# Does current drive voltage in a transistor?

I have this circuit, but I want to know how much voltage will be dropped between base and emitter of the transistor, and how much over the 1k resistor on the left.

Since a transistor does not have a fixed resistance, I guess I calculate the current using ohm law on the resistor? So if the source is 10v and the resistor is 1K, then this is 0.01 amps.

Then I calculate the voltage between base and emitter using the graph from the datasheet of the transistor?

So I have the check the Y axis first and cross reference on the x axis?

Then any "left over" voltage is on the resistor?

Yes. That is a very close approximation, because the 10 volts is very large compared to a diode voltage range of 0.5 to 0.7 volts for 1mA (estimated).

Notice you can use 0.060 volts estimated change in Vbe for 10:1 or 1:10 change in base current. Thus you can add 0.06 volts, and estimate the voltage at 1,000uA base current. Or you can subtract 0.06 volts, and estimate the voltage at 10uA or 1uA or 0.1uA base current.

Circuit Simulators use a similar iteration, hidden inside a matrix computation.

• For a lower voltage would it be the same? Say 1v? – user221241 May 20 '19 at 16:08
• For 10 volts (10v - 0.6v approximation) is 9.4volt, and 0.1 volt change in the Vbe is only 1% current error. For 1 volt (1 - 0.6), that 0.1 volt change becomes a big deal (that is up to you, of course), and you evaluate the impact of updating your estimate. – analogsystemsrf May 20 '19 at 16:43

In the case of a Bipolar Junction Transistor (BJT) yes, current is the mechanism for switching. One simplistic way of looking at a BJT is like a current controlled resistor (which also has a voltage drop like a diode).

In the case of a mosfet, it's the gate voltage that controls the transistor.

Since a transistor does not have a fixed resistance, I guess I calculate the current using ohm law on the resistor? So if the source is 10v and the resistor is 1K, then this is 0.01 amps.

The first thing to do is calculate the base current, then look up beta in the datasheet to find the collector current. There is a good tutorial here.

The base current is usually determined by a resistor like the one shown above that is 1k.

Once I_b is found, find beta, then find I_c or the collector current.

$$\\beta = \frac{I_C}{I_B}\$$

then find I_E: $$\I_E = I_C+I_B\$$

That looks like a BJT operating as a switch, to me. I can tell right away because you have $$\100\:\Omega\$$ in the collector circuit and $$\1\:\text{k}\Omega\$$ in the base circuit (about 10:1.) It's not a perfect way to tell, of course. But just "flying by" it's often good enough.

To design a BJT switch (assuming I already know if it needs to be high-side or low-side referenced), the first thing I need to know is the load current that will be switched. If it is within the capability of a small-signal BJT, that moves me in one direction. (If not, then I have to consider what kind of larger BJT I'll need and if I would prefer to use a MOSFET, instead.) In your case, the $$\100\:\Omega\$$ suggests some tens of milliamps to some low hundreds of milliamps so I can guess that a small signal BJT probably can probably handle it (as a first order guess.)

Let's say the supply voltage is $$\15\:\text{V}\$$ and the load is $$\100\:\Omega\$$. Then this suggests $$\150\:\text{mA}\$$ as a collector current.

If you always keep in mind that the base-emitter voltage is a function of the collector current, such that $$\V_\text{BE}\approx V_T\cdot\operatorname{ln}\left(\frac{I_\text{C}}{I_\text{SAT}}\right)\$$ and that you might experience at most a 10:1 variation in $$\I_\text{SAT}\$$ with small-signal devices, you and easily work out the range you might experience with any particular device you grab from a box. In short, perhaps $$\50-60\:\text{mV}\$$ variation in $$\V_\text{BE}\$$ between devices given the same collector current.

So, a typical small-signal BJT will have a $$\V_\text{BE}\$$ of about $$\700\:\text{mV}\pm 30\:\text{mV}\$$ when the collector current is about $$\4\:\text{mA}\$$. In this example case, the current is $$\\frac{150\:\text{mA}}{4\:\text{mA}}=37.5\times\$$ as much. So this means $$\V_T\cdot\operatorname{ln}\left(37.5\right)\approx 94\:\text{mV}\$$ higher. So I'd guess that the base-emitter voltage ($$\V_\text{BE}\$$) will be, in rough terms, $$\800\:\text{mV}\pm 30\:\text{mV}\$$.

That's the figure I'd use when working out the value of the base resistor or estimating the voltage drop across it, if I already knew the value of it.

Most of the time, these variations don't account for enough variation to be worth the above trouble. When designing for a small-signal BJT as a switch, you usually over-do it anyway by using $$\\beta=10\$$. And for most small-signal BJTs, that's over-kill. So the $$\100\:\text{mV}\$$ error introduced by ignoring all of the above calculations usually isn't worth worrying about. (But if you only have very low voltage overheads to work with, then it may matter more.)