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This is a follow up question from Emitter follower biasing with voltage divider or the emitter follower design example in The Art of Electronics.

Why is the recommended one-tenth ratio of base resistance to emitter resistance times \$\beta\$? The base current should just be approximated to the emitter current divided by \$\beta\$, with the base current changes against the varying \$\beta\$.

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  • \$\begingroup\$ You only getting a "stiff" voltage divider if Rth << (beta +1)Re where Rth = R1||R2. Or we can use another point of view. You can get the stiff voltage divider only if the load current (base current) doesn't change the voltage divider output voltage too much. Thus to get a stiff voltage divider we assume that the voltage divider current is much larger than the base current. And a common rule is to set the voltage divider current to roughly be ten times the base current. R1 = (Vcc - Vb)/(Ib + 10Ib) = (Vcc - Vbe)/(11Ib) and R2 = Vb/(10Ib) \$\endgroup\$ – G36 Apr 18 '20 at 7:18
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The factor of 1/10 is a crude, rule-of-thumb rule used to design voltage-dividers, where a load resistor may be connected to the output.

If you want your voltage-divider to maintain a certain output voltage, even when a load is suddenly connected, then the parallel-R of the two divider-resistors must be much smaller than the value of the added load resistor. (Or equivalently, the current normally flowing through the two resistors of the voltage-divider needs to be far greater than any output-current delivered to an external load resistor. That way the load-current will only "pull down" your desired value of Vout by a small percentage.)

The factor of 10X current, or 1/10th resistance, keeps the Vout of the divider from changing by more than roughly 10% when the load-resistor is added. (If you wanted the factor to be only 1%, then instead use 1/100th as your rule-of-thumb value.)

Think: if you measure the Vout of a typical voltage-divider by using a 10meg DMM voltmeter, the voltmeter will seem "invisible," and won't unexpectedly alter the Vout value. At least, not by much. Heh, now try doing the same using an old-school analog voltmeter with a 5K impedance! Depending on the two resistances used in your divider, the current being drawn by the added voltmeter may totally change the Vout, or even reduce it almost to zero!

Remember, with R1 and R2 disconnected, any AC ohmmeter "looking" into the transistor's Base terminal will not "see" the Re resistance. Instead, the meter will report a much larger value of Zinp = β * Re (or 750K in the AOE example, where β=100.) This 750K value acts as the "load resistor" to be connected to the output of any voltage-divider placed on the transistor Base.

Now design your voltage divider to create Vcc/2 + 0.6V = 8.1 volts.

The easy way: Chose R1 to be less than twice 750K/10, or 150K. Then calculate the value of R2. The resulting voltage at the Base won't be exactly 8.1V, but will be pretty close. (If you needed it much closer, then make your R1 and R2 roughly 100x smaller than 750K!) The hard way: calculate the exact values of R1 and R2 while the 750K resistance is already connected to the output of your voltage divider. (Maybe you only need one "pull-up resistor," where the 750K resistance serves as the "pull down" for the divider process.)

The end result here is a "voltage-input amplifier," where Base current is just a small and unimportant "leakage" current, and beta is irrelevant. (Beta must be high, far higher than ten. But its exact value is irrelevant, as long as Ib "input leakage" ends up being far smaller than Ie "output signal.")

And finally ...ask yourself what would happen if this transistor circuit was manufactured, and the beta for the part number being used may vary: soldering in transistors where beta can be anything between 80 and 300. That 750K value won't stay the same across a population of products! We want our voltage-divider to mostly ignore the presence of the transistor Base current. (We want to make the Ib into a small undesirable leakage, same as with FET stages.) In other words: transistor beta needs to become unimportant, and this is done by making R1 and R2 have fairly small values. (The same is done with op-amp input-networks, where the diff-amp stage is roughly the same as two common-collector stages, and the input to the two Base terminals is a voltage signal, not a current.)

PS

Are BJT bipolar transistors driven by base current? Or are they instead controlled by Vbe voltage? Well, go and ask Win Hill, author of "Art of Electronics," see...

https://cr4.globalspec.com/comment/720033/Re-Voltage-vs-Current

https://cr4.globalspec.com/comment/720374/Re-Voltage-vs-Current .

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  • \$\begingroup\$ Thank you for the conceptual explanation. Am I right that because input impedance sees emitter resistance to be (\$\beta\$ + 1) * Re hence if making base resistance large enough we can approximate resistance on the emitter side to be negligible, and hence we can ignore \$\beta\$ variability so long as Rb's resistance is large enough (e.g. 10 times larger)? \$\endgroup\$ – KMC Apr 18 '20 at 6:08
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    \$\begingroup\$ @KMC no, R1 and R2 resistance SMALL enough. Ten times smaller than 750K = [(β + 1) * Re], where R1 and R2 are each smaller than 150K. We want the divider's voltage output Vout to be "stiff," so when β is different, and therefore Ib changes, still the divider Vout doesn't change much, and the voltage across Re doesn't change much. In this way base current becomes unimportant (may vary widely,) and, as long as Vbe remains 0.6, then the voltage on Re remains close to 8.1V, even with wide changes in β. It's a voltage-input amplifier, where input current is unimportant, and β is unimportant. \$\endgroup\$ – wbeaty Apr 18 '20 at 22:43
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@wbeaty already gave an answer in regard to the voltage divider, and why you don't want to load it too much. But here's another way to look at it, and I think is relevant in BJT biasing. \$\beta\$ is not a constant (highly dependent on temperature) and thus it will make the collector (and therefore emitter) current, susceptible to the changes in \$\beta\$. This will disrupt the Q point of the amplifier.

Think about the following configuration, with an emitter resistor:

enter image description here

The collector current turns out to be:

$$ I_C=\beta I_B=\dfrac{\beta (V_{CC}-V_{BE})}{R_B+(\beta +1)R_E} \tag1$$

If you let:

$$R_B<<(\beta +1)R_E \tag2$$

And noting (as an approximation) that \$1<<(\beta +1)\$, then:

$$ I_C\approx\dfrac{V_{CC}-V_{BE}}{R_E} \tag3$$

That last equation is pretty much independent of the variability of \$\beta\$ so long as (2) is met. You could make that assumption, if for example, \$10R_B=(\beta +1)R_E\$, but that's just a rule of thumb (eg. you could make it a factor of 15 or 20). Note that this will work the same if you had an input voltage divider, instead of the single \$R_B\$ shown, those two resistors would combine into a Thevenin equivalent resistance, but the same concept applies.

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  • \$\begingroup\$ I think I might have got it now. Do you mean Rb >> (beta+1)Re instead of the reverse in terms of resistance value, and Ic ~ (Vcc - Vbe) / Rb instead of Re? Otherwise it worked the opposite since we want the current at the base to be fairly "stiff" if emitter load is subject to change. \$\endgroup\$ – KMC Apr 18 '20 at 6:12
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    \$\begingroup\$ @KMC No, wrong. We want the voltage at the base to be stiff not the base current, look at this example electronics.stackexchange.com/questions/451441/… In short, only if Ve is fixed for a given RE, we can achieve fixed emitter current (beta independent) And to be Vbe independent as well, we want Ve >> Vbe to also be true.This is why Ve = 1V or more is recomendet by AOE \$\endgroup\$ – G36 Apr 18 '20 at 7:30
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    \$\begingroup\$ @KMC and, if we ignore this rule-of-thumb, then our first 2N3904 circuit may seem to work, but then when we swap in a different 2N3904 transistor, it acts all screwy. The biasing will be all wrong, because the betas of 2N3904 can be anything between β = 100 to 300. This entire "voltage-input design philosophy" is extensively treated in AOA textbook, where BJT transistors are seen as "transconductance" elements, voltage-controlled current-sources. So, a BJT is the same as a vacuum tube or FET, with a control-voltage input, and where base-current is just an unimportant leakage. \$\endgroup\$ – wbeaty Apr 18 '20 at 23:03
  • \$\begingroup\$ Many thanks to the links and comments that correct some of my erroneous confusion. So this 1/10 ratio of Rb (or its Thevenin's) to Re is just a guide to pick the base resistor(s). If \$\beta\$ is large enough (e.g. 50-200) then even if Rb ~= Re or to the more extreme case (just for the sake of argument), if Rb is slightly larger than Re, the (\$\beta+1\$)Re would still dominate the impedance looking into the base - but 1/10 just make it "nicer.' \$\endgroup\$ – KMC Apr 19 '20 at 1:17

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