# How does a 'voltage drop' work?

It is my understanding that, in an electric circuit, voltage is divided amongst components in a ratio determined by their various resistances. That is, for example, if a 5Ω‎ and a10Ω‎ resistor were placed in series across a 12V power supply (assuming negligible resistance in wires), then the voltages across the 5Ω‎ and 10Ω resistors‎ would be 4V and 8V respectively - that being the EMF of 12V being divided in the ratio 2:1 as defined by the resistance values, correct?

However, some components aren't like this, and I'm interested to know why. Take for example the base of a transistor. The base is said to 'drop' around 0.7V between itself and the emitter, rather than being described as having a resistance value. It's easy enough to work with just by using a voltage divider, but it still seems a weird concept and I would very much like to know how this works.

• it's a quantum-mechnics / chemistry thing. kind of like how batteries make voltage by pushing electrons around semiconductors can cosume it by a subatomic process too. – Jasen Nov 22 '15 at 10:53
• This non-linear "resistance" explains (to me) some of the potential names for the device they named a transistor, itself a play on "transresistance" (vs. transconductance in valves.) – user65586 Nov 22 '15 at 14:49
• It might help to think of the transistor as a "negative voltage source" instead - for your example, assuming ideal power supplies (no internal resistance), what would happen if you connected another power supply (in series) to your current power supply, and then went on to add your other two resistors – user2813274 Nov 22 '15 at 19:49

This is because of what you might call 'differential impedance': the change in current per unit change in voltage. A resistor's differential impedance is linear: for every extra volt across it, a fixed number of additional amps flow through it, determined by that resistor's resistance value.

Semiconductors don't exhibit linear differential impedance; a unit change in voltage won't always create the same change in current. If you plot voltage vs current for a diode, you'll see something like this: This is what creates the diode's characteristic voltage drop. The curve is sharp enough that it can mostly be approximated by saying it has a constant voltage drop, but you can see that that's not true - the voltage drop will gradually increase with current. This is easier to see on a log scale graph: The upshot of all of this is that in a circuit with, say, a resistor and diode in series, they will find an equilibrium, where the current and voltage through each part forms a point on their respective IV curves. Since a diode's IV curve is very steep, the voltage across it shifts very little with current, and most of the voltage will be dropped across the resistor instead.

• you have not explained how it works, only how it behaves. – Jasen Nov 22 '15 at 10:57
• @Jasen I read the OP's question as asking about the impact of it in a circuit, rather than the physical principles. Perhaps he'll clarify? – Nick Johnson Nov 22 '15 at 11:21
• @Jasen sometimes, how it behaves is all you have. If you got a quantum mechanical explanation, would you then want to know how quantum mechanics worked? Or just accept that QM behaves like that? – Neil_UK Nov 22 '15 at 15:22

In the base of a transistor there is obviously more then a simple resistance. The so called voltage drop of 0.7V is due to the diode like action within the Base to Emitter connection.

In a partial approximation the Base to Emitter connection can be viewed as a small 0.7V battery and a resistor in series. This small battery opposes the voltage that is trying to turn on the transistor. When drawn as an approximation the direction of the battery (+ and - orientation) depends on the transistor type (NPN or PNP).

The 0.7V value is due to the band gap potential for the silicone materials inside the transistor, (old germanium transistors have a band gap of about 0.3V). As in physically traversing a "gap" there is an certain amount of effort (or voltage) needed to "jump" over the band gap. When the applied voltage on the transistor's Base reaches or exceeds this band gap potential current will begin to flow.

So with the approximation of a series connected battery and resistor you can start to calculated some of the other characteristics of a transistor circuit.

Other components can be approximated in this same fashion, for example a Zener diode can be thought of as containing a reverse voltage source equal to Vz. The gate of a Triac component (simulated with two or more transistors) may have a gate turn voltage of 1.5V. The forward voltage (Vf) of a simple LED has a similar turn on voltage requirement (band gap) that can be different depending on the color of the LED.

I think, to answer your question it is - as a first step - important to know the difference between linear and non-linear resistances.

For example, the base-emitter resistance of a BJT is strongly non-linear. In such a case, we always must discriminate between (a) the STATIC resistance Rbe (as a ratio of DC values only) and (b) the DYNAMIC (differential) resistance r,be at the corresponding operating point (defined by the DC values).

This dynamic or diff. resistance r,be is defined by the SLOPE of the non-linear V-I characteristic. The slope gives the small-signal conductance dIb/dVbe at this point - and the diff. resistance r,be=dVbe/dIb is 1/conductance.

Coming back to your initial qestion: The voltage divider rule still applies (also if a non-linear part is involved):

• for DC values - as long as you only take the DC resistance R for all parts into account;

• for ac signals - as long as the small-signal diff. resistances r are considered only.

(It is important to know that you must not mix dynamic and static resistances resistances; for a classical ohmic resistor we have R=r).