Connect with us

PNP bias/voltage divider

Discussion in 'Electronic Design' started by James, Oct 16, 2006.

Scroll to continue with content
  1. James

    James Guest


    I'm sure this is a repeat question, but searching hasn't revealed an
    answer to my specific issue. If there's a thread that addresses this
    already, please direct me to it. I'm not an EE, so forgive me if the
    answer is simple :)

    I have chosen a BC327 to use in a low power switching regulator and I'm
    trying to bias the base for -0.7V and 100mA of collector current. The
    emitter voltage will be anywhere from 6.5-8.9V. Assuming an Hfe of 150,
    the base current should be about 660uA. I'm calculating my voltage
    divider using 6.5V, anything more should only 'overbias' the base. I've
    calculated the R2 resistance to be about 10K ((6.5-0.7)/.00066 =
    9.66K). I used this to calculate R1 to divide 6.5V into 5.8V and 0.7V.
    This came out to be about 90K (6.5*9666/0.7 = 89.755K). Is this the
    correct way to apply the math in the circuit below if I want base
    voltage of -.7V at 660uA?

    | |
    \ |
    / R1 |
    \ 90k |
    / |
    \ |
    | /
    | <-660uA |V_
    +-----------| BC327
    | |\
    \ \
    / R2 |
    \ 9K |
    / | 100mA ->
    \ +-----------
  2. Tam/WB2TT

    Tam/WB2TT Guest

    I have problems vieving this, The view that looks the best appears to show
    the emitter of the transistor connected to the + supply. If that's wrong,
    you have other problems and ignore the first two paragraphs.

    You have some basic problems here. First, having a base voltage of -.7V
    (wrong number) does not mean an absolute -.7V, but rather -.7 with respect
    to the emitter. So, if the emitter was at +6.7V, the base would have to be
    at +6.0. Your resistor values are interchanged.

    You don't bias a bipolar trtansistor that way. At 100 ma, the base-emitter
    voltage is going to be some number between about .60 and .68. Haven't looked
    at your transistor, but you want it to work with all transistors having
    betas between about 50 and 300. You could probably get it to work somewhat
    by returning R2 to the collector of the transistor instead of ground. and
    assuming the collector voltage is 1/2 of VCC. You would have to make R2 a
    variable resistor, and hope the temperature does not change a whole lot.

    If this is a switching regulator, why is there *any* bias on the transistor?

  3. Tom Bruhns

    Tom Bruhns Guest

    To begin with, the base current is an exponentail function of the b-e
    voltage, and in addition it's quite temperature dependent. That means,
    for a particular base current, it's not a good idea to try to achieve a
    particular b-e voltage. The nominal b-e voltage just lets you know the
    base voltage above ground (that is, about 0.7V lower than the supply),
    so you can find the value of R2 to give you the required base current.
    Then R1 just insures that the base current goes to essentially zero
    when you are NOT pulling R2's bottom end down to ground potential,
    assuming you want to turn it off. R2 must supply the required
    transistor base current AND whatever current is in R2 when it has about
    0.7V across it.

    Second, the transistor DC current gain (HFE) is a strong function of
    the particular sample, and also of temperature. If you really want the
    collector to source 100mA to some fairly close tolerance, there are
    better ways to do it. If you just want the collector to pull up to the
    supply voltage (less a small drop) and the maximum current will be
    100mA, then you're OK. But it's usual to "overbias" the base, and go
    for a ratio of base to collector current of 20:1 or even 10:1.

    You didn't say anything about turning the transistor on and off, but if
    you want to do that, there may be some other considerations: speed,
    and how much voltage swing the thing that drives the on/off signal can

  4. Tom Bruhns

    Tom Bruhns Guest

    Resistor values interchanged?? 6 volts across his 9k R2 value does
    give about 667uA, and the current in R1 at 90k and 0.7V would be about
    7uA, leaving 660uA for the base. There is some question then about how
    fast it would turn off, if that's the desire, if you open the
    connection to R2, and even some question about just what the purpose of
    the transistor is, but looks to me like the calcs for the values are OK
    with the given assumptions.

    The assumptions, on the other hand, may be off-base. Should the
    transistor saturate, or should it act like a 100mA current source?
    Will it be turning on and off? Occasionally, or rapidly? Over what
    temperature range must it operate?

  5. Tam/WB2TT

    Tam/WB2TT Guest

    You are right. I guess it is a coincidence, but it struck out at me that if
    you reversed the R values, you would get the correct base voltage if you
    ignored the loadg. Still, current biasing is not the way to do it either.
    I've been there, and had the factory scream bloody murder when the nominal
    beta=40 transistors started arriving with betas of 385.


    There is some question then about how
  6. James

    James Guest

    First of all, thanks for all the comments, it really helps out and I
    appreciate everyone's time.

    The regulator is modeled after an idea that I read about a few years
    ago. That circuit could only source 20mA or so and required a higher
    supply voltage, somewhere around 12V I believe. Using the same concept,
    I've created a regulator that can be driven from a 9V battery (6.5V to
    8.xV), is fairly efficient (95% or so simulated) that provides 5.49V
    with <1.5% ripple. I'm trying to replace an LM7805 in a microprocessor
    application that requires up to 85mA and do it on the cheap :) The
    reason I don't want to use the linear regulator is that the application
    is portable and the LM7805 is wasteful, especially with higher
    voltages. Plus, the dropout is unacceptable for a 9V battery in my

    I am including a couple of links to both the entire schematic for my
    regulator as well as a plot of the voltage at the Q3 collector-L1

    R1 turns on Q1, which turns on Q3, causing voltage to build across L1.
    Eventually D1, a 5.1V zener, breaks down and turns on Q2 which turns
    OFF Q1 and, subsequently, Q3. The inductor rings through D1 and D2,
    with C2 holding the output voltage at approximately 5.49V. Voltage
    across D1 drops, Q2 turns off and the whole cycle starts over again.
    Q3's switching frequency is about 50KHz.

    I vary the load between 55 and 220 Ohms, simulating between 25 and
    100mA load at 5.49V. So far, it's fairly stable, but I haven't
    prototyped it yet.
  7. john jardine

    john jardine Guest

    Transistors are not precise devices. You've a cat in hell's chance of
    pinning down that 100ma.
    It's usual to swamp out the Vbe voltage variations by adding some emitter
    resistance. Changes in Vbe due to varying temperature and currents will now
    be small compared to the larger voltage across the emitter resistor.
    A la ...

    6.5 to 8.9 ---o---------o
    | |
    .-. .-.
    | | | |
    | | | |4.7
    '-' '-'
    |220 |
    | |6V
    | 1ma |<
    o-------| BC327
    |5.3V |\
    | |
    5ma| |
    | o
    | | 100ma load
    | | o
    '-' |
    | 1k |
    (created by AACircuit v1.28 beta 10/06/04
  8. First of all, the transistor will have a reduced gain as it
    saturates. The lower you need the saturation drop to be,
    the more base current you have to provide. So lets say you
    provide a base current that is 1/50th of the collector
    current, to drive a normally higher gain (say, beta 150)
    device well into saturation, to keep the on state conduction
    losses low. This would be 100 mA/50= 3mA.

    Your R3 has to supply that current ans also the current that
    bleeds through R2 while the transistor is on. The only
    purpose for R2 is to drain the base stored charge away, to
    quickly turn the transistor off when the current through R3
    is cut off. I usually try to keep the on state R2 current
    at least 1/10th of the on state base current. That would
    put it around .3 mA. If the on state base emitter voltage
    is about .7 volts, that means that R2 would have a
    resistance of about .7/.3 mA = 2.3k. Lets say we use 2.2k,
    a standard value. So R3 has to provide about 3.3 mA at the
    lowest supply voltage minus the Q3 base emitter drop and the
    Q1 collector to emitter drop. Lets say that is 6.5 - 0.7
    -0.3 = 5.5 volts.

    So R3 is about 5.5/3.3 mA =1.67 k. Lets say you use a 1.6k
    5% standard value.

    Unfortunately, at the fresh battery voltage, R3 supplies the
    base of Q1 with more current than is needed, and this hurts
    the efficiency of the supply a little. Replacing Q1 and R3
    with a switchable current regulator would reduce this effect.
  9. Tom Bruhns

    Tom Bruhns Guest

    In addition to what John Popelish wrote in reply to this, I'd add that
    it could help you in getting it to switch a bit more crisply to put a
    large value of resistance between Q3's collector and Q1's base. I'd
    suggest perhaps 100k ohms, or a bit more. Also, R5 can be eliminated;
    it does practically nothing. There's already ample path from the base
    of Q2 to ground through R4 and R6.

  10. Tam/WB2TT

    Tam/WB2TT Guest

    So, the 9K does not go to ground as you showed, but is a pulse input. I have
    two suggestions:

    Look at replacing the PNP with a P channel FET. This is going to save a lot
    of base current since you are concerned about power. You need a FET that is
    fully ON at 6V. You don't have to limit the gate voltage to limit the drain
    current; that is limited by the inductor and the switching frequency. What
    you are proposing to build is called a buck regulator. You can buy chips to
    do that from Digikey and Mouser, made by outfits like National, Maxim, and

    Or, forget about the switching regulator and look at LOW DROPOUT series
    regulators. The efficiency would be very close to V(out)/V(in). Probably
    less efficient than the switcher with a new battery, and more efficient with
    an old battery. The 7805 puts out 5V, and requires a minimum input of 7V;
    7.5 on a bad day (This is not low dropout)..

    I don't know what you mean by 9V battery, but if it is what I think, did you
    ever calculate how long it would last with a 60 - 85 ma load?

  11. Vans

    Vans Guest

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