# Does a transistor work by connecting the collector to emitter? If so, why do transistor regulators work?

I am wondering why a transistor, such as one used in this Zener diode + Transistor regulator , has the output controlled by the base voltage rather than the collector. My textbook (which has this image) says that the constant voltage being produced with the Zener diode (that is, its reverse breakdown voltage) is applied to the base of the transistor, and as a result, the Vo (voltage out) is the Zener voltage minus the transistor conducting voltage of 0.7V.

This largely makes sense, but wouldn't the transistor bridge the collector with the emitter once the base-emitter voltage exceeds 0.7V? And assuming the Vin (input voltage) is something like 10V, while the Zener voltage would still be something like 5.6V, why would the voltage at the emitter be (5.6V - 0.7V = 4.9V) and not 10V?

Think about the base and emitter - to get current through the collector to the emitter, the base-emitter region must be forward biased. Given that the base has 5.6 volts on it, the base-emitter becomes naturally forward biased when the emitter is at 4.9 volts. This places the base 0.7 volts higher than the emitter.

If more current tries to pass from the collector to the emitter, the emitter voltage might start to rise and this would shut-off the base-emitter junction and reduces conduction. Hence this cannot happen.

This circuit is called an emitter follower i.e. the emitter has to follow the base voltage minus the 0.7 volts (or so) that is required to forward bias the base emitter region.

And assuming the Vin (input voltage) is something like 10V, while the Zener voltage would still be something like 5.6V, why would the voltage at the emitter be (5.6V - 0.7V = 4.9V) and not 10V?

With 10 volts at the emitter (by some magical means) AND 5.6 volts at the base, the base-emitter region is completely "off" and thus no collector current can flow. In other words you paint an impossible scenario.

• I think i understand it better now - thank you. However, if this means that the collector is not bridged with the emitter (as it might if using the transistor as a switch) then is there a need to have the collector connected to the supply at all? Or does the emitter draw only 4.9V from the collector? Dec 1, 2018 at 17:13
• Actually, if you remove the collector and provide enough current through "R" to the zener (of which some flows through the base-emitter to the load), then it still works but then you are not utlizing the current gain of the BJT thus enabling you to use lower power zeners and a higher value "R". Dec 1, 2018 at 17:17
• So by keeping the collector connected to the supply, the transistor still limits the output voltage (well, zener is limiting it and the BJT emitter follows the base voltage minus 0.7) to 4.9V? Does this possibly indicate a voltage drop (10 - 4.9 = 5.1V) across the CE junction so the emitter doesn't get too high a potential? This would also tie into the consensus that the BJT acts as a current amplifier - would I be right in assuming so? Dec 1, 2018 at 17:22
• Yes, the collector-emitter drops (it has to drop) 5.1 volts across it and, of course, with a high load current this means a lot of heat in the transistor. It is acting as a current amplifier because the voltage gain of an emitter follower is slightly less than unity - really good transistors in this config might have a voltage gain of nearly unity (0.995). Dec 1, 2018 at 17:26

R has to be selected so the transistor operates at the top end of its linear region. That way, the zener diode will "steal" just enough base current to lower the voltage in the output loop to the zener voltage - 0.7V.

Transistor stands for "transfer resistor", and in principle that resistance (collector to emitter) can span a very large range, from microamps to amps. It all depends on the base drive.

If you try to control a transistor by explicitly controlling base voltage, as you have realized the effective resistance will change from very high to very low over a rather small span. How to get around this? Don't try to control the base voltage, but rather the base current. Or, if you do try to control base voltage, use feedback to keep the voltage at exactly the right point. Take your circuit, for example. You've left out a most important component - the load. Redraw the circuit as follows:

simulate this circuit – Schematic created using CircuitLab

The voltage across the load will increase with increasing current. I hope you see that the load voltage cannot get to more than about 4.9 volts. But let's assume that it does, say, 5.9 volts. Then the base-emitter junction will be reverse-biased by about 1.7 volts, the transistor will be turned off, and the load voltage will be zero.

So what actually happens is that, for example, as the zener approaches its operating point, so does the corresponding load voltage, but always about 0.7 volts lower. When the zener voltage equals 5.6, the load voltage equals 4.9, and everybody is happy.

This is all a gross simplification, of course, but at your stage of study take it as true. For instance, the transistor has what is called current gain, the ratio of base current to collector current. So in order for the transistor to provide load current R1 must provide enough current to supply both the zener and the transistor. Not only that, but beyond a certain current level the current gain actually decreases with increasing load current. So for very small load resistances (and "small" depends very much on the transistor) a large R1 simply will not provide enough base current to keep the load voltage at 4.9, and the zener will be starved for current and its voltage will drop until a new equilibrium is established. But that is probably a lesson or two in the future.

If the transistor turned fully on, raising the emitter voltage to the full 10V (as in your faulty scenario) then the base voltage would have to be 10.7V. Clearly, the Zener diode wouldn't allow such a high voltage at the base.

The Zener ensures that base voltage rises no further than +5.6V. If emitter tries to go higher than 4.9V, base current falls toward zero, and the transistor turns off by starving current flow between collector and emitter.
If the load at the output tries to pull more current, you might think emitter voltage would fall...but that simply increases base current, turning the transistor on harder.

Load regulation isn't wonderful with this simple regulator circuit. Some variation in the transistor's base-to-emitter voltage does exist - the range is roughly 0.6V -to- 1V, depending on how much current passes through from collector to emitter. Note that the base-emitter voltage for an example 2N3055 is also temperature-dependent (below) - all bipolar transistors act this way:
from Siemens 2N3055 data sheet

Transistors have this input-output plot

With 12 volts on the collector and 5.6 volts on the base (thus about 5.0 volts on the emitter), the bipolar has 7 volts across the Vce (the entire transistor).