# Silly BJT question

Discussion in 'Homework Help' started by blah2222, Aug 24, 2012.

1. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
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Say you have a simple NPN BJT switching circuit to control a light bulb (attached image) and say you rotate the potentiometer so that the BASE of the transistor moves towards +Vcc.

Now this may be a silly question, but why does the voltage between the BASE and ground not increase past ~+0.7 V even though it is physically connected to a voltage close to +Vcc?

I realize that the voltage between the BASE and EMITTER is ~0.7V due to the junction bias and the EMITTER is grounded but it just seems to me that it is magic that when you apply any voltage greater than 0.7V on the BASE of the grounded common-emitter NPN that the BASE voltage magically moves from whatever large voltage down to ~0.7V.

Maybe I am just not in the right mindset but I can't picture this one.

Thanks,
JP

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2. ### #12 Expert

Nov 30, 2010
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The base-emitter junction is a forward biased diode. The voltage actually increases a little as the current goes up but it's like pouring current through an 1N4001 diode. Each tenth of a volt increase requires about 8 to 10 times the current of the previous tenth of a volt increase in voltage.

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You could connect a 'stiff' DC supply directly to a base emitter junction and the current would increase such that the emitter-base forward voltage equaled the DC supply voltage. Depending on the source DC value the current could be very large and would probably destroy the transistor.

Think of this as effectively the same situation with a DC supply connected directly across a diode in forward bias mode. Lots of current with potential damage. That's why one usually places a series limiting resistance in the circuit to keep the diode forward current to a safe value.

4. ### wmodavis Well-Known Member

Oct 23, 2010
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And BJTs are current driven devices rather than voltage - not that you can totally separate the two.

5. ### ramancini8 Member

Jul 18, 2012
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This circuit is commonly used by newbies until they get tired of blowing transistors. Voltages always clamp to the lowest level.

6. ### wmodavis Well-Known Member

Oct 23, 2010
737
150

YUP! (plus a few spaces)

7. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
565
33
So if I am understanding this correctly, due to the exponential relationship of currrent and Vbe, if Vb is tied to +Vcc and Ve is tied to GND, it would force the supply to output a serious amount of current and if the supply can output this, the transistor would probably crap out as it is a heck of a lot of current.

And in the case of a regular supply that has a current limit, the supply (+Vcc) will actually be pulled down to ~0.7 V due to the current consumption of the transistor?

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Suppose the emitter base diode is modeled by its Shockley equivalent as

$I_B=I_se^{\frac{V_{BE}}{0.026}}$

If the reverse saturation current is Is=1E-16 Amp

and a stiff voltage of 0.65V is applied as VBE then the base current required to match the diode with the applied voltage would be 7.2uA. Change the voltage by a modest amount to 1.0V then the current requirement is the order of 5A!

Inserting a series resistance of 10kΩ would give a current of 2.6uA with a stiff 0.65V supply and 31.2uA with a stiff 1V supply. It's clear why a series resistance is required.

Last edited: Aug 25, 2012
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9. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
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For the very first circuit that I posted. If the wiper (connected to Vb) is turned the whole way from GND to +Vcc. Will it ever go above ~0.7 V assuming the source isn't stiff?

Like will it ramp up from zero volts to 0.1, 0.2, ..., 0.65, 0.66, 0.68, until it reaches a peak value close to 0.7V? I just want to know what happens when that pot is turned so that the wiper is directly connected to the source. Does the source's voltage change, does the source blow, does the transistor blow?

I feel like I'm making this complicated...

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Depends on several factors.

Unless the source is current limited then most likely the transistor will die.

As the wiper is moved up, the base current will probably reach the rated value and death will occur before the wiper makes it to the full supply voltage.

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11. ### WBahn Moderator

Mar 31, 2012
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On paper, tying the base of an NPN transistor to a voltage source creates a dilemma -- two contradictory expectations. In the case of an ideal voltage supply connected to an ideal diode (including a diode that is modeled as having a fixed voltage drop when forward biased) creates a mathematical dilemma that has no solution -- is the voltage 0.7V because of the diode or is it 12V because of the supply? (I'm using Vcc=12V just to have a firm number to bandy about).

But this is on paper. In the real world a real diode (or transistor base-emitter junction) and a real supply come to a solution.

A real diode (or transistor base-emitter junction) has a relationship between the voltage across it and the current through it that is monotonic and strictly increasing, which is just fancy-speak meaning that as the current in the diode increases so does the voltage across it, and vice-versa. In this respect, it is the same as a resistor. The difference (aside from the unidirectional nature of a diode) is the a resistor's relationship is usually pretty linear (multiply the voltage by x and the current gets multiplied by x -- do double the voltage and you double the current) while a diode's relationship is usually pretty exponential (increase the voltage by a x results in the current by multiplied by y). For silicon diodes, particularly BJT base-emitter junctions near room temperature, the relationship is approximinaly as follows: increase the junction voltage drop by 60mV and you get 10x the current.

A real supply generally can deliver a certain amount of current while maintaining the desired output voltage; beyond that, as the supply is asked to deliver more current the voltage output drops. The exact nature of how it drops depends on the supply, since a battery and a bench supply with current-limiting are going to behave quite differently, but the basic relationship will usually still hold: increased current draw results in decreased voltage, at least beyond some reasonable point.

So as you hook a real 12V voltage supply across a real silicon diode, the following happens (usually in about the time it takes me to type a single character or two at the keyboard): Initially, the voltage supply is putting out zero current and has 12V across it while the diode has zero current though it and therefore has 0V across it. The wire connecting the two, which has some small amount of resistance, let's call it 1mΩ for our example, therefore has 12V across it and would like to have 12,000A flow and so things start out on this path and the current drawn from the supply starts increasing and, as a consequence, so does the current in the diode and, therefore, so does the voltage across it. Since the supply is probably not going to be able to maintain 12V at anything approaching 12kA, the output voltage is going to start dropping. Hence, the voltage drop across the wire resistance is going to drop and the current level that the system is heading for drops. Assuming, for the moment, that all of the components involved (including the wire) can tolerate the final currents we reach -- and this is a really big assumption that we'll deal with shortly -- the system will eventually reach a current level that is consistent with the characteristics of all three components.

So what will that current be?

We've already put numbers to the wire (namely 1mΩ), let's put some numbers to the other components and assume that the diode has 0.7V across it when a current of 10mA is flowing and that the supply is a simple high-output battery that is well-modeled by a 10mΩ internal resistance. We don't need a simulator to get a very good estimate of what the final current level will be, we can do it iteratively very nicely, thank you.

 I Vdiode Vwire Vcc-12V 1mA 0.7V 1uV 10uV 10mA 0.76V 10uV 100uV 100mA 0.82V 100uV 1mV 1A 0.88V 1mV 10mV 10A 0.96V 10mV 100mV 100A 1.02V 100mV 1V 1000A 1.08V 1V 10V

The Vcc column has the nomonal 12V subtracted off, so the numbers in that column represent the drop in voltage at the supply output. Notice that, at 1000A, the total drops are 12.08V, therefore the final current would be just barely less than 1000A (as in, between 990A and 1000A).

How realistic are these results? Well, there are diodes (and transistors) that can handle 1000A. There are batteries that can deliver 1000A, there are wires that can handle 1000A. So it is certainly possible to build a real circuit that reaches these levels without causing damage to components.

But, in more practical situations, let's consider the power levels we are talking about. The supply basically has 2V across it at 1000A, so it is delivering 2kW of power (plus whatever internal heat generation is occuring). The wire is dissipating 1kW of power and so is the diode (just a tad more). The most common event is either for the supply to blow a fuse or the diode to release it's magic smoke and start looking suspiciously like an open circuit. It is also possible for the wiring to start glowing and letting the magic smoke out of it's insulation (if there is any) or the joints (be they solder or the metal-in-plastic of a breadboard or whatever) to overhead and fail, but usually the supply or the diode gives out before then.

But regardless of what actually happens, note that the paper dilemma is resolved and that the contradictory expectations are never realized -- nature will not allow it, even if it means resorting to physical destruction to prevent it.

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12. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I've done something similar to WBahn but with a BD139 transistor. I used a simulated non-linear load resistor to represent the lamp.

Attachment shows that the rated base current is exceeded before the pot wiper is fully rotated.

• ###### Transistor Killer.jpg
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13. ### blah2222 Thread Starter Well-Known Member

May 3, 2010
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Thanks a lot for the added detail. Very clear now.

JP