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I am facing problem in solving such mixed circuits with bjt, diode, opamp and Zener diode. In the above problem 1.zener can have 3 states (ON, OFF, Reverse ON) 2.diode can have 2 (ON, OFF) 3.BJT can have 3 (Cutoff, Active, Saturation) Should I go for trial and error approach or is there a better method to approach such mixed circuit problem ?


closed as off-topic by ThreePhaseEel, Voltage Spike, PeterJ, Lior Bilia, Bimpelrekkie Dec 12 '17 at 10:38

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    \$\begingroup\$ You already know which state the zener is in based on its direction, same for the diode labled D. Figure out what the voltage is between the 15k and 1k resistor first, then figure out the current through D. The rest should be simple \$\endgroup\$ – Voltage Spike Dec 7 '17 at 16:46
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    \$\begingroup\$ I'm voting to close this question as off-topic because homework needs an attempt at solution if posted. \$\endgroup\$ – Voltage Spike Dec 7 '17 at 16:47

You can think of it this way, the Base of the transistor will be 0.7V above the emitter which is at -12V, so the base will be at -11.3V, the diode in series with the base adds another 0.7V bump, so the voltage at the right of the 1K resistor will be -10.6V, the Zener voltage is 5V and its anode is connected to the -12V supply, so the voltage at the cathode of the zener will be -7V.

Now you have the voltages at both sides of the 1K resistor, and since this resistor is in series with the base, the current through the resistor is the same as the base current, so the current is:


The beta of the transistor is 30, so

$$I_C=\beta I_B=30(3.6mA)=108mA$$

Which is wrong, because that would imply that the colector voltage would be at -237V, so the current IC, is simply \$\frac{0-(-11.8)}{2.2k}=5.36mA\$ (I used -11.8 instead of -12 due to the 0.2V of Vce sat)

  • \$\begingroup\$ Right..it is saturated \$\endgroup\$ – Trevor_G Dec 5 '17 at 1:16
  • \$\begingroup\$ Thanks AJ. According to your solution, you assumed BJT to be in active region initially and the diode became ON with it. But how can you comment on Zener diode status that it will be Reverse ON ? \$\endgroup\$ – user170930 Dec 5 '17 at 1:25
  • \$\begingroup\$ @user3653788 good point.... \$\endgroup\$ – Trevor_G Dec 5 '17 at 2:02

Looking at your circuit, remove the zener for now and consider the base emitter junction as a simple diode. You end up with this on the base side.


simulate this circuit – Schematic created using CircuitLab

The voltage at the top of the zener then becomes

\$Vz = - (12-1.4) * 15/16 =-9.9375V\$

The voltage across the zener then is only \$12-9.9375 =2.0625V\$ so the zener is off.

The base current is then simply

\$Ib= (2.0625-1.4)/1000 =0.6625mA\$

If the transistor were not saturated that would make

\$Ic = 30 * Ib = 19.875mA\$

But we know the maximum current at the collector is \$11.8/2200 = 5.36mA\$

As such the transistor must be saturated and \$Ic = 5.36mA\$


Since you asked about the general approach to such problems let me explain my thought process on this problem, which is general.

The question asks for Ic. Looking at the transistor I can tell that the collector current can not be more than when it is saturated. With Vcesat at 0.2V that means 11.8V would be across the 2.2K resistor, so the max current must be 5.36mA.

So now I need to know if the transistor is saturated.

In order to know that I need to figure out the base current.

Looking at the base there is a diode and a resistor going to that zener with the resistor up to ground. I can't tell yet if the zener is off or not, but if it is acting as a zener the voltage on the left of the diode must be 1.4V. If the zener is off, the voltage left of the diode must still be 1.4V since there is no other current path. As such the diode and the base emitter junctions must be forward biased in either case. That is, they are definitely on.

That leaves me with one variable. Is the zener on or off. I know for the zener to be on, the voltage across where the zener is must be at least the zener voltage when the zener is removed. So that is fairly simple math.

I do that, and now I know the zener is off, and can calculate the base current from the voltage I just calculated. (Note: If the voltage I just calculated had put Vz higher than -7V, I would know that the zener would hold Vz at -7V.)

The final conclusion is just logic.

In other words, when solving a problem like this start with what you know and work back till you remove all the variables. It's not so much trial and error, as deductive reasoning.

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    \$\begingroup\$ this is the correct . trick question. never assume it is designed correctly. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 5 '17 at 2:10
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    \$\begingroup\$ Also remember that in reality \$h_{FE}\$ in saturation is about 10% of max hFE in linear region, except specs will usually be overdriven to 10% base current or Ic/Ib=10. Then when >10% max , you can estimate Vce rise using \$\Delta V / \Delta I= r_{CE} \$ slope during saturation, or ESR as I call it bulk resistance which is [Ω] ~ 1/Pd [W] power rating of device. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 5 '17 at 2:13
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    \$\begingroup\$ I forgot to check if the Zener was on, this is correct. \$\endgroup\$ – S.s. Dec 5 '17 at 2:21
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    \$\begingroup\$ I also hate it when they use negatives.... \$\endgroup\$ – Trevor_G Dec 5 '17 at 2:24
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    \$\begingroup\$ @Trevor Thanks for clarifying in such detail. This will help me alot. \$\endgroup\$ – user170930 Dec 5 '17 at 13:52