What are Small Signal Amplifiers

If you do not understand Ohm's Law then jump out now and come back. This is where we cover basic electronics and principles.

Amplifiers

Firstly for the purists let us attempt to roughly define a class A amplifier:

"An amplifier, with or without negative feedback, having the greatest fidelity in faithfully reproducing the input with the least distortion. It is however the least efficient, in as much the power delivered to the load is only a small percentage of the d.c. power used up in the amplification process".

That was not the most elegant definition admittedly, yet it covers the basics:

(a)    might use feedback.
(b)    good fidelity.
(c)    lousy efficiency.

Feedback - This is where part of the output signal is fedback to the input BUT 180 degrees out-of-phase (i.e. partially cancels the input). If it were in-phase feedback then we would have an oscillator - which in this case we definitely do not want.

Fidelity - This means many things to many people but to us it means the output must be an exact replica of the input but only magnified or amplified.

Efficiency - The theoretical limit to this amplifier's efficiency is 50%, meaning for every watt of output we will use up at least 2 watts of d.c. power input in to the amplifier. Depending upon the application this may or may not be significant. If we have a large power transformer available to us and power, literally to burn, who cares?.

On the other hand if miniaturisation were the keyword and precious battery supply is the only power source available thennnnnn! that's another matter.

Most experts will agree with me when I say you will only see this 50% efficiency in your dreams. So why use class A?. At low levels of signal and amplification, the losses through inefficiency are not significant and are far outweighed by the goal of fidelity or in the r.f. game, linearity, i.e. a linear amplifier.

Figure 2 - ac / dc amplifier and ac only amplifier models

Let's deal with the d.c. conditions first because we have to overcome one peculiar aspect of this type of transistor. It has to be turned on or have the base opened.

To do this, the base has to have a voltage applied to it which is approximately 0.65V higher than the voltage level at the emitter. In fact using this property allows a transistor to be used as a simple switch. Better still, arrange about 20,000,000 of them the right way around and you get a Pentium III.

With class A amplifiers, use resistors to   bias the transistor for d.c. conditions. Like this:

Figure 3 - schematic of a class A amplifier

Here we have our customary 12V d.c. power supply. We have a resistor from supply to base and another from base to ground. There is also a resistor from supply to collector as well as one from emitter to ground. What do these resistors do?.

Dealing with the pair associated with the base i.e. R1 and R2, they form a voltage divider network. The ratio of :

[R2/(R1 + R2)] * 12V = Base voltage (d.c.)

if we use 82K for R1 and 39K for R2 we would get a base voltage of

[39K/(82K + 39K)] * 12V = 3.87V (d.c.)

also the current flowing through these base resistors (ignoring any current into the base) is - from ohms law -

Ib = 12/ (R1 + R2) = 0.1 mA (rounding up slightly)

Now different people have different ideas but I'm a big believer in the theory that the current through the emitter should be between 5 - 10 times that of the base current. Let's pick a ratio of about midway,  7  i.e 0.7 mA for emitter current.

Of course if a higher emitter current was required then simply jack up your base current to maintain the approximate ratio.

Remember we said the base had to be 0.65V higher than the emitter?. Well if we have a base voltage (calculated above) of 3.87V and it is 0.65V higher than the emitter then the emitter must be (3.87 - 0.65) which is 3.22V. If we have also said that the emitter current is 0.7 mA then it follows the emitter resistor R3 must be - again Ohms Law -

Re = 3.22/.0007A = 4600 ohms or near enough to 4K7

If you are worrying about me jumping from 4600 ohms to the nearest standard, which is 4K7 or 4700 ohms, then keep things in perspective. Most resistors, capacitors in these circuits are about 5% or worse tolerence.

This then leaves the resistor from 12V supply to collector, R4.

This is called the load resistor. It could (at r.f.) easily be a choke, a transformer or a resonant circuit. For the moment we are hanging in at a.c., sort of working around audio frequencies but the same principles apply to r.f.

Figure 4 - final schematic of a class A amplifier

Now IF 0.7 mA is flowing (at d.c.) through the emitter then it must also flow through the collector. Also if our supply is 12V and the emitter voltage is 3.22V then the voltage at the collector must be:

Vcc - (Ic * RL)  Where RL is in fact R4 or

12V - (0.0007A * R4)

If we elect to have a load resistor of 6800 ohms or 6K8 for R4 what happens?. Well all we have concentrated on so far - in a long winded way - is the d.c. conditions. What about a.c.? Now here is the pointy end of the stick. Remember I said a.c., meaning the usual audio frequencies but the same principles apply to r.f.

Firstly we need a coupling capacitor from the previous stage as well as a coupling capacitor at the output. These are C1 and C3 respectively. Their sole function is to block d.c. so that d.c. voltages in this stage do not transfer to adjacent stages. Usually we look for a very low reactance (Xc) at the frequency of interest so that frequency is not impeded in the transfer. A capacitor will pass a.c. or r.f. but not d.c.

A capacitor of 0.82 uF has a reactance of about 650 ohms at 300 Hz and 65 ohms at 3000 Hz. These being the two limits of audio frequencies for communications purposes. These values are considered low enough to use this capacitor here. Of course a higher value can be used but then you get into electrolytic types and polarity must be observed.

If this were a hi-fi type amplifier then you would certainly use higher values. On the other hand with communications receivers anything which reduces 50/60 Hz mains hum should be used so we tend to go for the lower values.

Looking back to the beginning we note that for a.c. or r.f. purposes the emitter should be grounded. I say should be, because if at a.c. there exists a resistor to ground then the gain suffers because of "emitter degeneration". Later we will take advantage of this in another circuit employing feedback.

For the present we need a grounded emitter at a.c. How is this achieved? Simple, by-pass the emitter resistor with another capacitor, C2 of the same or similar value to C1 and C3. This capacitor has the effect of making the emitter resistor R3 invisible for a.c. or r.f. purposes.

The last capacitor, C4 (again the same or higher in value) is from the 12V d.c. supply to ground. This has the effect in conjunction with R5 in the power supply line of decoupling the supply. This resistor might have a small value of say 33 ohms.

This means any a.c. or r.f in our amplifier section gets shunted to ground and does not pass along the d.c. power bus line to contaminate other stages. Similarly no contamination from other stages should ever get into our stage (well we can always live in hope - only joking).

So just what happens now that we have surrounded our transistor with resistors and capacitors?.

Well first of all what is the likely input impedance (that "I" word again) of our little amplifier?. Some engineers could write a whole text book on that topic alone. For our purposes we will use the following approximation:

Z in or R in = [25 * Beta (ac)] / Ie (mA)

This means 25 times the a.c. or r.f. beta, divided by the emitter current in milliamps.

Now if this was an r.f. amplifier we could approximate the r.f. beta by dividing the Ft by the frequency of operation. Our 2N2222A transistor has according to the specs earlier, a d.c. beta (Hfe) of 100 but at a collector current of 150 mA. As I said I've measured wide variations. At r.f., the specs say an Ft of 300 Mhz and if the amplifier was used at about 7 Mhz we could use an approximation for r.f. beta of 300/7 or about 40. But our present example is at audio frequencies where I would approximate a figure of 90. Substituting we get:

Z in or R in = [25 * 90] / 0.7 (mA) = 3300 or 3K3

This figure would be affected by the parallel combination of R1 and R2 but being quite large in comparison, can be ignored in this particular example.

Let us agree on a net input impedance of 3000 ohms or 3K. What if a signal of 10 millivolts a.c. were applied to the input of our amplifier?. In the base we would have (using ohms law) an a.c. current of 0.01/3000 or 3.3 uA flowing (that is 3.3 millionth of an amp).

If our a.c. beta was 90 then we may assume our collector current would vary by 90 * 3.3 uA or 300 uA. Now this amplified current at an a.c. rate passes through our load resistor R4, or 6800 ohms or 6K8. Again using nothing more than ohms law we find 300uA * 6800 = 2.04 volt swing. The small signal voltage gain is 2.04/.01 or 204. This is a gain of 46 dB, a high and possible unstable figure.

Looking a lot more closely at R4, because we had (under no signal conditions) a standing collector current through our transistor of 0.7 mA there must have been (again ohms law) a voltage drop across it of 6800 * 000.7 or 4.76 volts. Therefore measuring the voltage on our collector we would have had (12V - 4.76V) or 7.24 volts.

Under our 10 millivolts signal conditions that voltage would have swung plus and minus the amplified signal of 2.04 volts. Being an alternating variation or a.c. it is able to pass through coupling capacitor C3 on to the next stage.

The main point is, if you have kept awake so far, you have now digested the basic principles of amplifier design.

Let's look at one example of a small signal amplifier,

A PRACTICAL EXAMPLE

Fig 1.

Here I've used a pretty standard and cheap transistor for our small signal amplifier. This transistor has some pretty impressive characteristics though.

The output circuit consists of a low pass filter network which also converts the desired output impedance we want Q1 to see to our standard 50 ohms output.

The 100 ohm resistor, RFC XL2 and the 0.01 uF capacitors are purely for decoupling purposes i.e., to keep RF out of the small signal amplifier power supply as well as other stages. Let's consider firstly the input circuit of our small signal amplifier.

Q1 is biased for DC conditions by R1, R2 and the emitter resistor of 270 ohms in this instance. Alert readers will be aware I like to bias the base voltage of my transistors to about 25% of Vcc (.25 * 12V) or 3V. It follows then that R1 will be about 3 times the value of R2 - think about it!. If the base voltage is around 3V then the emitter voltage is going to be 3v - 0.65V = 2.35V

If the emitter voltage is 2.35V approx. then the emitter current Ie through the emitter resistor of 270 ohms must be (from ohms law) 2.35 / 270 = 0.0087 or 8.7 mA. I've also said elsewhere I like base current to be about 1/7th of emitter current - alright these are my foibles and others would disagree. They're welcome to write their own papers.

So base current is going to be about 1 mA and seeing R1 + R2 are connected across 12V it follows that (from ohms law) R1 + R2 = 12V / .001 = 12,000 ohms or 12K. For biasing R1 is 3 times R2 so using simple maths R2 is 25% or 3K and R1 would be 9K which are not necessarily readily available standard values. We will make R2 = 3K3 and R1 = 10K which if you do all your sums is near enough and probably about a third of the values others might use.

So we have our DC conditions satisfied and the 0.01 capacitor in parallel with the emitter resistor means for RF purposes the emitter is at ground potential. This then leaves the output circuit to be discussed. The 22 ohms resistor in the collector circuit is there to discourage parasitic oscillations. RFC XL2 as I said before is only to decouple the power supply and I'd look for a reactance of around 20,000 ohms or at 2 Mhz something like 1 to 2.5 mH.

All this leaves is our low pass filter matching network. First question?? How much output power do we want? Huh? Yep that's how it all works.

Let's say we wanted +17 dBm for a mixer circuit. To the uninformed +17 dBm is a power relationship in milli-watts. Power is always (10 * log of power) so in this case in reverse we divide the 17 by 10 to get 1.7 which is the log of 50 so it follows that +17 dBm is in fact 50 mW of power. Learnt something?

Incidentally a power level of 50 mW into 50 ohms also equates to Erms = SQRT ( 0.05 * 50 ) or 1.58V RMS or 2.828 times that value to get pk-to-pk, which is 4.47V PK-PK.

Alright how do we design to get 50 mW out of our amplifier? By using the formula R = Vcc² / (2 * Po) or in our case [ (12V * 12V) / (2 * 0.05) ] = 1440 ohms. Want more power? Change the numbers! Obviously there are limits but you get the idea.

From the above the collector needs to see a load of about 1440 ohms which in turn has to be transformed into our 50 ohm load. By the way, if the amplifier doesn't see a 50 ohm load then all these calculations go right out the window. At the end I show my method of ensuring something like a 50 ohm load and more important the method helps the succeeding stage see a 50 ohm source.

This is a simple "L" network low pass filter designed in this case to transform 1440 ohms to 50 ohms. Follow these steps where SQRT signifies square-root-of:

1. XL = the SQRT of [(R1 * R2) - (R1 * R1)] = SQRT [(50 * 1440) - (50 * 50)] = SQRT [ 72,000 - 2,500] = SQRT of 69500 = 263.6 ohms

2. Xc = [(R1 * R2) / XL] = 72,000 / 263.6 = 273 ohms

Therefore the reactance of our inductor is about 264 ohms at our frequency of interest and the reactance of our capacitor is about 273 ohms at that same frequency. In the beginning I mentioned a requirement for a 1.8 to 2 Mhz small signal amplifier so we will nominally use 2 Mhz as our cut off frequency i.e. we want to pass all signals below about 2 Mhz but not above (filter out harmonics!).

Here I always see what capacitor has a reactance of 273 ohms at 2 Mhz using the standard capacitive reactance formula Xc = 1 / (2 * pi * Fo * C). Which when algebraically rearranged for our purposes becomes C = 1 / (2 * pi * Fo * Xc ). Slipping 273 ohms for Xc into that formula and 2 Mhz (2,000,000) should get you on your calculator 2.91.. ^-10 which should then be multiplied by exp 12 to arrive at an answer in pF. Doing that we get an answer of 291 pF which doesn't exist in the real world.

Now you have several choices here. (a) just plonk nearest standard components in for XL and Xc and don't worry about tuning - not recommended. (b) make part of Xc variable e.g. Xc comprises a fixed 270 pF capacitor with a 5 - 50 pF trimmer in parallel or (c) make Xc fixed and XL variable. You can only use the latter option if you have suitable slug tuned inductors available (they ain't cheap but could possibly be salvaged if you know what you are doing).

In the event you chose option (b) the required fixed inductor would be determined from the inductive reactance formula XL = (2 * pi * Fo * L). In ALL examples I use 6.2832 for 2 * pi. For our example we can again rearrange the formula as L = XL / (2 * pi * Fo) and plugging in this case 263.6 ohms XL from above and 2 for 2 Mhz we get L = 263.6 / (6.2832 * 2) = 20.98 uH. That is the inductance you would use, possibly with 60 turns of #26 wire on a T68-2 toroid as only one example.

If you elected method (c) - and this is really cool - I would look back at the capacitor required i.e. 291 pf, use the next lower value which is 270 pF and slot in a variable inductor which will tune through 20.98 uH. Feed a suitable signal to the amplifier, ensure the amplifier is terminated in a suitable fixed 50 ohm load (two 1/2 watt 100 ohm resistor in parallel = 50 ohms) and watch the output on a scope as the slug is adjusted. Wow! In fact you should get a similar effect with the variable capacitor method in (b). Certainly you will then understand why method (a) sucks.

I mentioned earlier how I ensure a 50 ohms load and succeeding stages see a 50 ohms source. I use a 50 ohm 3 dB attenuator. This is a resistive pi network attenuator which consumes 3 dB of power but represents a constant load. You put it in circuit after the last 0.01 uF coupling capacitor after the output.

Fig 2.

Now the downside. It consumes power. At - 3 dB that's half the power!!! What the hell just do your sums all over again to produce 100 mW from the amplifier. I would!

In this event your collector load is now 720 ohms, Xc = 197 ohms and XL = 183 ohms. At around 2 Mhz they translate into 403 pF (use 390 pF) and about 14.6 uH.

See - dead easy!

The above was written by Ian Purdie  and is an abstract from :

VK2TIP Ian Purdie's electronics tutorial radio design pages for amateur, ham radio, electronic project enthusiasts as well as for the electronic hobbyist and for home learning to design your own electronic projects