BC857 as a 1.8 regulator transistor for adau1701

Hello everybody, I have been making the schematic for my own 1701 board (yay yay). But since I am not an engineer I am awfully afraid of making any mistake. So I chose BC857 as a 1.8 regulator transistor since it is small and has a huge beta. It seems that this transistor should be fit for the task, but just in case, I wanted to ask, am I right? Is this transistor suitable? Again, I am just very afraid of making a mistake.

The datasheet is here.

transistor2sb1689.pdf
  • Sorry, some follow-up questions:

    1. What is the optimal value for the WB resistor, which resides on the collector of the WB transistor?

    2. And what is the optimal value of the pot, which is to be connected to an aux adc?

    3. Is 220uf capacitor enough for the WB operation?

  • 0
    •  Super User 
    on Aug 18, 2013 4:58 AM

         Hi Skfir,

        1.     The BC857 meets all the needed specs for the 1,8V regulator --

    • gain >100 at 60mA
    • Power dissipation -- rated 200 mW, will handle the needed 90mW

         2.     The WB input has a high impedance, so a 10K pullup will work fine.  Note, however, that the WB trigger circuit from the EVAL1701MINIZ board doesn't work as shown, because the transistor's base drive remains while the power supply drops.  See http://ez.analog.com/message/23440#23440

         3.  Once you switch power OFF, the filter capacitor must supply the current for long enough to allow the writeback.  Its voltage drops at the rate of     dV/dt = I / C.   Thus the voltage drop  V = I * t / C

    For example, assume your total system current is 220mA (the -1701 alone uses about 110mA).

    The writeback takes 73uS, but let's allow 100uS.

    With a 220uF cap, the voltage drop is,  V = (0.22 A) * (1e-4 s) / (2.2e-4 F) = 0.1 V

         This is within the usual 5% voltage tolerance for Vdd of 3.3V, so 220uF would work in this case.

         4.  The auxiliary ADC's input impedance is 30K ohms.  If you drive it with a 10K pot, the pot's worst case output impedance is 10K / 4 = 2.5K.  The pot's output impedance is highest at the center of its rotation, going down to near zero at either end.  This adds a slight non-linearity between the pot's shaft position and its loaded output voltage, but since 2.5K is much less than the 30K load, no one will notice it.  Thus a 10K pot will be fine.

         Bob

  • 0
    •  Super User 
    on Aug 18, 2013 11:19 AM

         Hi Skfir,

         I'm just an ordinary soul, and glad to hear that you're moving your blind-assist device along to its final form.  When it's all working it would be fun to see it in action!  PCB layout can make or break a product.  It's actually something I rarely do because where I work there's an engineer who does it so much better than me.  Generally in PCB layout the value of resistors, etc. is not especially important -- you can (and probably will!) change a few in the debug process.  Far more critical is the layout itself -- pin-outs (3-terminal devices are a real pain!), trace runs & grounding of high-frequency portions, etc.  These are messy to fix in debug.

         Bob

  • Hi Bob! Yes, I read as much as I could about the layout and making it fine is really hard. The good news is that I have already built the prototype of the whole device on a breadboard and it works somehow, though, with some noises and other "unpleasantnesses".  But I hope for the best. Of course soldering all those 201 etc. parts would be quite problematic, but I have no choice, so it must be done. The good news also is that the layout for the power supply is ready and the supply, especially lo-to-high voltage (I need about 400v there), which keeps it stable independent of the battery fluctuations is quite difficult to make especially with no any experience in the area and virtually no ready examples.