Connect with us

Microchip's _MCLR pull-up cct

Discussion in 'Electronic Basics' started by MarkMc, Mar 11, 2005.

Scroll to continue with content
  1. MarkMc

    MarkMc Guest

    Hi

    I've a relative newbie to electronics and PIC microcontrollers in
    particular. I'm trying to design a cct using PIC16F628A.

    I'm looking at the Microchip mid-range PIC document
    (DS31028A-page28-4), and this shows a diagram of how to isolate VDD
    from VPP when the ICSP connector is connected.

    The diagram consists of VDD connected to a resistor (R1) and a
    capacitor (C1). Between the capacitor is a diode(D1) which is
    connected (normal bias) to _MCLR/VPP and the ICSP connector. A second
    connection is made from VDD to the VDD pin via resistor(R2) on the PIC
    and ICSPconnector.

    (anybody care to show me some ASCII cct drawing tools?)

    I understand that D1 protects/isolates the main cct from the +13v VPP
    when programming.

    My questions are;

    1) D1 is referred to as a Schotky diode in the document. Can somebody
    explain what one is, and recommend a model number for me to use.
    2) Also, no values are given for R1, R2 and C1. I assume 'cos they
    can't possibly know the capacitance of a third party cct, but can
    anybody recomend good starting values for these? I can give you a
    pointer to my circuit on the web if you'd like to see the cct. Should
    C1 be a tantalum cap or electrolytic, ceramic or what?
    3) I don't use any of the other programming pins for my application to
    make life easy, but my cct has another component which fires off a data
    burst to my PIC periodically. I assume this won't affect the PIC while
    it's being programmed? Or should I isolate the entire cct from VDD via
    a dip switch or something to be sure?

    Thanks
    Mark
     
  2. PeteS

    PeteS Guest

    A Schottky diode is one of a class of 'hot-barrier' diodes and they are
    used because their forward voltage is typically 0.2-0.3V for small
    signal type devices. This is to make sure MCLR discharges quickly.
    There's no protection required if MCLR is connected only to a reset
    signal.

    You should not need to isolate this pin beyond the typical schematic
    shown in the manual.

    Typical devices for this sort of application might be the MMDL770T1
    from On Semi (nee Motorola SPS) [http://www.onsemi.com/]
    Get a parametric table of devices by searching for schottky diodes. You
    want low Vf, Low If (forward voltage, forward current) and fast
    recovery is preferable.

    You only need the full RC circuit for a slow rising Vcc system.

    This is from DS31003A (Mid range PIC reference manual)
    **Figure 3-3 shows a possible POR circuit for a slow power supply ramp
    up. The external Power-on Reset circuit is only required if VDD
    power-up time is too slow. The diode, D, helps discharge the capacitor
    quickly when VDD powers down.**

    A slow rising input is defined in that manual as < 0.05V/millisec.

    If you have a slow rising Vcc, then MCLR should stay low until Vdd is
    valid. That occurs at around 2.5V, but I would suggest making sure it
    holds off until Vdd is about 3V.

    You need to calculate the RC pair, and for this you first need to
    define your power. Let's say it's 1V/millisec. That means we don't want
    MCLR to go above ~75% Vdd until at least 3 milliseconds later. A delay
    of 10msec would work for a good conservative circuit. (It also
    simplifies the mathematics).

    1xRC is about 63% Vdd, and we'll use that as it's close enough if we
    extend the time a little. The listed VIH for MCLR (the input voltage at
    which a high is guaranteed to be detected) is 0.8VDD.

    So RC >= 0.01
    Choose a resistor (Microchip suggests <40k to minimise voltage drop due
    to input current on the pin).
    So let's go with 20k
    The C = 0.01 / 20k = 0.05uF

    Note that this gets you about 63% Vdd after 10msec, but that's ok.
    A typical value to use would be 0.47uF

    Cheers

    PeteS
     
  3. MarkMc

    MarkMc Guest

    Thanks very much for enlightening me Pete.

    Very much appreciated.

    I'm using the _MCLR pin straight to VDD (supplied by a 7805 vreg), but
    also it connects to an ICSP +13v supply for programming. Sorry for
    being dense here, would this mean that I do or don't need the diode
    set-up?

    If not, should I still have, say a 0.1uF tant. cap between VDD and gnd,
    right next to the PIC IC?

    Cheers,
    Mark
     
  4. You need the diode to block the Vpp(+13V) from powering up the rest of
    your circuit and potentially killing some parts by applying excessive
    voltage to them. I suspect that they used a Schotky(sp?) or "hot
    carrier" diode just for the low voltage drop feature (.2 - .3V) just to
    keep MCLR as close as possible to Vcc.
    If you'd like, it shouldn't harm anything. You would probably only need
    that if glitches/spikes were likely on the Vcc that might result in a
    device reset.
     
  5. PeteS

    PeteS Guest

    Well, the diode is there to make sure that the MCLR pin doesn't exceed
    Vcc during power down. If you have a fast falling Vcc, then the cap/res
    pair will hold MCLR high without the diode, exceeding the rating of the
    device and powering the rest of the chip through the input protection
    diodes (this is generally not a good thing [tm]).
    I would normally have the diode there for protection against transients
    anyway, but ICSP requires that you do not have the diode that way.

    Look at the application diagram on page 28-4 of the previously
    referenced manual. There is a diode there (make it a schottky again so
    that when MCLR is powered by Vcc you aren't 0.6 - 0.7V below Vcc, but
    0.2 - 0.3V below) but this time to isolate the power supplies (we call
    this diode OR'ing, by the way). In this particular circuit, assuming
    Vpp is not present, then the diode will still discharge the cap at
    normal Vcc power down, not exceeding about Vcc + 0.3V at the MCLR pin,
    so you still get the device protection.

    You should bypass the device, but I would suggest a 1-10uF ceramic as
    esr dominates ripple performance, and ceramics have outstanding esr
    ratings (typically below 10 milli ohm - I have seen them as low as 1
    miili ohm).
    I don't see a need for bulk bypass beyond that - the maximum current is
    Icc + I (IO), so you should be in the order of 10s of milliamps. Note
    that using a ceramic like this means you don't have to also have a
    small value ceramic (which you would need with a tant, unless it's one
    of the later ones), and tantalum has it's own problems (highly toxic if
    they overheat for whatever reason). During programming, the current is
    higher, of course, but not that high.

    Hope that helps

    Cheers
    PeteS
     
  6. Bill Carson

    Bill Carson Guest

    2) Also, no values are given for R1, R2 and C1. I assume 'cos they
    You might find the information in this free PIC book useful:
    Getting started with PIC microcontrollers, by Fred Stevens
    a PDF can be downloaded from here:
    http://www.the-electronics-project.com/

    see page 18 for a disussion of component sizes for the RC circuit.

    HTH,

    B.C.
     
  7. MarkMc

    MarkMc Guest

    The diagram on page 28-4 was the diagram that first prompted my post.

    Forgive me for asking, I'm not an electronics professional, but what
    does "BYPASS" mean? is this the normal capacitor for protecting
    against ripples and spikes going in to an IC? I've always been told to
    use a 0.1uF tant. capacitor for this due to their speed to react. I'm
    pretty sure I read this in a PIC book once too. I'll happily learn
    some other way, but I'm not sure I fully understand.

    Cheers,
    Mark
     
  8. PeteS

    PeteS Guest

    Bypass is, as you assume, to deal with local voltage variations due to
    the chip itself. As the chip operates and switches it's outputs (and
    its inputs are changed), the device uses more or less current, and this
    shows up as noise on it's power pins (there are lots of reasons for
    this, including the nductance of the bond wires from the die to the
    pins of the device).

    Bypassing is used to combat this noise, and we usually talk of it in
    two ways:

    Bulk bypass. This is a fairly large capacitor used as energy storage
    for those sudden increases (and decreases) in current. Large variations
    in current a long way (electrically) from the regulator sense point can
    have dramatic voltage fluctuations - bulk bypass is used for this.

    Signal bypass: This is to eliminate (well, minimise) the noise on the
    power lines due to high speed switching of the device.

    Note that these two do overlap considerably; high speed switching can
    also have large current variations associated with it, although we try
    to minimise that in really high speed stuff by using (almost) constant
    current devices (and that's another story entirely).

    I am sure tantalums were recommended because their performance is
    superior, electrically, to electrolytic capacitors. As I noted above,
    tantalums have their own issues though. Modern ceramic capacitors have
    up to 100uF at 6.3V in a 1210 package (I know, I've used them - check
    out this panasonic part : ECJ4YB0J107M), have superior performance to a
    tantalum (lower esr, better ripple current performance, better
    temperature characteristics) for general bypassing.

    This is not to say tantalums are not widely used - I use them myself as
    part of the loop filter in some LDO supplies which require my output
    ESR not be *too* low (Look carefully at the datasheet for the National
    LP3961 for an example).

    That's the quick explanation.

    Cheers

    PeteS
     
  9. MarkMc

    MarkMc Guest

    Thanks Pete, that helps clear things up. I'm enjotying learning about
    this stuff. Let me see if I'm getting this. I should use ceramics
    (100uF?) in close proximity to all of the IC's (PIC being one of them)
    in my cct.

    Should I put a tant. (0.1uF) on the 5v regulator too? (78L05)?

    Cheers,
    Mark
     
  10. PeteS

    PeteS Guest

    You should have a bypass cap for the 7805 (I assume it's powering a few
    things, such as the I/O devices) - about 10uF is a good rule of thumb
    for light loads.

    Use a 0.1 - 1uF cap for the PIC.

    If you use ceramics, that's all you should need.

    Cheers

    PeteS
     
  11. MarkMc

    MarkMc Guest

    That's great, thanks for all your help Pete
     
  12. niftydog

    niftydog Guest

  13. MarkMc

    MarkMc Guest

    That's perfect, thanks niftydog.

    Cheers,
    Mark
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-