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I'm reading Art of Electronics, and in the section Transistor Current Source, they mention "The base voltage can be provided a number of ways. A voltage divider is OK as long as it is stiff enough. As before, the criterion is that its impedance should be much less than the DC impedance looking into the base (Beta*R_emitter)"

Why is this ?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ I would use an (in my view) easier description: The current flowing through R1, R2 should be significantly larger than the current flowing into the base of Q1. Imagine if Q1 needs 10 uA base current to make the desired Ic (=current through R_load). If the current through R2 is normally only 20uA, that extra 10 uA to the base makes a huge difference. Now we make 200 uA flow through R2 and R1, then an extra 10 uA for the base makes much less difference. \$\endgroup\$ – Bimpelrekkie Oct 31 '17 at 8:18
  • \$\begingroup\$ To make more current flow through R1, R2, they need to have a lower value right? What is the output impedance of the voltage divider R1, R2? And the input impedance at the base of Q1, is \$\beta*R_{emitter}\$ right? Since we do not want to disturb the voltage coming out of the voltage divider, the impedance from the transistor loading it should be much higher than the output impedance of the voltage divider. \$\endgroup\$ – Bimpelrekkie Oct 31 '17 at 8:25
  • \$\begingroup\$ If you're reading AoE also read page 10 and 11 (Edition 3). It explains why the load impedance should be way larger than the source impedance. \$\endgroup\$ – dirac16 Oct 31 '17 at 9:16
  • \$\begingroup\$ A general 'rule of thumb' is to have at least 10x the required base current through the biasing resistors (R1,R2). (source impedance < load) This allows for a simple voltage divider calculation by ignoring the base current. \$\endgroup\$ – JIm Dearden Oct 31 '17 at 10:16
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The current sink will work regardless of their rule. It just won't provide the easily predicted value of current for the load, if you don't follow the rule closely. Let's see why.

After converting the base pair to its Thevenin equivalent:

schematic

simulate this circuit – Schematic created using CircuitLab

You can apply KVL and get:

$$I_B=\frac{V_{TH}-V_{BE}}{R_{TH}+\left(\beta+1\right)R_E}$$

Now you can figure the following:

$$V_B=V_{TH}-I_B\cdot R_{TH}$$

Given that the collector current (aka the load current) is \$\frac{\beta}{\beta+1}I_E\$, it must be the case that the load current is:

$$\begin{align*} I_{LOAD}&=\frac{\beta}{\beta+1}\cdot\frac{V_E}{R_E}=\frac{\beta}{\beta+1}\cdot\frac{V_B-V_{BE}}{R_E}\\\\ &=\frac{\beta}{\beta+1}\cdot\frac{V_{TH}-I_B\cdot R_{TH}-V_{BE}}{R_E} \end{align*}$$

Substituting in \$I_B\$ you get something like this:

$$I_{LOAD}=\left[\frac{\beta}{\beta+1}\right]\cdot\left[\frac{V_{TH}-V_{BE}}{R_E}\right]\cdot\left[1-\frac{R_{TH}}{R_{TH}+\left(\beta+1\right)R_E}\right]$$

Which can also be written out as (to make the ratio stand out and to emphasize that it is the ratio of two certain resistance values that is important in the following discussion):

$$I_{LOAD}=\left[\frac{\beta}{\beta+1}\right]\cdot\left[\frac{V_{TH}-V_{BE}}{R_E}\right]\cdot\left[1-\frac{1}{1+\frac{\left(\beta+1\right)R_E}{R_{TH}}}\right]$$

Note that the first factor is almost always very close to 1. So it can be ignored. The second factor is the current we'd expect when we designed the resistor divider at the base, in the first place. As you would expect that the emitter would be \$V_{BE}\$ less than the Thevenin voltage and of course this voltage across \$R_E\$ would produce the expected current there. That is, if you use the unloaded divider voltage!

Now, the third factor is the issue here. You want this to be 1, since that means your unloaded divider voltage is the right one to use in predicting your current sink value. But if it isn't 1, then the actual value will be different than the expected one (given no load on the divider.)

If you look at the third factor, I think you can see that if \$R_{TH}\$ is small compared to the value of \$\left(\beta+1\right) R_E\$, then the second term of that factor is close to zero and so the third factor will be close to 1. But if \$R_{TH}\$ isn't small in comparison, then that fraction (the second term of the third factor) will significantly reduce the third factor's value from 1 to something smaller. And so the predicted value won't be nearly as close to the actual value as hoped.

You can also see this as: "If the base current is small compared to the available current flowing through the base pair of divider resistors, then the predicted voltage at the divider will be close to the actual voltage present there and therefore the base will obtain that nearby value and reality will be closer to prediction." That's the qualitative hand-waving that also gets you to the same place.

But it all becomes quantitatively clear in the math, itself. The math not only tells you the same thing as the hand-waving does, it also tells you by exactly how much you might be off if you don't follow the rule by some amount. So it provides both the insight as well as quantities you can use if you choose not to follow the rules.

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  • \$\begingroup\$ Super helpful explanation! My only question is where did B/(B+1) come from in the 3rd equation you listed? Shouldn't I_load just be Ve/Re ? \$\endgroup\$ – VanGo Oct 31 '17 at 21:08
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    \$\begingroup\$ @VanGo The collector current is slightly less than the emitter current. That's where that term comes from. I just added a short note to that effect in the answer to help clarify. \$\endgroup\$ – jonk Oct 31 '17 at 21:25
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You can also "regulate" the operating point, by sensing the Vce, using this circuit

schematic

simulate this circuit – Schematic created using CircuitLab

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