Help with Voltage Divider driven NPN emitter follower

Question

I am trying to solve the Exercise 2.5 in Art of Electronics. Use a voltage divider and emitter follower with 15V supply to make a 5V output, and within 5% of 5V at a 25mA load. This has been solved before in this forum: Designing a stiff voltage source using an emitter follower

They seemed to use a guess-and-test method for choosing the voltage divider resistors and I am wondering if there is a way to work backwards, saying that the emitter voltage is 4.75V with the 25mA load (worse case, 5% below the 5V target, and say a worst case beta of 30), to algebraically figure out the exact and most efficient resistors to choose to meet these requirements.

Answer 1

There is a wide range of resistors you can use. We usually use rules of thumb like 'at least 10x current flowing through the divider to current supplied by it'. But it could be 20x, or 50x, or 5x, with only slightly different performance. Nailing the exact resistor values, or 'most efficient' resistor values, is not possible.

The main thing is to use resistors low enough to swamp the 2:1 beta variation that the transistor will throw at you, and still keep within output voltage spec.

For instance, let's put a lower limit on R1/2 for a beta variation of 30:100 at 25mA output current. The base current will vary between 25mA/30 and 25mA/100 which is 830uA to 250uA. You have a voltage output spec of +/- 5%, let's assign +/-1% to beta variation, as there are other error terms like load variation, VBE tempco, line variation that need to fit into that 5%. A 2% swing on 5v is 100mV. So with a deltaI of 580uA, you can tolerate a deltaV of 100mV, so R1/2 needs a tap point resistance of less than 100m/580u = 172 ohms. As the resistors are roughly a 2:1 ratio, that puts them in the ballpark of 250 and 500 ohms. They can be less, which will result in less voltage variation. Fiddle about with the exact values to get the right ratio, while maintaining their parallel resistance less than 172 ohms.

That illustrates to some extent why we don't use as simple a circuit as this for a regulator, and/or use transistors with a higher minimum beta.

Once you've factored in load variation, which also puts a constraint on minimum R1/2 values, and the other error terms, you can then see how close you are to your +/- 5% spec, and if well clear, perhaps increase that +/- 1% error allocation to beta variation to allow those resistors to be bigger, and so use less current.

That's why it seems like we 'guess and test' to get these values. There are so many assumptions and tradeoffs that we're rarely happy with the first values we arrive at, when we see the consequences of the assumptions we have made to get to them.

Answer 2

You are the designer, which means that you are going to choose the NPN transistor to be used. As the load is around \$25 mA\$, a small signal model can be used which normally has a \$\beta = 100\$ or more. Take your pick of a transistor with say \$\beta = 150\$. Good specs give a range of beta, say \$\beta = 100..200\$.
Then there's the \$V_{be}\$ that's of importance, look them up for the transistor you chose and in the operating points of \$0..5V/25 mA\$, assuming the current through \$R_3\$ neglible with respect to \$I_1=25 mA\$.

Now establish 'worst' and 'best' conditions. It's clear that the output voltage will be highest if \$I_1=0\$ and \$\beta\$ is maximal and \$V_{be}\$ is minimal. The other way around, \$V_1\$ will be lowest if \$I_1\$ highest, \$\beta\$ is minal and \$V_{be}\$ is maximal.

Now write the output voltage as function of the parameters \$R_1..R_3\$, \$V_{be}\$, \$\beta\$, \$I_1\$: \$U_1 = f (R_1, R_2, R_3, V_{be}, \beta, I_1)\$.
Establish the total differential of this function with respect to \$V_{be},\beta, I\$, like: \$\partial U_1 = \frac{\partial U_1}{\partial V_{be}} . dV_{be} + \frac{\partial U_1}{\partial \beta} . d\beta + \frac{\partial U_1}{\partial I_1} . dI_1\$, 'simplify' to \$\Delta U_1=\frac{\partial U_1}{\partial V_{be}} . \Delta V_{be} + \frac{\partial U_1}{\partial \beta} . \Delta\beta + \frac{\partial U_1}{\partial I_1} . \Delta I_1\$ and put in the \$\Delta\$s the minima and maxima that lead to the worst and best results respectively.

Now solve \$R_1, R_2, R_3\$ to stay within your desired 95..100% of 5V. This last step is the tricky part, and that's where electrical engineers developed a preference to a mixture of 'educated guesses' and iteration ('trial and error') that you seem to want to pass-by at the cost of higher efforts in solving sets of mathematical equations.
Actually your approach might be better in very complex situations which most engineers would try to avoid during their design by chopping up the problem is small partial problems, which then can be solved with the methodologies of 'guess and iterate'.

Answer 3

Whatever your intended output voltage is, your base voltage should be about 0.65V higher. If you're trying for 5V, create a voltage divider that will hold the base at 5.65V. If the 25mA load accounts for most of the current (R3 will take some) and you have a gain (hfe) of at least 50, 3K would be a good total resistance.

That should get you close. If you want it more exact, then do the stuff Neil_UK said.

Answer 4

Unless this is just an academic exercise you need to include hFE and Vbe variations with temperature (including self heating).

If it's an oversimplified academic exercise (ideal source, ideal resistors, real or semi-real transistor, junction temperature constant at some default value), you can easily determine the maximum R1||R2 if you consider minimum hFE and Vbe constant. We know Vbe will not be constant with current though, so R1||R2 will have to be lower to account for the change.

Determining the exact value involves a nonlinear equation (because of Vbe) so it will be a hassle to solve it in closed form. Probably possible though. This is the kind of problem University professors love to set- just possible to solve algebraically- and is not necessarily how you should create real designs.

It's trivial to set up the equation and solve it iteratively though. Or just use SPICE and manually do a binary search if you don't feel like setting up and running a solver.

In reality, with some judgment, one or two iterations is close enough (particularly when our rules of thumb get us into the ballpark), but if you wanted to get it to within 6 decimal places, it's not that much harder (just even more useless).