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    Oops, Anotherforummember you are right about that goto $-1 and about the TRMT. ;D Mike, I agree that you have to check the TXIF flag before filling the buffer, but since the XMIT_RS232 does not return until the transmission has completed (ok, untill TXREG's cotents are transferred to the TSR), this probably won't be necessary when calling this subroutine. Good luck scottnj!

    Hi scottnj Well, I am not very familliar with the F876 but I'll do my best! Your hardware seems ok. There seems to be a small problem in your code though. In the XMIT_RS232 subroutine you wait for the TXIF to be set. This flag is set when you have palced data in TXREG and the data is sent. Maybe you could try writing your code like: XMIT_RS232 MOVWF TXREG btfss PIR1,TXIF GOTO XMIT_RS232 RETURN What I mean is that you should first send the data and return when finished (TXIF flag set)
  3. and smaller!

    Very interesting article Ante, thanks. I'm sure Bardeen and Brattain would't have imagined this in their wildest dreams! ::)
  4. laser lights

    Hi You may have already built that 5V supply but... I recently found some ready made very easy to use switching regulators made by Texas Instruments (Powertrends to be more specific). They do operate at 24V, have outputs of 5V, as you need, and adequate output currents. What is even better is that you can have free samples of them. Check out PT5100 (5V/1A) or PT4142 (5V/4A), You may look at TI's site under: Analog and Mixed signal -> Power management -> Plug-in power modules and find the regulator that suits you best.
  5. Barrier potential discrepancy

    Sure there is current before 0,7V! As you see in the equation the built-in voltage depends on doping (and temperature) so you can play with it to some extent by controlling the doping. Not to mention that electron-hole recombination is a statistic phenomenon and therefore when you say 0,7V it does't mean that there is no current at all at lower voltages. As for the depletion region, it can not be significantly reduced or disappear. It exists there because you put two differently (p and n type) doped semoconductors together. And yes, it can be extended when you reverse bias the junction You can take a look here to find a well written and more descriptive approach:
  6. displaying decimal output with binary inputs

    Hi black_wing, if I have well understood, you want to use a counter and display the result in 7 seg displays. Normally, you use BCD to 7 segment decoder/driver ICs such as 4511, 74LS46, 74LS47, 74LS48. You choose one to match your display (common anode/cathode, current requirements) BCD is much like decimal but for 4 bits it counts only up to 9. The rest combinations wich are valid in binary, are invalid for BCD (though there are some chips that will dsplay hex on the 7 seg display - look at the datasheets) If you want to feed directly binary to the BCD inputs, you need to do a little combinational logic... If you just want to feed the output of a counter to it, you can use a BCD counter. You can find the implementation in the Digtal stopwatch project: You just have to omit the oscillator (CD4060 and crystal oscillator) and the rest is acounter that displays the result in 7 seg displays. To count up to 30 you have to make a proper circuit to reset the counter at 30
  7. Barrier potential discrepancy

    Hi Kevin In a pn junction in equilibrum, where p and n regions "touch", majority carriers from each region recombine with majority carriers from the other region. That is: free electrons from the n region recombine with holes in the p region and holes from the p region recombine with free electrons in the n region. A depletion region is thus created. In the depletion region there are no "mobile" carriers. The diffused holes leave behind uncovered fixed negative charges and similarly there are fixed positive charges left behind by the diffused electons. In this way, a separation of charges occurs, causing an electric field (- is in the p region and + in the n region). This electic field is called built-in potential and has the opposite polarity of the battery when you forward bias the junction. It is given by: Vbi=VTln[(NdNa)/ni2 ] where: Vbi is the built-in voltage VT is the thermal voltage for Si (26mV/deg C) Nd and Na is the concentration of doping atoms in the n and p region respectively ni is the intrinsic carrier concentration in Si This potential is constant, has a value of 0,7V approximately for Si diodes and normal doping (depends on doping and temperature), and you have to overcome it in order to have a current flow through the junction. This is practically what you measure and call forward voltage drop. When you bias the diode you do have an electron current and practically you create an electric field opposite to the built-in voltage. When the electric field you apply equals or exceeds the built-in voltage, your diode conducts.
  8. Silicon Photonics

    Hi all We all know that due to its indirect band-gap, silicon is not a suitable semiconductor to manufacture optoelectronics. Having a Silicon light source (LED or LASER) though, would come in very handy in today's technology, especially in telecommunications. There is a decade or so that electroluminescence and photoluminescence have been observed from porous Silicon. Reseach has been re-stimulated quite recently when initially photoluminescence and later electroluminescence was observed in structures containing Silicon nanocrystals. Contrary to porous Silicon, silicon nanostructure devices ARE FULLY compatible with today's CMOS processes. This means that the cost of combining the drive electronics (silicon) and light source (composite semiconductors) is practically eliminated, allowing in a few years the full integration of these two components and other optics as well (waveguides, modulators, etc.) to be possible, giving a new direction in the evolution of electronics. ST has already produced a LED based on Er-ion doped Si nanocrystals. A nice presentation of the subject, is by Intel here: and you may as well perform a google serach, you'll find many relevant pages.
  9. Fet and mosfet

    Here is a nice presentation of how MOSFETs, JFETs and BJTs are implemented and briefly how they work: you can also check out: and if you still want more, try a Google search 8)
  10. Fet and mosfet

    Hi tweak232 The term "FET" (Field Effect Transistor) covers all transistors functioning in the same way. In this sense a MOSFET is a FET. If you mean the difference between a JFET and a MOSFET, it is a structural one: In a JFET (Junction FET), the isolation between the chanel and the gate is practically a reverse biased pn junction In a MOSFET (Metal - Oxide - Semiconductor FET) the channel is insulated from the gate by the oxide layer. The transfer characteristics are very similar between a JFET and a depletion mode MOSFET. JFETS always operate in "depletion mode" ie. they are normally on devices like depletion mode MOSFETS.
  11. Unable to identify this part...

    Hi all! These components are not ceramic resonators! They are EMI (ElectroMagnetic Interference) supressors and are used to limit the interference your device causes or accepts. I also had problems identifying them in the past... You might want to check out a datasheet as an example: The equivalent is shown on the attachment
  12. urgent (counter question)

    Hi all. well Kevin it is not exactly like that. Normally you build counters using flip flops. According to how you will connect the clocks of your flip flops, you can get a sync. or an async. counter. You have a syncronous counter when all clock inputs of the flip flops are connected to a single, common clock. An asynchronous counter uses each flip flop's clock differently. You may want to take a look at this page for more details and some examples: As for advantages and disadvantages, each counter fits in different applications and therefore I don't think you could make a comparison between them. In general you put sync. counters in synchronous systems and asynchronous in async. systems (these are the faster ones btw., but are more complex to design)
  13. Leds

    Oops, sorry, it's
  14. Leds

    Hi Codyhtml You just need to calculate the resistor as AJB2K3 said, but you will need to know the forward voltage drop of the LED (VL) which depends on the colour of the LED. As for the current, it should be no less than 1-2mA and no more than 20mA. This .pdf should answer all your questions: As for the resistor colors, you can doenload the "resistor color coder" in the downloads section:
  15. LCD graphic controllers (EPSON)

    Hi dtsormpa If you haven't found it already, all the necessary information to do this is here: There are many interface examples with 16-bit microprocessors. Ok, it's for the S1D13504 but I guess driving the rest of the controllers of the family will be more or less the same.