When in university I used to get confused how a transistor (NPN) can draw current from ground. I'm now studying power amplifiers and again, this same configuration appears. How does Ir draw current from seemingly nothing/ground? enter image description here

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    \$\begingroup\$ Ground isn't nothing; it's just the voltage that you've arbitrarily chosen to call zero. Notice that the base-emitter junction is connected on the other side to a negative voltage. What matters is the voltage at the base minus the voltage at the emitter; voltage is always relative. \$\endgroup\$
    – Hearth
    Apr 24, 2018 at 15:29
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    \$\begingroup\$ Assume that Vcc > Gnd > -Vcc \$\endgroup\$
    – Colin
    Apr 24, 2018 at 15:33
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    \$\begingroup\$ Put the minus probe of a voltmeter to -Vcc and the plus probe of the voltmeter to ground. What voltage do you measure? \$\endgroup\$
    – JRE
    Apr 24, 2018 at 15:34

4 Answers 4


Simply redraw the circuit and you will see from where this base current is coming from.


simulate this circuit – Schematic created using CircuitLab

As you can see the \$Q_2\$ base current is coming from \$\textrm{Vee}\$ voltage source.

And flows this path:

+Vee--->R1---> Q2 base-emitter junction--->-Vee

And remember that in the electronic ground is just a reference point, not a real Earth ground. And we are measure all our voltage with respect to this reference point called ground.

enter image description here

To measure the voltage we need two points in the space. One of this point is treated as a reference point.

We have a very similar situation when we try to measure a height of an object. We need a reference point. The most common reference point is "above mean sea level".

But when you measure the height of the table in your house the floor now becomes your reference point.


Current flows from the higher potential to the lower potential and since the base is connected to ground '0V' and the emitter is connected to a negative voltage which is less than 0V(because its negative!!).

The emitter base junction will be ON if the emitter voltage is sufficiently below ground.About 0.6V below ground for silicon transistors.

Since the base-emitter junctions is ON.Current will flow from ground through the base resistor to the base-emitter junction.This current constitutes the base current and hence a current from the collector of the lower transistor will flow to its emitter.


There are two things to consider when thinking about "how could it work, having the base connected to ground?" - one is the DC question - how could the transistor ever be biased "on" to work, which others have tackled (basically: "ground" here is part way between the two power supply rails so the emitter is negative with respect to the base, so Q2, an NPN transistor, will be happily acting as a constant current sink - albeit with the slight complication that the current will be hard to control unless the diode is at the same temperature as the transistor!)

But the second part of the question is to look at the AC (signal) viewpoint - the emitter is connected to negative, the base also is getting no signal, so (you may ask) why should there be any signal on the output? The answer to that is the top transistor, fed with the input signal in an emitter-follower configuration, will determine the output voltage... if Q2 draws a constant (say) 100mA then if Q1's iE varies from 0 to 200mA the output current will vary between -100mA and +100mA.


If you do your calculation with -Vcc = -15V and for example R=10k and diode voltage is 0.7V then you get this result;

Vcc=Ir*10k+0,7V then Ir = 1.43 ma

so connecting to ground doesn't mean it will not draw any current.

(KVL from left ground to -Vcc)


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