secure anything inside the transformer with normal glue. This would be destroyed by the heat and the windings might move or fall apart. You can use small blobs of bathtub silicone to secure the windings, or bind with cotton sewing thread. The arrangement is left in the wax, until there are no more bubbles rising, which indicate evaporating humidity or air. Then it is slowly pulled out and left to cool down.

The secondary winding has now 260 / 390/520 turns accessible. This allows tapping the right one for the lamp. The 390 turns worked well, but 520 turns gave a little bit better starting performance, so I used the full winding. There are no separate filament heater windings, as we use starting method 1.7 here. If we would use method 1.2 we would use notably less secondary turns, but add two very well insulated heater windings for ca 7-10V each.

The transformer is generously insulated, as you see from the table above. It was the aim to make it proof against idling operation over a longer time, for example a non-connected or defect lamp.

5. Some calculations for fun

Peak induction of the ferrite core?
The voltage on the primary winding of 24 turns can be coarsely assumed to be a sine wave slightly over 10V peak = 7,8V effective. The frequency in operation is ca 12 kHz. This equation here is valid only for sine-wave, but for all frequencies:
U = 4,44 * Bmax * A * f * n
(U= effective applied voltage, Bmax= peak induction in the iron core, A= cross-section of the iron, f= frequency and n= number of turns). Turned around, we find:
Bmax = U / (4,44 * A * f * n)
Bmax = 7,8 / (4,44* 0,00007 * 12000 * 24) (dimensions in V, m2, Hz)
Bmax = 0,09 T
rounded to Bmax = 0,1 T
This is a rather low value for ferrite material, these antenna rods could be loaded with roughly 0,2 T, so it is in the conservative but acceptable range.

Iron losses?

These can only be very roughly estimated, as the exact ferrite sort is not known and the tables are usually for rectangular waveform. Some extrapolations of diagrams of similar materials indicate a specific loss around 20mW/cm3, and as we have 3,5 cm3, it would be an iron loss of Pe= 70mW or 0,070W

Resistive losses in the copper wire?

Average length of one secondary turn L = 48mm, number of turns ns = 520, total length comes to
Ls=24960mm or roughly 25,0m
Cross-section of the wire of D = 0,2mm is As = 0,0314mm2
Resistance Rs = 0,019 *Ls / As this rule of thumb assumes the cross-section in mm2 (!)
Rs = 0,019 * 25/0,0314 = 15,1 Ohm
estimated effective current 30mV / 0,25Ohm = 0,12A (see oscilloscope picture)
Resistive losses of the secondary winding
Prs = I^2 * R = 0,22 W
Similarly we find for the primary winding:
Average length of one winding = 36,1mm, total length = 867mm, cross-section = 0,502 mm2, resistance = 0,033 Ohm, current while conducting 2,8A, losses while conducting 0,26W, losses averaged 0,074W or
Prp = 74 mW.

Transistor voltage drop losses?

The average current is 0,8A. The transistor conducts in 24 usec within a full period of 83 usec. In the moment of conduction, it thus carries I = 0,8 A * (83/24) = 2,77 A
During this time, the Collector-Emitter voltage drop is in average 0,45 V and the power loss is 1,25W.
Again, averaged over the whole time, it comes to Pce = 0,36 W

Transistor switching losses?

Too difficult to derive from these simple measurements. It would need an analysis of both switching on and off, then correlating the momentary current and CE-voltage drop.

Transistor base current losses?

In this configuration they are passing the “useful” primary winding, and thus can be neglected.

Sum of the above estimated / calculated losses:
Iron losses     70 mW
Copper losses primary side     74 mW
secondary side     220 mW
Transistor voltage drop     360 mW
Transistor switching     360 mW (just guessed)
Capacitor dielectric losses     500 mW (estimate based on temperature rise)
Base resistor losses     280 mW
Sum     1,86W

If the circuit takes 0,8A at 12 V, the input power is 9,6 W.

Assumed that the inverter takes 1,9 W for itself, the lamp gets 7,7W which is close to the rated 8W.
The efficiency would be 0,80 or 80%. This is all very coarse but shall show the way how to approach.
A push-pull inverter with FET’s could be more efficient, but would need more parts and a special trick for starting the lamp.

The best method of measuring the efficiency of such devices is to calculate the overall thermal inertia of the whole device and put it into a styrofoam box. Then operate it, for example, rising from 20C to 40C, measuring the time. The loss power can then be calculated. On the DC input side, it can be accurately measured. Based on both values you can calculate the efficiency.

6. The “Manhattan” method of making a PCB

This is a nice method if you don’t want to make the whole etching process and have relatively few parts. You just make a layout of where to put which part, then make pads and glue them on the copper side of a fresh PC board. Advantages: you have a ground-potential shield over the whole area, and you can connect to ground with relatively short impedances.


inverter circuits


I dont recommend nibbling, which causes cracks and delaminations. I made a clean sawcut, deburred and bevelled the pads with sandpaper, and glued with heat-resistant glue (maybe overdone).
7. Oscilloscope pictures
(drawn by hand, old analog scope)
Betrieb-1
If you are young and have good ears, you might hear 12 kHz, coming from the transformer. In this case the circuit needs to be tuned to over 16 kHz or needs to be put into a casing.
The Base-Emitter voltage is about 1 V, indicating that the transistor is fully saturated and really in switch-mode.
The transformer output is somehow close to a sine wave, but the capacitor and lamp form a kind of high-pass and cause a rather odd lamp voltage.
Looking closely I could detect a 900 kHz oscillation, just at the limit what the oscilloscope could show. I think it is not critical, but if you operate the device close to radios, TV’s, mobile phones, maybe good to put the device in a metal casing and use a grounded reflector behind the lamp.
Betrieb-2
The lamp current does not correlate in shape with the lamp voltage. This is normal, as the lamp is a strongly non-resistive element, and it has its own impedance.
The transistor is switched fully into saturation. Usually everybody expects a voltage drop of 0,7V but this is a value based on the maximum permitted current of the transistor. If you stay below and apply exactly the correct base current, the CE voltage drop may be around 0,3V. This usually speaks for using a larger transistor than necessary for thermal reasons, and to load this rather large one with a moderate current.
The simple circuit does not provide really fast switching. 1usec would be ok, here it seems to be around 2.
Leerlauf
The idling voltage is 700Vpp and the transformer voltage in operation was ca 400 Vpp, there is a voltage drop on the transformer of ca 300Vpp. But this is not “lost” voltage, it is just voltage drop over the inductivity of this open-core transformer. Its leakage flux inductivity is high, the voltage is “soft”. Unloaded, the frequency is 31 kHz, dropping down to ca 12kHz when the lamp has started.
inverter circuits

fluorescent on an inverter circuits

fluorescent on an inverter circuits



Wednesday, September 10, 2014