Even in a modern circuit with integrated analog, digital and computer technology, there are times when a simple Transistor Amplifier is all you need to complete your project, boost an output, or invert a level.
The transistor celebrated it’s 70th birthday last year, being definitely now regarded as a mature technology. Yet many hobbyists still regard the transistor as something more mysterious than the Op-Amp, and therefore use an op-amp even where a simple transistor might do a better job with even more output power.
Others use the transistor as nothing more than an electronic switch, or relay, which although useful, is an over-simplification of its utility. Remember - the transistor is an analog current amplifying device.
Transistors are still the basic building block within the ICs we use, still a good choice for power handling, and still do some tricks that are easier with a transistor than an IC, unless you can find an IC specifically designed for that odd application.
Caution: This article requires some mathematics! but before you run off, it’s mostly only simple grade 8 maths.
While there are some complex maths techniques that can be applied, for the simple amplifiers we will be looking at here, you will only need Ohm’s Law; V=IR, and some “Rules of Thumb” used over the last 70 years to simplify the design process. These rules of thumb are simplified calculations based on the Voltage Divider circuit and Capacitive Reactance calculations. Although you might occasionally see the symbol ‘π’; you can simply replace it with 3.14. You should not need a Scientific Calculator, and I can leave my Slide Rule in the drawer!
SELECTING YOUR TRANSISTOR
Just as there are many types of light bulbs there are many types, sizes, and specialties for transistors. Before designing your circuit you must first select a transistor, but before you can do that, you need to define your amplifier application.
As an example project, let us look at designing an amplifier to accept the audio output from an electret microphone element, and amplify the microphone signal over a frequency range of 300Hz to 3000Hz, which covers the human voice as required for speech.
We will need to select a power supply, so for argument sake let us use a simple 9V battery. You might prefer to run the amplifier from a car auxiliary socket (12V) or off a USB port (5V) if you prefer to. Just change the values and do the maths over again. Whereas op-amps tolerate a range of supply voltages, transistor circuits must be designed for a set supply voltage.
Very briefly, the electret mic has a simple mic element connected between the ground pin and the input of a JFET transistor, which is self-biased, so they are the only two elements inside the “Electret Mic” case. Lucky for us they are easily wired to an amplifier by adding a “bias resistor” which is actually a load resistor necessary for the electret device to work against to develop the output voltage. More expensive electret mics may have a built-in op-amp, but we are only using a commonly available type, as found in most electronics catalogues.
The impedance of the electret can be estimated by placing an Ohmmeter across the electret terminals, which actually measures the resistance between the FET source and drain connections. Yes, we can look deeper into both Electrets as well as FETs in another Classroom session.
My Electret/FET measured a little under 2.2kΩ, so to make the output voltage at rest to be half the supply voltage, the resistor should almost match the FET impedance, resulting in a 2.2kΩ being chosen.
The only other component is the capacitor that avoids the DC voltage on the electret output from changing both the electret bias, and the voltage of the amplifier bias. It can also limit the bandwidth of the electret by forming a filter in concert with the 2.2k load resistor, and the 2.2k of the electret.
The capacitor should be equal to the 2 @ 2k2 resistances in parallel, as the power supply is considered a dead short at ac frequencies. Therefore by transposing f = 1/(2πRC) and taking ‘R’ to be Rload//Relectret, the resistance of the load resistor in parallel with the electret, we get C = 1/(2 × 3.14 × 3kHz × 1.1 kΩ) = 48nF, approximately 47nF which is a standard value capacitor.
The output of the Electret is given in datasheets at somewhere between -35dB and -55dB in Volts per milliBar! Good for scientific instruments but a little deep for Makers. Therefore, we chose to simply wire the circuit, add some voltage, and measure it on the oscilloscope with a transistor radio talking away in front of it at around voice loudness. (Severely Technical!)
The oscilloscope gave us a changing value, naturally, but we estimated the envelope to be typically around 20mV peak-to-peak (p-p).
Remember that we might want 1Vp-p, so we need an amplitude gain of 1Vp-p/20mVp-p, = 50. That’s why we need an “Amplifier”, to raise the ‘Amplitude’ by a factor of 50. (Hence the term!)
An amplifier gain of 50 is “doable”, but it may result in more noise than we want. For this exercise, we’ll use a single transistor amp, and see what gain we can make, but should we like to make something with better audio, we might have used two, three or more stages, and the best transistors we could find.
If we did use two stages, what gain would each require? Half of 50 equals 25 each, doesn’t it? Well, that’s one reason why we like to use dB. A voltage gain of 50 is the same as a voltage gain of 34dB. If we had two stages they would share the gain equally, therefore 17dB each, which comes to a gain of just over 7 each. The gain of each of the two stages is simply the square root of the total gain (50), giving a per stage gain of 7.071. Of course, we can simply use the square root method for two stages, but what about four stages? That’s why dB is the choice for more complex circuits.
Of course, the power level of the single transistor amp won’t be high, so a small transistor called a “Signal Transistor” is all that is required, and the most common small signal transistors often come in a case that is called the TO-92 case (see below).
The TO-92 is easily recognised if you have looked into any common transistor circuit, but don’t be fooled into believing that all TO-92 transistor cases contain the same transistor. In fact, the case is also used for diodes, varicaps, DIACs, and even a 7805 low power regulator IC. The case style is just a set of dimensions, and does not define the transistor or whatever you might find inside it.
In fact, my suggested transistor, the BC548 transistor, for example, also comes in an SMD (surface mount device) version, as do many other transistors. The two outlines and pinouts are given in the following figures.
The case style is chosen by the manufacturer to allow sufficient cooling to the transistor. If the transistor gets too hot, a state known as Thermal Run-away can result, which essentially melts the transistor silicon, allowing at least parts of the N and P layers to flow together, altering the transistor specs at the very least, but in the worst cases forming the transistor into an insulator or conductor, or a hole in the PCB!
We at DIYODE often mention “Letting the smoke out”, which is a common talk reference to a failure on a component, but in reality, electronics can be very sneaky in hiding the death of a component, leaving it appearing as good as new. Maybe you never saw any smoke because there wasn’t any!
Datasheets contain a lot of information of which we are only interested a few numbers, and for our amplifier, the BC548 which is readily available and useful in a wide range of applications suits nicely.
|IC||Collector Current - Continuous||500||mA|
|TJ, Tstg||Operating and Storage Junction Temperature Range||-55 to +150||°C|
|fT||Current Gain Bandwidth Product (@VCE = 20 V, IC = 20 mA)||250||MHz|
Running through the specs listed in the table...
VCEO is the maximum allowable voltage across the Collector to Emitter when the transistor is turned Off. i.e. V C E O.
VCBO therefore, is the maximum voltage across the Collector and Base when turned off, although you can be certain that 75 volts would very likely cause the base to turn on!
VEBO is the voltage across the Base and Emitter when the transistor is turned off, but as a diode, it should be ~0.6V forward biased and should not be reverse biased, and as you can see, never by more than 6V!
IC is the Collector current, and limited to 500mA continuous, perhaps more for very short bursts, but also limited by the temperature dissipation of the device.
TJ is the temperature across the junction, or Tstg is the temperature of the whole device in storage.
FT is the Current Gain Bandwidth Product, which is the product of the gain required and the frequency of highest operation. Therefore, the BC548 transistor would have a current gain of 1 at 250MHz, or 250 at 1MHz, assuming other values are suitable, as each condition effects other limits. Transistors are limited by four related maximum values, Vmax, Imax, Pmax, and finally Tmax.
Our amplifier has been chosen to be a Common Emitter (CE) Amplifier, which means that the emitter is common to the input and output circuits. The base will become the input and the collector becomes the output.
The transistor has a nominal current gain of greater than 100 under most conditions, so every mA into the base should cause 100mA at the output, or perhaps more to our scale, every µA into the base should cause 100µA increase at the collector. The emitter takes both the base current and the collector current to ground, and as such, the emitter current is the sum of the base current and the collector current.
Note that there are also Common Base (CB) and Common Collector (CC) amplifiers, each with their own advantages and disadvantages.
We can go deeper at a later stage, but as a single transistor analog amplifier, the circuit will be operated in a “Class of Amplification” known as Class ‘A’, which requires that the output voltage should be centred at half the expected output voltage limits, roughly half the supply voltage for our circuit, or 4.5V.
We should normally use a “Load Line” chart to select the best operating values or use a simulation program such as SPICE, we would need another article to introduce each. They also both require some idea of the load to be driven before any accuracy can be expected.
We are to use a high impedance input on an Arduino, so we can get away with less collector current.
QUIESCENT VS SIGNAL
Amplifiers are designed in two stages, ‘Quiescent’, or no-signal condition, and Signal conditions. So, we begin with the resistors and finish with the capacitors, in this amplifier.
The supply voltage was chosen as 9V from a 9V battery, which is a fairly constant voltage when the load is reasonably constant, so no filtering is necessary. If you choose a “rectified and regulated” supply, make sure that it is itself properly filtered.
The output of the amplifier (VC) should be middle of the voltage range, and if it were just the resistor RL and the transistor between the supply lines, the output voltage would be 9V/2 = 4.5V. The reason for this is so the amplifier output voltage can swing as high and low as possible between the supply rails, thus allowing the maximum undistorted output. Our amplifier isn’t intended to go more than 1Vp-p, from 3.5V to 5.5V, so we can simply go with 4.5V at the output.
The emitter voltage (VE) can be the ground voltage, and for a simpler amplifier we can leave it out, so you might be wondering what it does and why we choose to have a resistor, ‘RE’ at all.
RE is a negative feedback resistor that increases the emitter voltage as the combined base and collector currents pass through RE. In Quiescent conditions, RE is simply a small voltage drop, which we will come back to soon.
The base voltage is set to provide an almost constant ~0.6V above the emitter voltage, so VB = VE + ~0.6V. Note we say ~0.6V as the voltage VBE varies with load current in a logarithmic relationship between 0.6V and 0.7V, and is sometimes said to be around 0.622V at 25°C.
Therefore, 0.6V is a “Rule of Thumb” that is considered to be on the safe side of reality. In simple terms, the voltage VBE must be VE + 0.6V, but without deeper math, may require adjustment when the circuit is built.
RE would therefore normally cause us to change the collector voltage slightly as the emitter voltage is above ground. The swing voltage is, therefore, less than rail to rail, and the collector voltage should technically be halfway between the swing limits. i.e. They all interact!
For a voltage gain of 10, RE should be approximately 1/10th of RC. That’s not perfectly correct as RC carries the collector current and RE carries the collector current plus the base current, but as we’re using a range of fixed value resistors, we will probably not be able to use the exactly calculated values anyway.
No, not those little black things that look like dead flies in buns! Transistors work because of current amplification, in that the collector current is some value many times the base current. We can talk about ‘A’ as amplification, ß, which is IC/IB, or hfe all of which tend to get rolled together and confused. I’ll leave hfe for the engineers, use ‘ß’ for current gain and leave ‘A’ as the amplifier voltage gain.
The collector current should be kept small for less heat generation and therefore, less noise. The datasheets suggest a gain of 100 even at 100mA, but to keep the noise down, we can try our luck with 20mA, and do our calculations based on a gain of 100, but with some gain control added to have a controlled gain of 10.
Therefore, if the ß for the BC548 is 100 or better, we can say that the base current is likely to be found by IB = IC/100 = 20.45mA/100 = 0.2045mA or 204.5µA.
The resistors RB1 and RB2 form a voltage divider to set the base voltage VB, and as voltage dividers are loaded by the circuit they feed, thus changing their tapped voltage value, the voltage divider is usually calculated for a “Bleed Current”, (Ibleed) to take 10 to 20 times the base current, with the lower value taking less current from the supply and the higher value providing better regulation of the voltage.
We will opt for 20 times making the resistors smaller in resistance, and making the “Bleed” current, Ibleed = 20 x IB = 20 x 200µA = 4mA.
The Collector Resistor (RC) is often called the Load Resistor (RL or Rload). From the circuit details now defined, RC = (VCC – VC)/IC = (9 – 4.5)/20e-3 = 225Ω. We could add standard value resistors in series and/or parallel to make 225Ω, but our numbers at this point are not fixed in concrete, so let us select a 220Ω resistor, and transposing the formula to see what this does to the collector current; IC = (VCC-VC)/220 = 20.45mA, which is quite acceptable.
We are happy for a gain of 10, so VE = VC/10. RE should be found by Ohm’s Law, remembering that RE also passes the base current to ground, so RE = (VC/10)/(IC+IB) = 0.45V/20.6545mA = 21.7Ω, 22Ω!
A common and easier “Rule of Thumb” method is to simply divide the value of RC by the Voltage gain. i.e. RE = RC/10 = 220/10 = 22Ω.
Now RB1 and RB2 can also be calculated quite easily based upon the quiescent base voltage being estimated at 0.45V + 0.6V = 1.05V, and Ibleed being 20 times IB, which in turn is 1/100th of IC. Ibleed therefore is IC/5 = 20.45/5 = 4.09 mA.
RB2 = VB/Ibleed = 1.05V/4.09mA = 256Ω, ~270Ω.
RB1 = (VCC-VB)/(Ibleed + IB) = (9V – 1.05V)/(4.09mA + 204.5µA) = 1851Ω, ~1.8 kΩ, BUT, there may be a problem with increasing the value of RB2 from 256Ω up to 270Ω, yet reducing the value of RB1 from 1.851kΩ down to 1.8kΩ, and in fact, this is where the typical “Rule of Thumb” method can fall apart.
A better “Rule of Thumb” solution is to “always fudge in the same direction”. Another is to keep the ratios of the voltage divider resistors the same, to get the correct output.
Therefore, let us check the ratios; RB2/RB1 = 256/1851 = ~ 1:7.2. First chosen values = 270/1800 = ~1:6.7, so the collector current will increase causing the output voltage to fall. Increasing RB1,= 270/2200 = ~1:8.15, or decreasing RB2, = 220/1800 = ~1:8.18 , so in both cases the collector current will decrease causing the output voltage to rise.
So whether we increase RB1, or decrease RB2, the ratio is greater and the difference to the calculated ratio is also greater. In this case, the better option is to go with the rounded values of RB1 = 1.8kΩ and RB2 = 270Ω.
We could investigate every pair of standard value resistor until we find something closer, but in the end, we don’t really care if the output voltage is a little high, or low. We simply prefer it to be between VCC and VE, and as close to the middle as we can easily get it.
In all cases the gain should still be controlled by the ratio between RC and RE, so that should not be an issue.
At this point, we have a circuit with quiescent values, which we can build and measure. The values are given in the diagram here.
Now is the time to consider what happens when an ac signal is applied to our amplifier. To avoid changing the resistance values and altering the quiescent state un-intentionally, we typically use capacitors in three roles:
- to supply the signal to the amplifier
- to sample the signal and provide it to the next stage, and
- to provide an ac gain that is better than the DC gain.
C1, was chosen as a 47nF when we worked on the electret, and there is no need to change it except to say that being connected to VB is not the same as being connected to ground, and the maths for an exact calculation of a value for C1 should include a much larger circuit than used in finding 47nF. The value depends on other values in the circuit including the electret load resistor, Rload, RB1 and RB2, and the “amplified resistance of the emitter circuit” being equal to (RE + Re); not something we can cover here. Simply accept that calculating an exact value would take a lot more maths.
The intended load has a reasonably high input impedance, the input to another amplifier or an Arduino. Essentially, the capacitor is connected to the junction of the Collector and the load resistance (RC), and one “Rule of Thumb” is that it should have a reactance one-tenth of the load resistance. We haven’t done reactance so simply accept that C = 10/(2πfRC) = 2.41µF, ~2.2uF.
That should be a non-polarised electrolytic capacitor, with a voltage of at least 150% of the supply voltage, e.g. 16VDC capacitor.
The capacitor from the emitter to ground, CE, effects the gain at signal frequency. Being in parallel with RE, it can only increase the gain as the frequency rises, as the parallel combination reduces feedback. There are many opinions on the best method to use here, and some conflict. In general, the circuit works without CE, but if you need more gain at ac, then any capacitance will help, the larger the capacitor, the greater the gain, except falling off exponentially as frequency increases. Many older HiFi amplifiers were improved by increasing the capacitance.
One method, again based upon filter thinking, is to make the reactance of the capacitor equal to RE at the lowest frequency of interest, e.g. 300Hz. CE = 1/(2πfRE) = 24µF ~22µF.
There’s no such thing as the final circuit for as long as there is a soldering iron handy and a supply of alternative parts! However, as we’re running out of time and space, here is the sum of our discussions in a circuit form.
All experiments should be completed by a conclusion statement. Please make your own project, experiment, and come to your own conclusions. For me, I need more than one stage, and that’s as simple as connecting our single stage in as many stages as you want, using only one “coupling” capacitor between stages. Some filtering would be nice, and a volume control, and a driver and power output stage.
In fact, as a simple amplifier, I do prefer an op-amp, and even op-amp based power stages, and filtering, etc. I guess op-amp devices are easy to work with, easier to design with, more tolerant of supply fluctuations, and you can replace one IC with the same part number without the circuit conditions changing.