How can I calculate the gain of a class C amplifier with no bias? These amplifiers often have little or no emitter degeneration, but I don't see how to figure the intrinsic emitter resistance when there is no quiescent current. I understand how r_e is derived from Ebers- Moll, but this assumes the base voltage is held constant.
I'm guessing it's going to depend on transistor beta given the operating conditions? Is it not correct to speak of gain in this situation, because of the nonlinearity? How about output power for a given drive level?
What kind, collector or emitter output? An emitter follower is obviously slightly less than 1, except for that little +/-0.6V offset where nothing happens. A collector output depends on what's driving the base and how much collector current that in turn produces.
For small signals, the gain is of course zero, but that's not too useful. It might be more useful to speak of voltage or power gain at optimal input / output conditions, in which case you'd probably be reading up on RF amps to determine that.
Tim
-- Deep Friar: a very philosophical monk. Website:
Collector output. I'd like to be able to calculate the gain if I know the voltage and source impedance at the input. The collector load is some (possibly complex) impedance up to the positive supply. Emitter has a small or zero resistance R to ground.
If you can recommend any good books, that would also help. I have the Art of Electronics and Experimental Methods in RF Design. The problem is there seems to be a gap between what AoE explains regarding this sort of thing and what EMRFD assumes from the reader. I also have Introduction to Radio Frequency Design by Hayward, but it is a pretty difficult book for me.
A Class C amplifier certainly has a power gain and an efficiency, but the active device is usually configured to operate as a switch.
When amplifying RF signals of limited bandwidth, Class C is invariably used with some kind of filter to reconstruct sine waves from the pulses of current that flow through the Class-C 'switch' - the simplest form being a (coupled) parallel tuned circuit. At the very least, ITU regulations limit the extent to which harmonics of signals can be radiated. The apparent power gain, and then the efficiency, of a Class-C amplifier is determined by the loaded Q of the filter amongst a few other factors.
With each input cycle, the active device biased to work in Class C passes current for a fraction of the cycle, less than 50%. The amount of current depends on the input drive, the device characteristics, what the device is connected to (i.e. the filter) and the power supply. The pulses of current applied to the filter result in a sine-wave component of current that can be fed to the system output port (plus harmonic components that may need to be further filtered out). The strength of the resulting fundamental sinusoidal component can be found from the Fourier component of the current pulses at the fundamental frequency. Higher Q reduces loss in the system so it increases the efficiency, and reduces the strength of harmonic components in the output signal. Push-pull is very common, not least because it cancels the second harmonic which could otherwise be expensive to clean up sufficiently to meet the ITU rules. It also allows cheaper devices to be used.
On a sunny day (Fri, 21 Aug 2009 21:39:35 -0700 (PDT)) it happened Kyle Cronan wrote in :
In my view a class C driving a tuned circuit is basically a switch with less then 50% 'on' time. For switching you will like to drive the transistor (or MOSFET) into saturation. This will also have to be done fast enough. So you are faced with beta (current gain) and hole storage time in case of a transistor. Ccb wil also try to reduce switching time and lower impedance. You will need to 'overdrive' the thing. I think 'gain' is not the right word here, just treat it as a switcher.
I assume this is RF amplifier with the sine wave input, the filter source and the filter load. First, you should determine the angle of conduction of the transistor. Then, assuming the collector current is a piece of the sine wave, apply the Fourier to find the power of the fundamental component at the output. The angle of conduction is determined by the input drive amplitude and the bias. The collector current depends from the input drive level, supply voltage, source and load impedances and the C-B capacitance; other parameters are less important.
Vladimir Vassilevsky DSP and Mixed Signal Design Consultant
The gain depends on the input because the stage is very nonlinear. The transistor (or diode) equation will tell you the current at each instant in time. You can fill a spread sheet with many points along the curve to get the current at many times in the cycle. You can then multiply the current by sin() and cos() functions to get the two parts of the fundamental and then combine the averages to get the magnitude.
You can also plug the design into spice and have it do the work.
The strong feedback through C-B capacitance determines the behavior of the stage. Since the load of the stage is LC, the collector voltage is close to the full sine wave. So the inherent nonlinearity of the transistor is not that important; what matters is the angle of conduction. You have to solve the equation including the source and the load impedances, Ccb and Vcc.
You can fill a spread sheet with many points along
Archimedes, Newton and Einstein perfectly dealt without Spice and Matlab :-)
Vladimir Vassilevsky DSP and Mixed Signal Design Consultant
Okay, so it sounds like the analysis is not nearly as simple as for an amplifier with a voltage divider for base bias! I'm going to try simulating some circuits or just trying different collector loads experimentally.
I get that the transistor is basically acting as a switch. I guess I was trying to figure out how much gain would be possible before driving it to cutoff, assuming that there would be clipping. But now I see that you just let an inductive collector load ring up while the transistor is cut off. Am I describing that right?
There's something I don't understand: EMRFD is always saying that the collector "likes to see" a load equal to the supply voltage squared over twice the output power. Certainly, that's the value you get for an output sine wave with a maximal peak-to-peak amplitude, equal to Vcc. But in this case, there are two complications: the harmonics from nonlinear operation of the transistor, and the ringing up of the inductor past the positive supply voltage. So how can you justify starting the design process with this equation? Is it that the two factors complicating things always exactly cancel each other out?
Oh, okay, I think I was looking at a section of the book that presented a "stress test" in which the drive level is very high and the LC network is detuned: you get very high collector voltages over only a fraction of the cycle (because of charge storage, apparently). These are BJT amplifiers.
r
Wait, if the collector load is resonant, won't the peak-to-peak amplitude be twice Vcc?
Ok, so the harmonics are negligible in terms of power if the collector voltage is nearly sinusoidal?
Not generally true. It is advantageous for efficiency in a Class-C amplifier to have the device operating either at saturation or cutoff. There is no benefit to 'linear operation' when the inherent degree of distortion is so great that it requires a tuned circuit to reconstruct the output signal.
formatting link
has some of the details, and also attempts to illustrate differences between classes C and D. The latter also involves switching devices but in Class D the input signal is modulated using PWM (or PRF modulation, or suchlike) and the output signal is low-pass filtered - effectively to demodulate it.
Class C is often used to amplify an unmodulated signal (e.g. in RF multiplier and driver stages) and is fine for amplifying constant-envelope signals such as FM/phase modulated signals. However, amplification of signals like SSB and AM that require linearity to avoid generation of in-channel intermodulation products usually demands the quasi-linear characteristics of Class AB, or Class B in push-pull. Some AM transmitters use Class-C output stages with high-level modulation: the power supply voltage to the switching Class-C stage(s) is varied by the modulating signal.
Sure there is. Driving that capacitor when it's not at exactly 0V takes an awful lot of power. Switching hard into that load burns up your switch.
A simple and easy to construct example is a blocking oscillator (which, in the tuned case, is basically a self-excited class C oscillator). Set it up with a resonant load and you'll only get terrible efficiency -- the transistor burns so damn much power trying to drive that capacitor, and the capacitor needs to be fairly large to get a moderate Q at the low impedances involved. However, operate it in blocking mode (no cap, time constant determined by bias current and inductance) and you can basically run without a heatsink. Beautiful square waves, high efficiency -- high harmonic content.
The point of class C is to use a minimally maximum drive: just enough to saturate *on the peaks*, not so much that you burn all your efficiency on the flanks of conduction.
Tim
--
Deep Friar: a very philosophical monk.
Website: http://webpages.charter.net/dawill/tmoranwms
The OP didn't say anything about the frequencies involved. I left out the C-B capacitance. Now you are leaving out the E-B capacitance and the fact that both are lossy and the fact that there is inductance in all three electrodes and the extra phase shift in the collector current.
The inherent nonlinearity is important. It is what causes there to be an angle of conduction in the first place. Also the OP was talking of the zero bias case. In this case the angle of conduction is large enough that the transistor can't be assumed to be only conducting at the point of bottoming as is the typical simplification for the class C case.
n
:-)
Yes but just imagine what they could do today with it.
I assummed the textbook case of the RF power amplifier with the tuned input and output.
You are correct. The "grid-leak" resistance in the bias path is the essential parameter.
Computer is just a tool of trade; it is very addictive. Probably, Archimedes could build somewhat bigger catapults, Newton would be caught in the loop of the further optimization of the accuracy of the navigation measurements, and Einstein would be probing yet another numeric models of the wave functions.
Vladimir Vassilevsky DSP and Mixed Signal Design Consultant
Okay, I sort of get what you guys are talking about but I still have no idea how to do the design. This is my situation: I have available about 26 dBm from the previous stage at a low impedance, less than 10 ohms. I would like to take advantage of this and avoid transformer matching at the input. I need an output power of 4-5 W. I would like to avoid the need for any high Q inductors that would need to be hand wound, but I would like reasonably high efficiency. Ability to operate over a wide frequency range of 3-7 mhz would be a bonus, but a bandwidth of several hundred kilohertz would be acceptable.
There you go assuming. I have fixed a class-C push pull audio power amplifier. It sounded just good enough that people could understand what was said but didn't pass current at idle.
e
f
Unless it is the case where the bias path has a low resistance and the RF drive has a finite impedance and power at the frequency of interest. Since the stages input impedance decreases with the amplitude of the input, a class-C stage can effectively be driven by a constant power.
b :-)
I liked his idea for the hook and winch system to defend against ships. You hand these things off the wall of the fort. When a ship get too close, you hook it and start to wind up on the winch. This lifts one end of the ship. The other end sinks. You then let go and down it goes.
The early flame thrower was also a clever idea. It scares the heck out of the enemy and actually burns the ships to the water line too.
You could compensate for the distortion of the output stage by feedforward from the driver stage. There is a brilliant bridge-type solution for that; IIRC first used in Quad-405 amp.
Long while ago I've burned (literally) with the VHF Class C BJT amplifier with R = 0 in the bias path. The R creates automatic bias, which stabilizes the operation wrt the output load. If the output is overloaded or underloaded, the operating point of the stage is shifted so to compensate for it. If R = 0, the stage is more sensitive to the proper impedance matching.
Vladimir Vassilevsky DSP and Mixed Signal Design Consultant
This can be built from few discrete components, however the simplest solution would be using an IC from MiniCircuits. Just follow the application manual.
Vladimir Vassilevsky DSP and Mixed Signal Design Consultant
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.