How to correctly bias an NPN transistor without allowing base voltage to be too high

Question

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I am trying to correctly bias a BC337 NPN whose collector is connected to a 9V source. I wish to use it as a switch in response to a separate signal (which is either 0V or 9V) applied to its base.

I think I'm right in understanding that the current Ib applied to the base must be limited to that which does not allow excessive current Ice. I I believe I can calculate that.

But on the one hand I keep reading that Vbe should not exceed 5V, on the other I keep reading that Vbe is simply a diode drop ~0.7V. I cannot see how both can be true.

What is the correct way to understand how Vbe behaves?

EDIT: In response to @unimportant I have included a schematic. SW1 is there to simulate incoming signal (0/9V). R1 and R2 I put in case voltage divider is required in order to control VB. My concern is the conflict in what I have read about the behaviour of VBE. It is not clear to me how there is risk of VBE rising above 5V (which apparently destroys transistor) and if there is no risk then surely voltage divider is unnecessary, only R1 or RB if at all.

EDIT 2: per @greybeard, ground connections added

Answer 1

Vbe, if negative, should not be greater magnitude than -5V for the BC337.

For example if you have your collector connected to +9V, base at 9V, and a capacitor parallel with a resistor connected from the emitter to ground (so emitter voltage about 8.3V), when you pull the base to 0V the transistor can be damaged as the load capacitance discharges through the reverse-biased base-emitter junction.

You may be confused about the sign since datasheets will sometimes state the parameter as Vebo (emitter with respect to base) and 5V rather than Vbe or Vbeo and -5V.

It sounds like you want to make a “high side switch” which is better made with a PNP transistor or a P-channel MOSFET.

n.b. Your added schematic does not match the description (collector is not connected to the supply) so the example may not be exactly applicable.

Answer 2

How the transistor output behaves when saturated

I think I'm right in understanding that the current Ib applied to the base must be limited to that which does not allow excessive current Ice.

This statement would be true for a transistor amplifier stage where the base current controls the collector current. In your case, however, the transistor works as a switch with two states - "on" (closed) and "off" (open).

For the transistor to be in the "on" state ("saturated"), a sufficiently high current must be applied to the base. From this moment, the collector current does not depend on the base current, but only on the supply voltage and the collector resistor (load), according to Ohm's law - Isat = Vcc/Rc.

So even if you keep increasing the base current, the collector current will not change because the transistor has "saturated". If you think of it as a "variable resistor" (very useful viewpoint), its "resistance" has gone to zero and cannot decrease any further.

How the transistor input behaves

What is the correct way to understand how Vbe behaves?

Think of the transistor base-emitter junction as a diode that can be forward biased (then the transistor is "on"), "zero biased" or backward biased (then the transistor is "off"). In your case, only the first two can be observed; so there is no point in bothering about the third one and trying to understand it by reading hard-to-understand data sheets (the purpose of manufacturers is to sell you devices, not to explain how they work).

How to control the transistor switch

Turning it on

So, to turn on ("bias", as you said) the transistor, you have to pass sufficient base current Ib > Isat/beta through the base-emitter junction. For this purpose, as you may have guessed, you need only a humble resistor.

Turning it off

Conversely, to turn off the transistor, you must interrupt the base current, or even more safely, make Vbe equal to zero. The simplest way to do it is to connect a resistor in parallel to the base-emitter junction.

Another (preferable) way is if your switch not only breaks but also grounds the circuit input (i.e., it is an SPDT switch). This is how electronic circuits are made by means of the complementary output stages.

Building the circuit

To realize the philosophy behind the transistor switch, let's follow its evolution by building it step by step. I have shown below in six steps how a mechanical switch (button) can control the load through a transistor. The schematics are conceptual; try each of them in both states (closed and open) of the button. You can set them from the SW parameters window of CircuitLab; then hover the mouse over the interesting nodes and elements to see the voltages and currents.

Of course, the most interesting is the voltage across the load. I have used a little trick to simplify the circuit - combined the load and the voltmeter into only one "bad" voltmeter with low resistance (the suitable 1 kohm value); thus it can serve as a "voltage-visualized 1 k load". If you want to monitor the current in the same way, you can set 1 k internal resistance of the ammeter to "worsen" it.

STEP 1: Ideal switch

To control the voltage across and the current through a load, first we can connect an "ideal" switch in series. It has zero resistance when closed and infinite resistance when open.

As we expected, the output voltage across the load is exactly equal to Vcc when SW is closed...

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... and exactly 0 V when open.

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STEP 2: Real switch

But real switches have some small resistance Ron when closed and some high but finite resistance Roff when open. From this viewpoint, they can be considered as variable resistors with two extreme values of that resistance. In the schematics below, I have set in the R parameters window suitable values (Ron = 10 ohm and Roff = 1 Mohm).

As you can see, the effect of these resistances on the output voltage is small. It is close to Vcc (8.9 V) when SW is closed and reached the lowest resistance. Figuratively speaking, it is "saturated" because no matter how hard we press the button, SW cannot lower its resistance more than 10 ohm...

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... and close to 0 V (8.9 mV) when open.

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STEP 3: Transistor switch...

Now we have only to replace the mechanical switch by a transistor that acts as the variable resistor above. But it is a much more versatile "variable resistor" because it can smoothly change its "resistance". As they say, it is an analog device, but here we make it work as a digital one. It can even change its current "resistance" when the load changes its resistance... but that is another story.

In order to ensure the maximum collector current through the load when the transistor is "on", we must pass the corresponding base current through its base-emitter junction. We can do it with just one resistor R1 acting as a voltage-to-current converter. So there is current flowing through and voltage across the base-emitter junction; so the transistor is "on". It acts as a closed real switch with low resistance.

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When SW is open, there is no current flowing through and no voltage across the base-emitter junction; so the transistor is "off". It acts as an open real switch with relatively high resistance.

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STEP 4: ... with leakage...

Unfortunately, in addition to the useful ones, undesired voltages and currents try to affect the transistor base. For example, it could be a "leakage resistance" Rleak between Vcc and the base.

When SW is closed, it does not really matter...

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... but when SW is open, the effect can be significant (look at the voltmeter to see).

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STEP 5: ... neutralized

That is why we include another resistor R2 to divert the leakage current from the base. It is not useful (even harmful) when SW is closed...

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... but will work when SW is open.

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STEP 6: Final circuit

After all that, here is what your schematic would look like.

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