I am in the 6th week in Navy school and we are on voltage regulators or circuits that have components that maintain a constant voltage output. I need an explanation on how to work out this problem here:

Find \$V_o\$ and the zener current where \$R_L = 1 k\Omega\$:

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The answer is 11.3 v for the output voltage and 36 mA for the zener current. This is my attempt:

\$I_e = I_b + I_c = (1 + \beta)I_c\$ and \$V_o = I_e \times R_L\$

I wanted to say that the base voltage is 12 volts but that doesn't seem to work out with this problem.

  • 1
    \$\begingroup\$ Would Electrical Engineering be a better home for this question? \$\endgroup\$ – Qmechanic Nov 23 '14 at 1:31
  • \$\begingroup\$ Have you learned about emitter follower circuits yet? \$\endgroup\$ – The Photon Nov 23 '14 at 1:53
  • 1
    \$\begingroup\$ newbie, why do you say a base voltage of 12V doesn't work out? \$\endgroup\$ – Dan Laks Nov 23 '14 at 2:04

You have an error in your equation. The calculation of \$I_e\$ is: $$I_e=I_b+I_c = (1+\beta)\color{red}{I_b}$$ Not $$I_e=I_b+I_c = (1+\beta)\color{red}{I_c}$$

To solve this particular problem, however, rote application of memorized equations is not the best path. Here's the thought sequence that will guide you to the solution.

  1. The 12V zener diode on the base of the transistor probably means \$V_b\$ is being held at 12V. So check that the voltage at the cathode of the zener is greater than 12V. Looks like there's a 20V source connected to the cathode with a 220\$\Omega\$ resistor, so it's probably safe to say the zener is able to clamp at 12V (this may not end up being true if it turns out that more than 8V is dropped across the resistor).

  2. Under normal circumstances, the emitter of a BJT will be approximately 0.7V less than the base. We already know the base is 12V, so the emitter must be 11.3V.

  3. The current through \$R_L\$ can be found immediately using Ohm's Law: $$I_{R_L}=\frac{V_e}{R_L}=\frac{11.3V}{1k\Omega}=11.3mA$$

  4. Since there is no other load in the circuit, we know that all of the current through \$R_L\$ must come through the transistor. Therefore, \$I_e=11.3mA\$

  5. Now we can use the equation you started with (with the correction I mentioned above), where \$\beta=50\$. Rearranging to solve for \$I_b\$, we get: $$I_b = \frac{I_e}{1+\beta} = \frac{11.3mA}{1+50} = 221.6uA$$

  6. We can do a quick check here to confirm our assumption in step #1 above. The base current of 221.6uA is the minimum current through the 220\$\Omega\$ resistor that will allow this voltage regulator to work. A quick look at Ohm's Law shows us that the theoretical minimum voltage drop across the resistor is then \$V_{min}=221.6uA*220\Omega = 49mV\$. That's far less than 8V, so we're good. Of course, the resistor will end up dropping exactly 8V because the necessary additional current will be consumed by the zener diode.

  7. Now that we know how much current is going into the base of the transistor, we can solve how much is going through the zener diode. The only path for current to get to the base and the zener diode is through the 220\$\Omega\$ resistor. The voltages on either side of the resistor are known: 20V and 12V. Therefore, using Ohm's again, we can find how much current is going through that resistor: $$I_R = \frac{20V-12V}{220\Omega} = 36.36mA$$

    So 36.36mA is going through that resistor, of which 221.6uA is going into the base of the transistor. That leaves \$36.36mA - 221.6uA = 36.14mA\$ left for the zener diode. Looks like your solution guide apparently rounded to just 36mA.


First, there is a mistake in your initial equations. Ic = Beta * Ib and therefore Ib + Ic = (1 + Beta)Ib, NOT (1 + Beta)Ic - which will make for a huge difference quantitatively !

Furthermore, I would like to answer this with some added commentary that applies to real world design. First, the voltage at the base must be the rated zener voltage (12V) so long as the zener diode has enough ballast current through it. This is the purpose of the 220 Ohm resistor betwenn Vi and the zener/base node. Without it, there would be no current in the zener because for the transistor to be in forward-active mode (which it needs to be in for this circuit to work) current must flow INTO the base, NOT out of it. And therefore the 220 Ohm resistor ALSO acts to bias the transistor (i.e., provides current INTO the base). The bulk of the current in the 220 Ohm resistor - but not quite all of it - is biasing the zener diode. The remainder is the base current. Using KVL & KCL, the current in the 220 Ohm Resistor (call it Rb; 'b' for 'bias' or 'ballast'):

I[Rb] = (Vi - Vz)/Rb = 8V/0.22Kohms = 36.36mA

The current in Rb that diverts into the base is:

Ib = Ic/Beta = Ic/50 = (Ie - Ib)/50; therefore,

Ib = Ie/51.

Now, Ie is just V0 / RLoad = 11.3V / 1.0Kohms = 11.3mA. . . . because . . . . .

When a transistor is in forward-active mode, assuming no component power ratings are being exceeded, there is a FORWARD diode drop from Vb to Ve, i.e.,

Ve = Vb - 0.7V. . . . . therefore . . . .

Vo = Ve = Vb - 0.7V = Vz - 0.7V = 12V - 0.7V = 11.3V and therefore:

Ie = ILoad = IRLoad = 11.3V / 1Kohms = 11.3mA.

. . . . . and now, by the way, we can calculate Ib:

Ib = Ie/51 = 11.3mA/51 = 222uA which as expected is a tiny fraction of the ballast current, or the current in the zener diode. Finally . . . . .

Iz = I[Rb] - Ib = 36.363mA - 0.222mA = 36.142mA

A couple of interesting points, again practical design matters. First, note that the power we are asking Rb to dissipate is:

P[Rb] = Rb * V[Rb]^2 = 220 ohms * 64 V*V = 0.29W . . . .

which is more than a quarter watt. So if you are going to build this you'd better
choose at least a half-watt resistor for Rb. Power dissipated by the transistor is approx

Px = (Vc-Ve) * Ie = (20-11.3)V * 0.012 A = 0.1W = a tenth of a Watt.

Does something seem weird here ? There is more current in the ballast resistor than there is in the load ! And more power dissipated in the ballast resistor than in the transistor. I am not saying this is wrong, it will work, but as a practical matter, it is wasteful; it would be better to use a zener that operates well with only about one-tenth of the current in this circuit, or about 3.6mA. Such things do exist. That way, you can use a much lower value AND power Rb than 1/2Watt ("an exercise for the student" - what would be the new Rb value ?) and it will work fine because Ib is still a small fraction of Iz (about 0.064 or one-sixteenth - please confirm) so the transistor will still regulate the output voltage just fine. Rload = 1K is not much of a load in this circuit for discreet components. In fact you can keep this new lower Rb value, and lower RLoad substantially, thereby INcreasing ILoad, and so long as you do not exceed the power rating of the transistor, that would be fine. ALSO, ILoad is limited by Beta, because with excessive ILoad, Ib requirement will become too large, robbing too much current from the zener, and the zener will no longer regulate. Those two differences brings us closer to a real world practical example. For a circuit presented in a textbook to illustrate the principals of analysis, it's OK, but from a practical viewpoint, it should have the adjustment I mentioned.

One more important point. Note that Vi = 20V "unregulated" means that Vi varies somewhat. In the real world this could be quite a bit, in fact. As Vi varies so do a lot of other things like many of the currents we calculated earlier. The trick is that the zener must have a very flat shelf in its V/I transfer curve, which means Vz = rock solid 12V no matter what Iz might be. It will never be perfectly flat, so Vz and therefore Vo will vary somewhat with varying Vi. But the variance will be much attenuated. If you look at specs for power supplies, you will see this attenuation factor. The more attenuation, the better.

One more question. Why not just use a zener without a transistor (assuming you want Vo = 11.3V and you can find a zener at that rating) ? In fact, there are some real-world applications that do just that, when the load requirement is not much. You probably already saw that in a previous lecture/book chapter.

Ans: For load current requirements in the neighborhood of the zener ballast current or higher, the answer is because the transistor, with its remarkable current amplification property - it's all about the Beta - isolates the load from the zener. Do you see that ?

  • \$\begingroup\$ I have GOT to learn to use a nice equation editor like you guys and so many others on this board are doing. How are you doing that ? With an external app like MS Word ? Or is there a way using the editor in this app ? Thanks in advance ! \$\endgroup\$ – LateNiteOwl Nov 23 '14 at 3:49
  • \$\begingroup\$ Good eye Dan, quite right ! Thanks for pointing it out. I edited it. \$\endgroup\$ – LateNiteOwl Nov 23 '14 at 3:52
  • \$\begingroup\$ Quite welcome. I'll remove my comment in a minute. By the way, the awesome equation formatting is done using LaTeX. If you search online, there are a bunch of tutorials. Here's a quick primer: everything between double dollar signs ($$) will be interpreted as an equation. \$\endgroup\$ – Dan Laks Nov 23 '14 at 3:54
  • \$\begingroup\$ @LateNiteOwl: The equation editor is part of LaTeX and is integrated into this site's online text editor. Here is a wonderful resource. :-) \$\endgroup\$ – EM Fields Nov 23 '14 at 6:40

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