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I was playing with a BJT based circuit and I learned about the breakdown voltage of BE junction.

I came out with a solution of connecting a diode in series, with higher breakdown voltage. But I am not sure if connecting in the bottom and top are both same

enter image description here

where the zener BE is the model of the base-emitter junction. If I connect the protection diode in the emitter (D2) I will certainly avoid the stress in BE junction. Does it act in same way if I connect the dioe to the base? (D3)?

The bjt has a breakdown of 6V. so if I apply like 12V (limited with a high resistor) I see a voltage of around 7V, which makes sense since 6V is just the minimum specified. The current through the gate is very low, since I work with 12V and limiting resistors of 270k. when I breakdown the junction and 47k as pullup. The BJT in the bench is a BC846ALT1G.

EDIT: since the complete circuit could depends on the collector state, here is a sketch of the circuit:

enter image description here

The emitter of Q2 is the one which I try to connect to 12V (VCC) with 270k, while the collector of Q2 is always connected to 12V. Basically I want to bypass a resistor with a BJT where the OUTPUT goes to another circuitry which uses a pull down of the same size. The load driven by the Q2 emitter is 270k/2 Ohm, since the other pulldown is 270k as well, there is nothing else.

I was trying to understand the situation more deeply in order to apply less changes as possible.

Q1 is kept conducting or open (applying a current limited voltage on base, like 5V or 0V) and works as it is supposed to work.

The reverse junction which I am referring to is on the Q2 BE junction.

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  • \$\begingroup\$ Start with input specs min max V,I , rise time and look how it’s done before making design guesses A series diode and Pull down maybe \$\endgroup\$
    – Tony Stewart EE75
    Feb 27, 2018 at 15:47
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    \$\begingroup\$ A solution without a problem that becomes an ineffective solution if the added diode leakage current is greater than the EB leakage current. Put a diode across the BE in the opposite direction to protect BE. \$\endgroup\$
    – Andy aka
    Feb 27, 2018 at 16:08
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    \$\begingroup\$ Usually the best method to protect it is not to use it somewhere it will see that effect. "The output of Q2 is the one which I try to connect to 12V with 270k" Huh? Why would you be doing that? You might want to expand that a bit so we can properly understand what your issue is here.... \$\endgroup\$
    – Trevor_G
    Feb 27, 2018 at 16:36
  • \$\begingroup\$ Updated with a complete circuit \$\endgroup\$
    – thexeno
    Feb 28, 2018 at 13:45

2 Answers 2

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As you draw it, there's no difference assuming you've either not connected the BJT's collector or shorted it to the base.

In a real circuit there might be a difference as then the collector might come into play.

So depending on the complete circuit it could make a difference if the extra diode is in series with the base or the emitter.

In most circuits the low base-emitter breakdown voltage is not an issue. In some circuits it can be though.

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  • \$\begingroup\$ I edited the OP \$\endgroup\$
    – thexeno
    Feb 27, 2018 at 15:59
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    \$\begingroup\$ You're only showing part of the circuit. You have to show what makes the BE of Q2 have a reverse voltage. Also connecting the "output" (you should write: emitter) of Q2 to 12 V: like this the emitter of Q2 will never reach 12 V, the best it can do is 12 V - Vbe = 11.3 V. I suspect that your circuit has other issues as well, that makes it extra important to show everything. \$\endgroup\$
    – Bimpelrekkie
    Feb 27, 2018 at 16:39
  • \$\begingroup\$ Updated the schematic in the OP \$\endgroup\$
    – thexeno
    Feb 28, 2018 at 13:46
  • \$\begingroup\$ As long as there is never a battery of power source connected to the ouput, there's no need to "protect" the BE of Q2. When Q1 conducts then indeed Q2 will get about 6 V reverse voltage but that isn't an issue as almost no current can flow, R3 limits that current to less than 40 uA so there can be no damage. Ergo: adding "reverse voltage protection" is pointless. It is not needed. \$\endgroup\$
    – Bimpelrekkie
    Feb 28, 2018 at 14:02
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    \$\begingroup\$ The reason for the datasheet to list that Vbe max reverse voltage is because it is the voltage which the manufacturer guarantees the BE junction can handle without breakdown. Despite the name, breakdown doesn't break anything in the sense that damage will occur. Zener diodes use the same principle, these are actually used in breakdown mode constantly. As long as the current and power does not exceed the ratings, the zener diode will not be damaged. Same for a BE junction. The extra diode is simply not needed as nothing will break without it. \$\endgroup\$
    – Bimpelrekkie
    Mar 1, 2018 at 6:39
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We dont know your load but you can put a reverse diode across Vbe if Collector can pull down load when output emitter follower is off. VI*t may be a factor on turn off.

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