Let's Make Robots!

Transistors –[page 2] Common Base or Collector uses, & Beginning Switching Theory



This is a continuation from http://letsmakerobots.com/node/28841


       Continuing into transistor design, in the previous page, we saw that there is not just one type of circuit (common emitter) but also two others as well (common base and common collector). These are not so commonly used, so many people may not have even known they existed. Regardless of that, how would we use these circuits anyway?


       The other two types are not useless trivia, but can be very important when you have a particular need. Let us take an example for each.  [There are other examples, but these examples are what came to my mind.]


       Robot builders often want to use radio links either to control the robot or to send data back from the robot to a central computer.  What if we want to hook up a small antenna to a high impedance input receiver or amplifier? (In the example, I used a ground plane antenna, but it could be any type. The point is in matching the impedance.)


       Looking at the three types of transistor circuits mentioned, we quickly see that a common base configuration is perfect for the job. (Input impedance around 50 ohms and a high impedance output).


       In the next circuit diagram below, you see two resistors and a capacitor hooked to the base and you may wonder why. In order to do away with the separate battery biasing the base in the generic sketch in the previous lesson, we have to clamp the base to a certain voltage within its range, so the transistor will conduct properly.

       [Note that "common base" does not have to mean "grounded base"]


ADDENDUM: I guess I did not explain common base properly, so allow me to expand on that.

The formula for finding reactance (Xc) is:


Xc = ----------    where F is the frequency in Hertz, C is capacitance in farads and π is the

          2πFC        constant, Pi (3.14159265358979…etc.)

1)      The antenna coax connection is 50 ohms

2)      The signal passes through the 100 pF capacitor, which has a reactance (like resistance but at a certain frequency) of 5.05 ohms at 315 MHz. and 3.7 ohms at the upper radio frequency. --Low enough that it does not noticeably degrade the signal.)                            [and one error - It says 100 pF or less, -it should say 100 pF or MORE.]

3)      Likewise the 1 KΩ resistor is enough higher than the 50 Ω that not much signal is lost to ground, but the emitter is anchored to ground. (Since the collector-emitter current must also pass through a 100 KΩ resistor, the emitter is still essentially at ground level.

4)      Now the signal has arrived at the transistor via the emitter.

5)      To pass through from emitter to collector, the base must be in the conducting range. This is done by the R2-R3 voltage divider, with the .001 μFd capacitor helping to regulate the base. (The size is sufficient at these frequencies.)

6)      R4 is high enough to avoid drawing down the output impedance too much and will drain very little from the signal strength.

7)      Finally the signal leaves the stage through the (DC-blocking) 5 pF capacitor which shows only 101 Ω at 315 MHz and less than that at 433.92 MHz. Compared to the high output impedance the loss through the output cap is negligible.

8)      Also, when I mentioned the "gimmick" capacitor, it is just "extra information". It shows you a way to make such a small capacitor as 5 pF without having to locate one. In other words, the gimmick cap. would be in place of the 5 pF cap. [I would use about one inch (2.5 cm) of 22 to 24 guage insulated wires wrapped snugly around each other but without damaging the insulation. DC will not pass through, but the RF signal output will jump the gap and continue on to the next device or amp stage.]. 



       All right, common base may have some uses, but what about common collector?  Let us take another problem.


       Let us say we are using a crystal microphone that has a high impedance and very low current output (down in the microamps). We will send this to an amplifier via a coaxial cable, but we are worried that with such a low signal current, it will be susceptible to picking up noise and hum.  What can we do? We will need to amplify the signal current and at the same time match it to a 50-ohm coax.  This can be done with a single common collector transistor stage.










       You decide to power it with a 4-pack of rechargeable batteries that read about 5 volts when fully charged, so you set your Vcc equal to 5 volts. Since the signal is down in the microamps, you pick your collector-emitter current at 10 milliamps. (An arbitrary number, but you see that the highest current gain will be available around that value.)


Design solution:

       In designing a common collector amp, you typically set Ve (emitter voltage) at half the supply voltage. (Vcc=5 volts, so Ve will be 2.5 volts.) Knowing both the voltage across R3 and the current through it, you decide R3 = 250 ohms.


       With a current gain expected at ~100 times, the base current needs to be 10 milliamps divided by 100, or 100 μAmps. We know further that with ~0.6 volts base-emitter differential, Vb should be about 3.1 volts.  We must pick R1 and R2 to give us that voltage at the base. You should also know that the total current through R1 & R2 ought to be about ten times the needed base current, so R1 & R2 will pass about one milliamp. With the supply voltage at 5 volts and passing one mA, then R1 plus R2 will be 5000 ohms. We want to pick R1 and R2 so they add up to about 5000, but remember we also need the base to be 3.1 volts, so this gives us the ratio for the resistors. 3.1 v./ 5.0 v. => R1 / 5000 ohms. Solving this we see R1 should be 3100 ohms and R2 will then be 1900 ohms. Since these are not commonly available values, we pick nearby values that preserve the same ratio (and thereby preserve the 3.1 volts at the base).


       We pick R1=2 KΩ and R2=3.3 KΩ, also written 3K3.


       For now I will skip the calculations on the capacitors other than to say they need to be values that are only one tenth the impedance of the stage, -either input or output, -depending on whether we are looking at C1 or C2. This reactance (essentially the same as resistance at a certain frequency) is picked relative to the lowest frequency your microphone will be sending –perhaps 20 Hz. (C1 works out to be 63.77 μF. You could get away with 50 μF or higher.  For C2, I would recommend 500 μF or larger.)


        All right then, it looks like there will indeed be times you will want to use either common base or common collector transistor circuits. Therefore, you should keep them in the back of your head for those odd occasions when you need to match high to low or low to high impedances. For everything else, the common emitter circuit will likely suffice.



Let's take a look now at



       While roboteers at LMR may need audio circuits occasionally, the more common use of transistors is going to be switching circuits, ­–turning things on or off.  Let us look at a switching application and try to design a circuit we can use.


       Quiz question: Let us say we located a tiny motor for our little robot and want to drive it. We found a geared pager motor that draws 24.6 milliamps normally or as much as 184 milliamps (mA) when it stalls. We check our transistors on hand, and find a 2N3904. We know it is a common transistor, but do not know anything else about it, so we google it and find a spec sheet at: http://www.fairchildsemi.com/ds/2N/2N3904.pdf  Looking at the specifications, we see the collector current can be 200 mA continuous and 300 mA on peaks. Great. It looks like this transistor will work, but what resistor do we need at the Base to protect our micro-controller while still driving the motor on and off?


       Answer: Somewhere we remember seeing that the gain (hfe) of a transistor is the increase from the base current to the collector current.  Base current times gain equals the collector current that will flow. Thinking about this a moment we realise this would also work in reverse, meaning that since we know the collector current we need (184 mA), if we divide that by the gain we should get a value for the base current the transistor will need from the micro-controller, but which hfe? There are several listed.


       Hfe is the hybrid gain in a common emitter circuit. Okay, our test circuit (see drawing) is drawn as a common emitter circuit. However, none of those hfe figures listed is for 184 mA. The highest we see is for 100 mA. Should we use 30 as our gain? No. Regular hfe figures are used in analogue circuits such as audio or radio frequency amplifiers. For turning things on or off, we need to use the "saturation" figures.


       Look at the VCE(sat) or VBE(sat) lines. These show figures when the transistor is saturated. That is what we need to use for switching circuits. We see IC = 10 mA, IB = 1 mA and IC = 50 mA, IB = 5 mA and realise that in both cases, IB (the base current) is one-tenth the value of IC (the collector current), so that is the answer. Hfe(sat) is 10.

       If we divide the 184 mA stall current by 10, we find that we need to supply 18.4 mA at the base to switch 184 mA. This is within the allowed current range of our micro-controller. (20 mA per lead with a PICAXE, and 40 mA per lead on an ARDUINO.)


       We are using a 5-volt supply but there will be a voltage drop across our transistor base-emitter junction. Look again at the VBE(sat) line in the spec sheet. It says there is a drop of 0.65-volt minimum and 0.85-volt maximum. The minimum value would allow more voltage to drop across the resistor, so let us go with that as a "worst case" in figuring our resistor value. There is also some drop through the micro-controller, but let us presume it is only about 0.2 volts (similar to the VCE voltage drop on our 2N3904). This gives us (5.0) – (.65 + .20) = 4.15 volts. Using that and 18.4 mA for the current, by Ohm's Law we find the resistor value of ~225.5 ohms. This is not a common value, so we pick the next smaller resistor value. It happens that 220 ohms is quite common, so that is our answer.


       Case solved: We need a 220 Ω resistor and our micro-controller will easily switch the little pager motor on and off.

BTW, the specs on the geared pager motor are a real motor I found at:

http://www.hobbyengineering.com/H1206.html It is available other places as well.


Transistor tutorial continues: http://letsmakerobots.com/node/29007


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Lots of great knowlege, Dan. I'm trying to absorb it a bit at a time Let me start with your common base circuit and see if I understand.

Resistors R2 and R3 form a voltage divider, holding the base at 5V / 2 = 2.5V. I'm not clear on the fuction of the 0.001uF capacitor at the base. Am I right in thinking this 2.5V base voltage will turn the transistor on as long as the voltage difference from base to emitter is greater than the Vbe saturation voltage (1.2 - 2.0V)?

At the emitter, a relatively small voltage signal would be expected from the antenna. This will produce a signal referenced to ground, and filtered through the 100pF cap and R1.

At the collector, the output will be pulled high by R4, but when the transistor is on, you'll get current through R4 and R1. The voltage at the output will follow the signal level at the input.

The 5pF cap at the output is for noise filtering? Also, is the gimmick capactor you mentioned to be used in place of this 5pF cap?

I appreciate you posting this stuff. I'll probably have more questions, but I'll post separate comments for each topic to keep the post more readable.

You have it basically correct, but let me expand on it a little. I have added some further description in the body of the text above & labeled it, "ADDENDUM". The 100 pF cap allows passage of the RF signal, being only about 4 to 5 ohms at the radio frequencies. (A larger capacitor could be used to lower that further but should not be needed.) With the emitter base junction in an on-state, signal can pass through the transistor to the collector. I would not say the 5 pF capacitor was for noise filtering, but rather is there to pass the signal through to the next stage, while blocking any DC flow between this stage and the next one.

Also the mentioned "gimmick" capacitor was just "extra information". It was only mentioned as an easy way to create such a small capacitor as 5 pF without having to locate one. It would be used where the 5 pF cap is now. At that frequency even such a small capacitor should pass the signal through just fine.