simulate this circuit – Schematic created using CircuitLab

I am really struggling to understand how this circuit works theoretically. What will be the currents through the resistors? My understanding (which I suspect is wrong) is as follows:
Ir1 = (7-0.7)/100 = 63mA (B-C is forward biased)
Vemitter = 7 + 0.7 = 7.7V
Ir2 = (12 - 7.7)/1k = 4.3mA
Vcollector = 7 - 0.7V = 6.3V
Ir3 = 6.3/1k = 6.3mA

Now, what confuses me the most is the zener diodes. What is their purpose in the circuit? I have not made use of them in my calculations. Could someone please help me understand this circuit? Thanks!

  • \$\begingroup\$ According to the data sheet the 1N3347A is a 160V, 50W zener. The 2N3906 is just a small signal transistor. The zeners in the circuit make no sense to me at all. \$\endgroup\$ – JIm Dearden Feb 13 '17 at 10:55
  • \$\begingroup\$ @JImDearden because it is actually 1N4733A :) \$\endgroup\$ – C K Feb 13 '17 at 11:05
  • \$\begingroup\$ @ÇetinKöktürk Lol - must go to specsavers - its 5V1 1W, now that does make sense \$\endgroup\$ – JIm Dearden Feb 13 '17 at 11:06
  • \$\begingroup\$ Where did you find this circuit? What is it supposed to do? \$\endgroup\$ – Oskar Skog Feb 13 '17 at 15:59

enter image description here

With 5V1 zeners, the base would be at 7V0 + the tiny voltage drop due to the base current. (Ib * 100)

Assuming a 0.7V Vbe drop the emitter will be at 7V7. This gives an emitter current of (12-7.7)/1000 = 4.3 mA

Taking beta = 100 the base current, Ib, will be 4.3/100 mA or 43uA

This takes the base voltage to 7V0 + 0.0043V which is approximately 7V0 (1dp).

Collector current will be Ie + Ib which is approx. Ie = 4.3mA, this gives a collector voltage of 4V3.

This is below the 5V1 of D2, so D2, R4 will have no effect.

If V1 increases over 7V the base voltage follows and the current through the transistor drops. The emitter voltage rises and the collector voltage falls.

If V1 decreases D1 clamps the base voltage at 6V9, the emitter voltage falls to 7V6 and the collector voltage rises to 4V4.

This is still under the 5V1 of D2 so this part of the circuit still has no effect.

R4, D2 seem totally redundant.

What does the circuit actually do - no idea.

  • \$\begingroup\$ Thanks so much for the explanation. I still have a question though: How is the voltage across R1 0.1V? From what I see, V1 is at 7V, and the Vbase is at 7V. So the voltage across R1 should be (7-7)V = 0V. Am I missing something here? \$\endgroup\$ – abruzzi26 Feb 13 '17 at 12:11
  • \$\begingroup\$ Oops, my mistake - you are correct - At 7V the base current of 43uA will only produce a voltage drop of 43 * 100/1000000 = 0.0043V across the 100 ohms which is naff all, this puts the base voltage at 7.0043V - given the approximation of Vbe to 0.7V (0.65 +/- 0.5V) we can safely ignore it. Will amend answer. \$\endgroup\$ – JIm Dearden Feb 13 '17 at 13:00
  • \$\begingroup\$ "What does the circuit actually do - no idea." Waste electricity. It will dissipate a little power as heat, nothing else, assuming both the 12V and 7V sources are actually sources. There aren't any outputs in this circuit, no LEDs, speakers, connections to other circuits, light bulbs, high wattage resistors, etc. \$\endgroup\$ – Oskar Skog Feb 13 '17 at 13:56
  • \$\begingroup\$ If 7V is a load rather than a source, this circuit might buck 12V to 7V in some extremely specific circumstances. The E-B junction for Q1 will act as a diode (voltage drop ~0.7V), let's call it D3. The stuff connected to the collector will only draw some current from the 12V source. D1 will prevent the voltage drop across R2-D3 from exceeding 5.1 V. If R2*I+0.7V <= 5.1 V then out = 12V - 1000 ohm * I - 0.7V - 100 ohm *I, but if R2*I+0.7V > 5.1 V then out = 12V - 5.1V - 100 ohm * I. U(I) = 11.3 V - 1100 ohm * I if I <= 4.4 mA and U(I) = 6.9V - 100 ohm * I if I > 4.4 mA \$\endgroup\$ – Oskar Skog Feb 13 '17 at 14:14
  • \$\begingroup\$ @OskarSkog I never deal with "what if..." senarios as It then depends upon making too many assumptions (never good in an engineering context). I simply deal with the question that was asked, as it was asked. As far as I'm concerned its really not worth spending any more time on, I appreciate your input and if you wish to post up a second answer please feel free to do so. \$\endgroup\$ – JIm Dearden Feb 13 '17 at 14:35

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