What is the maximum voltage applied to NPN transistor base? (like for MMBT3904)

I cannot really find out what the maximum voltage one can apply to the base of the NPN transistor in a circuit like this one is. I highlighted the base in question.

Can I apply more than the typical 6V in the datasheets that refer to emitter-base breakdown voltage, as that one refers to reverse voltage in this case, am I right?

Link to an example datasheet for convenience.

• Absolutely not. Commented Feb 25, 2022 at 21:59
• With the emitter grounded, the base will be almost impossible to raise. Commented Feb 25, 2022 at 22:16
• Normally R5 can be is 10X bigger than R19 if R5 comes from a voltage similar to Vdd not like that which is shown and assume Vbe will be around 0.6V here depending on base current Commented Feb 25, 2022 at 23:35
• This was a rookie question but great answers, thanks guys! Commented Feb 26, 2022 at 9:53

The base-emitter junction behaves like a diode, so the voltage at the base can only be a diode drop above the emitter voltage (i.e. ~0.6 V). Attempting to apply a higher voltage like 6 V would cause the BJT to (try to) conduct far too much current.

This is evident from the chart in the given datasheet which shows the collector current vs. the base-emitter voltage:

6 V from the base to the emitter is off the chart, and the collector current would be massive (in reality, the BJT would be destroyed).

The 6 V maximum emitter-base voltage refers to when the base-emitter junction is reverse biased -- just like a diode, the junction does conduct significantly with a reverse voltage until that reverse voltage exceeds the breakdown voltage. The polarity matters.

In the circuit you show (where the emitter is at ground), the voltage at the base can vary from -6 V to ~0.8 V. The -6 V minimum comes from the maximum emitter-base voltage, and 0.8 V comes from the fact that the maximum collector current is 200 mA and the above chart shows that the collector current is ~100 mA at a base-emitter voltage of ~0.8 V.

The 6V is the reverse voltage, Veb.

The forward voltage, Vbe, is roughly maximum of 0.8V.

The base voltage is not so important in this configuration, the base current is.

Applying 6V as Vbe voltage will result into damaged transistor.

In forward voltage it’s not really a useful question. If you want a numerical answer, over a wide temperature range about +300mV is safe, however not a very useful number in most situations.

In reverse, the 6V figure from the datasheet is often useful. The actual reverse breakdown voltage will be somewhat higher, perhaps 9V-ish, but 6V is safe.

The B-E junction behaves like a diode, so applying a relatively high forward voltage such a 1V will cause a lot of current to flow which will destroy the transistor. The junction also has a low reverse breakdown on transistors optimized for normal use.

First and foremost a BJT transistor is a current amplifier. Yes, you could utilize a BJT as a voltage amplifier because it's possible to put a resistor at the output (collector) and the amplified current will manifest as a voltage by the simple calculation V=IxR. A BJT will absorb a minute current and manifest a current "leak" almost linear to the current input by a factor of hFE value, which is, in essence, the current amplification factor, usually in the range of 50 to 300. the BJT as this particularity that its input act like a diode with a maximum forward voltage and maximum reverse voltage. For an NPN BJT the typical forward voltage V(BE) ( notice the order in which the BE is presented) is in the range of 0.65 volt (germanium transistors V(BE) is lower).

For the same NPN BJT the typical reverse voltage V(EB) ( meaning now you are feeding a positive voltage from the Emitter to negative Base ie: reverse) is in the range of about 5 or 6 volts, though this value may vary a lot, some NPN V(EB) may go up some 50 volts in special transistors ( or even more).

A diode will not accept ( or a transistor Base-Emitter junction ) will not accept more voltage than specified without actually destroying the internal PN junction. This is why transistors are in most circuits polarized with resistor arrangement to regulate the adequate input current necessary to achieve the circuit task. So, it is not typical to worry about the Base-Emitter voltage, rather Base-Emitter current is consider. Some circuit arrangement will also take advantage of an Emitter resistor to also achieve adequate input current control.

As for the reverse NPN V(EB) voltage is usually taken into consideration when the transistor is likely to receive a reverse depolarization voltage due to circuit intricate arrangements. In such case it is important to prevent that voltage not to exceed the maximum ( typically 6 V ) otherwise the transistor will be irreversibly destroyed.

Summary:

• NPN V(EB) is the maximum reverse polarization voltage before destruction (range 5 or 6 v)
• NPN V(BE) is the typical voltage that will manifest onto the Base-Emitter when proper current is applied to the base ( range 0.6v to 0.8v )