pulldown resistors question

Discussion in 'General Electronics Chat' started by fran1942, Oct 1, 2011.

  1. fran1942

    Thread Starter Member

    Jul 26, 2010
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    Hello, as an electronics beginner, I am trying to understand how the pulldown resistors in this circuit work.
    I have written a description as to how I understand it.

    Can someone please take a look at the attached schematic and my text, to confirm if my understanding is correct.

    Thanks kindly for any help.


    "MY UNDERSTANDING"

    The source of the logic circuit is a pair of high or lows from two Picaxe outputs on pins 6 and 7. These two signals feed respectively into the bases of two NPN transistors (Q3 and Q4), therefore enabling them to act as switches controlling the VCC source. The emitters of Q3 and Q4 are directly connected to ground in order to 'channel to ground' any floating signals (created by susceptibility to electrical noise) when Q3 and Q4 are supposed to be turned off, i.e. a floating signal could potentially cause Q5, Q6, Q7 or Q8 to activate unintendedly.
    The 2 pull-down resistors R8 and R9 become necessary when Q3 and Q4 are open in order to prevent a short circuit. Care was taken to ensure they were not too high in value otherwise when either of the two PNP transistors, Q7 or Q8 were activated, their base current would pass through R8 or R9 and potentially generate a voltage of higher than 0.6V. If that point came up more than about 0.6V there is a danger of the other NPN transistors, Q5 or Q6, being turned on when they shouldn't, e.g. if the pull-down resistor was too high and the current from a PNP transistor was elevating that to 1V then you would also have an NPN transitor inadvertently turn on.
    By having the pulldown resistor smaller than the base resistors, it stops the voltage from 'lifting it's feet'.
     
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  2. SgtWookie

    Expert

    Jul 17, 2007
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    Hi again Fran,
    I'd been wondering why I had not heard any more on the prior threads. I didn't know that you had questions/doubts. It is good that you are asking questions; I will try to explain them.

    If you still don't understand something or are unsure about it after this, please go ahead and ask more questions, or at least try to describe where your doubts/confusion lie.
    Yes, and for now let's assume that those "high" levels will be the same as Vcc, and the "low" levels will be for practical purposes, 0v (same as ground).
    That is right.
    This is not correct.
    Q3 and Q4 are connected in a common collector configuration; also called a "voltage follower" configuration. The collectors are connected directly to Vcc. The current available on the base is multiplied by the gain of the transistor, and the voltage at the emitter measures ~0.6v to 0.7v lower than the voltage on the base; but it won't go lower than ground.

    So, when the voltage on the base is 0v, the emitter is also 0v. When the voltage on the base is high (Vcc; 5v) the voltage on the emitter will be ~4.2v-4.3v; basically the same as the forward voltage drop of a diode when current is flowing through it.

    They are not connected directly to ground; R8 and R9 are there as pull-down resistors. If the emitters were connected directly to ground, then when current started flowing through the base, the transistors would burn up due to excess current, as nothing would limit collector current except for the limitations of the power supply and the transistor itself. That would also be very hard on the PICAXE, as the I/O pins need to be limited to ±25mA current as an absolute maximum.

    Now, R8 and R9 will act as "base return" resistors for the rest of the transistors.

    Well, in electricity and electronics, an "open" is a circuit that is disconnected; usually as a fault condition nowadays, as in "the fuse burned open", "the breaker is tripped open circuit", etc.

    R8 and R9 are performing these two basic duties:
    1) Limit the base current; thus the PICAXE pins 6 & 7 output current to under 10mA.
    2) Act as base return resistors for the remaining transistors.

    I don't see that happening. However, it would have been more effective to use higher-value base return resistors on the bases of the individual NPN transistor.
    Well, you wind up with other problems by doing it that way.
    You need to start off by determining how much current your loads require; in this case your loads are LEDs.
    Right now, the only LED that has a current limiting resistor is D2; R13 is 180 Ohms.

    I don't know what color the LED is, or what it's Vf or the desired current is; however if it's a red LED with a Vf of 2.1v @ 20mA, then 5v-2.1=2.9v; 2.9v/180 Ohms = 16.111mA, so you're in good shape there.

    Now you know that you need 16.111mA current through the collector of Q6.
    You need to calculate the value of the base resistor.
    Rbase = (Vin - Vbe) / (Ic / 10)
    where:
    Vin = the voltage that will be on the end of Rbase opposite of the base.
    Vbe = the voltage on the base relative to the emitter; you can usually use 0.7v for ~2mA through ~50mA collector current for common transistors.
    Ic = the desired collector current.
    So, let's substitute:
    Rbase = ((5v-0.7) - 0.7v) / (16.111mA / 10) = 4.3/0.0016111 = ~2234.5 Ohms. 2.2k is the closest standard value of resistance; about 1.4% lower than what's needed, which is plenty close enough.

    The first part of the substitution went ((5v-0.7) - 0.7v) because the (5v-0.7) is the base-emitter voltage drop on Q3 or Q4, and the 2nd - 0.7v is the Vbe of Q6.

    Now if you wanted to use a base return resistor for Q6, you can use roughly 3 to 5 times what you calculated for the base resistor, and connect that from the base to the emitter. 10k would be just fine.

    You have more problems though, because D3, D4 nor D5 have current limiting resistors, and the base resistors are much too large.
     
    fran1942 likes this.
  3. fran1942

    Thread Starter Member

    Jul 26, 2010
    58
    0
    Thanks for the great reply. I am learning a lot from you and the others.
    I edited part of my description as follows:

    "The 2 pull-down resistors, R8 and R9, limit the base current for Q3 and Q4; thus the PICAXE pins 6 & 7 output current to under 10mA (a safe amount for a Picaxe chip). These resistors also act as base return resistors for the two PNP transistors Q7 and Q8. PNP base current flows in an opposite direction to it's NPN contemporary.
    R8 and R9 are positioned where they are rather than directly on the Picaxe outputs (immediately before the base of Q3 and Q4) for the reason that if the emitters were connected directly to ground, then when current started flowing through the base, the transistors would burn up due to excess current, as nothing would limit collector current except for the limitations of the power supply and the transistor itself."

    Regarding your comment that R8 and R9:
    2) Act as base return resistors for the remaining transistors.

    Just assuming theoretically that there was no issue with preventing a short circuit through Q3 and Q4, then these 2 'base return resistors' for the PNP transistors would not be necessary, would they ?
    So how exactly do the 'base return resistors R8 and R9' contribute to the operation of the PNP transistors ?
     
    Last edited: Oct 1, 2011
  4. SgtWookie

    Expert

    Jul 17, 2007
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    Here's a simulation of your PIC outputs being sequenced, and the LEDs turning on and off. Notice that the LEDs which don't have current limiting resistors are getting > 60mA current! :eek:
     
  5. fran1942

    Thread Starter Member

    Jul 26, 2010
    58
    0
    Thanks for that simulation Sergeant, and I will address those current limiting resistors on the LED.

    If you could just confirm I on track now with this blurb and last question:
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    I'm starting to wonder if you are documenting someone elses' circuit for your own understanding, or if you are planning on turning this in for credit as a school assignment?
     
  7. fran1942

    Thread Starter Member

    Jul 26, 2010
    58
    0
    SgtWookie, very surprised to read your latest post.
    Can I respond my telling you that as a 41 year old changing careers, I am busting my guts trying to get on top of this subject. Perhaps you have misinterpreted my enthusiastic posts as having ulterior motives ?
    If you implying that I am a cheat, then I at least want the opportunity of publicly denying your implication.
    If I have overstepped the mark somehow, then send me a PM so that I can more clearly understand the protocol required here.
     
    Last edited: Oct 2, 2011
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Please accept my apologies. There seems to have been yet another rash of students as of late that are attempting to sneak in and get their homework done for them by posing questions as if they were hobby projects. It has something to do with the way you worded your request that triggered my response - on top of a distinct lack of coffee this morning.

    Please don't take it personally. I have a pretty good batting average, but I don't claim to be infallible.

    Back to your questions:
    This part is OK.
    This is not correct.
    They two resistors act (more or less) as base return resistors for Q5 and Q6, which are NPN resistors, turning them OFF when Q3 & Q4 are not sourcing current from their emitters.
    At the same time, if Q3 is off, R8 will sink current from the base of Q8 via R15.
    If Q4 is off, then R9 will sink current from the base of Q7 via R16.
    This is correct.
    The resistors are on the emitters to cause a voltage drop when current is sourced to the base, causing current to flow from the emitter. The voltage drop across the corresponding emitter resistor will cause associated NPN transistors that are wired in a common emitter configuration to turn ON, and cause the associated PNP transistors that are wired in a common emitter configuration to turn OFF.

    Yes, I did say that - but I should have qualified it with "NPN", as if no current is being sourced via Q3 or Q4's emitters, the PNP transistors that have base current paths to these resistors will be turned ON.

    This circuit is not a particularly easy one for a novice to understand. It has both PNP and NPN transistors in both common collector (voltage follower) and common emitter (saturated switch) configurations. It is also missing some key elements that will cause the early demise of three of its' LEDs. I am not particularly fond of it.

    The circuit would need to be re-designed. Base resistors would be necessary, as would resistors on the collectors. This would also invert the logic; common collector voltage followers are non-inverting logic, where common emitter saturated switches invert the signal input on the base.

    They prevent excessive PICAXE I/O pin current;
    They prevent excessive base current;
    They sink current from the PNP base resistors when Q3/Q4 are turned off;
    and they keep the other common emitter NPN transistors turned off when Q3/Q4 are turned off.
     
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