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Having done the exercise 2.2 from Horowitz's "The Art of Electronics," I feel uncertain and would like somebody familiar with the matter to check my solution. With this aim in head I've searched the forum and found this topic: Designing a stiff voltage source using an emitter follower, which gives a solution. But, even if this is useful and solves the exercise, it uses another approach than the one intended to be used in the textbook (this exercise is given in the section talking about follower's input and output impedance). Furthermore, my solution gives lower values for the resistors and I don't understand where my error is.

Briefly, my solutions are (for simplicity I don't use emitter resistor—just the load):

1st version:
Voltage divider equation is $$V_{in} = V_{source}\frac{R_2}{R_1 + R_2}$$ where \$V_{in} = 5.6V\$ (input to the follower) and \$V_{source} = 15V\$. From this we can express \$R_2\$ from \$R_1\$: $$R_2 = \frac{5.6}{9.4}R_1$$ And for simplicity we denote \$\frac{5.6}{9.4}\$ as \$k\$

Next, the resistance of the divider lower leg is actually parallel resistance of \$R_2\$ and \$r_{in}\$ (the follower input resistance). So, the actual voltage at the lower leg is $$V_{div} = 15 \frac{\frac{R_2r_{in}}{R_2 + r_{in}}}{\frac{R_2r_{in}}{R_2 + r_{in}} + R_1}$$ solving this and using \$k\$ we get $$V_{div} = 15 \frac{kr_{in}}{(k + 1)r_{in} + kR_1}$$

Dividing the nominator and denominator here by \$k\$ and denoting \$k' = 1/k\$ we have

$$V_{div} = 15 \frac{r_{in}}{(1 + k')r_{in} + R_1}$$

\$V_{div}\$ should not be lower that \$5.35V\$. So,

$$15 \frac{r_{in}}{(k' + 1)r_{in} + R_1} \geqslant 5.35$$

Finally, resolving this regarding \$R_1\$ and using the relation \$r_{in} = (h_{fe} + 1)R_{load}\$ from the book gives: $$R_1 \leqslant \frac{(15 - 5.35(1 + k'))(1 + h_{fe})R_{load}}{5.35}$$

Choosing the value of \$R_{load}\$ from condition of \$I_{max} = 25mA\$ at \$5V\$ which gives \$R_{load} = 200\Omega\$ and using \$h_{fe} = 10\$ as proposed in the topic mentioned above I've got \$R_1 \leqslant 275.4\Omega, R_2 \leqslant 164\Omega\$

2nd version
We can also consider \$R_{source}\$ which is the divider impedance:

$$R_{source} = \frac{k}{(k + 1)}R_1$$

The \$R_{source}\$ formula is derived using one more time the denotion \$k = \frac{5.6}{9.4}, R_2 = kR_1\$

Then we use \$Z_{out} = \frac{Z_{source}}{h_{fe} + 1}\$ from the book substituting \$Z\$ with \$R\$ which gives $$R_{out} = \frac{k}{(k + 1)(h_{fe} + 1)} R_1$$ As the load forms voltage divider with the follower output impedance and as the voltage drop at the load shoud be \$ \geqslant 4.75\$ we have $$V_{load} = V_{out} \frac{R_{load}}{R_{load} + R_{out}} \geqslant 4.75$$

Resolving this with \$V_{out} = 5V\$ and using earlier derived formula for \$R_{out}\$ we have $$R_1 \leqslant \frac{0.25(k + 1)(h_{fe} + 1)R_{load}}{4.75k}$$

Using one more time \$R_{load} = 200\Omega\$, this gives \$R_1 \leqslant 310 \Omega, R_2 \leqslant 184 \Omega\$

So, as one can see, the values are about 10 times less than in the topic mentioned above and, furthermore, they are different in my 2 versions. I can't see where is my error.

Could someone help me with this?

Thank you very much

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Wow, I have the first edition of this book - I guess I got it 25 years ago ;-). One of the approaches I "learned" from this bookset (=including workbook) is that you should either build it to try or simulate it and try to understand what you see from there. O Without referring to the book, here is my approach.

At 50mA, the minimum DC gain is rated at 60 (not 30). I'll use 30 anyway to stick to your analysis. We allow a voltage change of 5% which is 0.25V. B-E voltage drop has a big impact on this. If the quiescent collector current is 10mA, the voltage drop can be 0.1V @30mA.

So we have about 0.15V left for the load change versus the bias resistors. The load looks about 30 times bigger (no need to be too precise), lets call it Re. First we suppose that R3 is part of the source impedance (Rs). Vout/Vopen = (Re/(Re+Rs). Hence for a drop of .15V (3%) we have \$0.97 = 6k/(6k+R_s)\$. \$R_s=6000/.97-6000=180\Omega\$. 15V is about 3 times 5V so \$R_1\$ is about two times \$R_2\$ and \$R_s\$ is about \$\frac{1}{1/2*R2+1/R2} = \frac{1}{1.5/R2} = R_2/1.5 = 185\Omega\$.

So R2=277.5Ohm, which we round to 270 - a value we can find in the manufactured values. We need 5V+0.6V on the base, so $$R1 = \frac{15V-5.6V}{5.6V} * 270 = 453 Ohm$$. The closest is 430 Ohm (keeping in mind that the voltage will drop because of R3). So bias voltage is $$15V*\frac{270}{430+270}= 5.78V$$. With 470 Ohms, it would be 5.47V, which is too low.

If I set 10mA as quiescent collector current, then I have R3=500Ohm, or 510 Ohm for a standard value. The effective R2 is then 270 Ohm//(510 Ohm*30)= 265 Ohm. The bias voltage is 15*265/(265+430)= 5.72V and I get about 5.1V at the output.

Adding the load of 200Ohm, the effective R2 is 265 Ohm//6kOhm=253.8 Ohm. The bias voltage is \$15V\frac{253.8}{253.8+430}=5.57V\$. Our B-E voltage drops itself from -0.6V to -0.7V, the output is at 4,87V. The total drop is 5.1-4.87=0,23V, just under 5%.

Another check of the drop is to check how much the voltage change in R1 is due to the change in current. The current change is 25mA/30=0.83mA. Hence the voltage drop is 430*.00083=0.38 One could think that the extra current of 25mA results in an extra base current of 0.83mA and hence a drop over the 430 Ohm resistor of 0.36V, higher than the 0.23V. However, with a drop of 0.13V, the current in the 265 Ohm equivalent resistor drops by 0.5mA. Hence the current in R1 increases by only 0.33mA which corresponds to a drop of 0.14V (roundings, etc. will explain the difference of 0.1V). So this check using currents also verifies.

Therefore, R1=430, R2=270 and R3=510Ohm looks like a good solution. One could tune R1 and R2 a bit to find an output value closer to 5V when there is no load, but 5.1V within reasonable limits of the design goal (2%) The estimated voltage drop is just under 5%.

In this process, I've used only the standard resistor values. I did not do a very precise calculation, but followed a 'hands-on' approach because practical components add variation anyway.

That's where the Art is: knowing how to get to a result without resorting to the full mathematical formula which gets you a 0.001% precise solution.

Also, once you get a solution, you have to verify it. I did it by checking the actual result without and with a load. I kept in mind the variation of the B-E voltage by checking the datasheet while already choosing a reasonable quiescent current (10mA looked reasonable). In the endd, it all adds up to a good solution.

You can notice that my solution does not respect your constraints - R2>184!!

Now this does not indicate where your error is, but it does indicate that the difference is not an order of magnitude.

Your second approach says that R1*(Vlow/Vdrop)=Rload/Rout. There is an issue with units where. On the left I have a resistor value, on the right a ratio. I can't explain this formula. It is not clear how you get to the inequality of the first version either. It looks probable that your mistakes are there.

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  • \$\begingroup\$ Hello, Thank you for your answer. Unfortunately, I could not fully grasp it on my beginner level. For exapmple, I can't figure out where 30mA value comes from. Moreover, Horowitz don't give the notion of quiescent current so far (I'm at the section 2.03 using 2nd edition). More important for me is that I cant see how to use your solution to check if and where the mine is wrong. Any guidelines would be great. For a while, I'll update my original question to show in more detailed manner how my calculations were done. \$\endgroup\$ – Ruslan Jun 12 '16 at 14:51
  • \$\begingroup\$ The quiescent current is the current when there is no load. \$\endgroup\$ – le_top Jun 12 '16 at 15:35
  • \$\begingroup\$ The 30mA comes from the datasheet. When you look at the B-E voltage versus collector current, you can estimate the voltage change with collector current change. \$\endgroup\$ – le_top Jun 12 '16 at 15:48
  • \$\begingroup\$ Assuming from your first result that R1=275.4Ω and R2=164Ω and Rload=200Ω. Then Rload*(1+hfe)=6200Ω. And 6200Ω//164Ω = 159Ω. With Vdiv=159/(275.4+159)*15=5,49V representing a drop of 0.11V versus the 0.25V you expected. This confirms that your result is not ok [first requirement before checking other formulas] \$\endgroup\$ – le_top Jun 12 '16 at 16:29
  • \$\begingroup\$ I assumed hfe=30 in my calculations, and now I see you write 10. Assuming from your first result that R1=275.4Ω and R2=164Ω and Rload=200Ω. Then Rload*(1+hfe)=2200Ω. And 2200Ω//164Ω = 152.6Ω. With Vdiv=152.6/(275.4+152.6)*15=5.35V. So the first case is ok when assuming hfe=10. \$\endgroup\$ – le_top Jun 12 '16 at 17:17

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