Please 'Boom' Responsibly As most of you have noticed, the noise ordinances have become much tougher lately. Most of this is due to idiots, yes IDIOTS, who drive through residential areas with their windows down while their system is playing at full power. To make things worse, the music they listen to has all sorts of foul language that's not suitable for small children, (who may be playing outside). There are even a few people, who are even beyond idiot status, that play their systems at full power through residential areas after 10:00 PM (when many people go to bed). I don't believe that this type of behavior is good for the industry. If the fines get too stiff, people will stop buying large systems. If this happens, more people will get out of car audio (who wants a mediocre system). People get interested in things because they're exciting. A deck and four 6.5" speakers are not going to interest many of the younger car audio enthusiasts. If car audio enthusiasts keep annoying more and more people, the fines will keep getting tougher. All of this will only reduce interest in the equipment that fuels the industry. If you want to listen to your system at full volume, get out on the highway where there's little chance of bothering anyone. When you get to a red light, turn it down. If the only thing attractive about you is your 'system', you have some work to do. Bottom line... Think about what you're doing. Think about other people. It's not the end of the world if you have to turn the volume down for a little while. |
In the following
demo, the lines inside the window are the collector, base
and emitter voltages (relative to the power supply
voltage). The white line is the collector voltage. The
black line is the emitter voltage. The yellow line is the
base voltage. The top of the window represents the full
supply voltage. The bottom of the window represents 0
volts (ground). You should notice a few things:
In the following calculator, if you set the base voltage to zero, the transistor will be turned off and the collector and emitter resistors will pull the transistor's collector and emitter up to the power supply voltage and down to ground. As was noted above, the ratio of Rc/Re would change the rate that the collector would change with respect to the emitter voltage. If you change only the value of one of the resistors, you can see that the collector voltage moves varying amounts from the positive power supply. If the input signal were a sine wave, you'd see that the signal gain would change with a Rc/Re change.
To start the calculations, we need to start with things that we know as fact. We know that the base voltage is 5 volts. Remember that Vbe (voltage from the base to the emitter) is approximately the same as the voltage drop across a forward biased diode. If we take Vbe to be 0.6 volts, we can say that the emitter voltage is 4.4 volts. Since we also know the value of the emitter resistor. Using these 2 facts, we can calculate the current flow through the resistor. If we use Ohm's Law... Ire = E/R Ire = 4.4/15 ohms Ire = 0.293 amps As we said before, some current flows through the base of the transistor. The current through the collector is equal to the current through the emitter minus the current through the base. We can calculate the current through the base if we know the current through the emitter and the transistor's beta (Hfe). The beta can be found in the manufacturers datasheets. The DS will give a range of values. Select a value somewhere in the middle of the range. Since we only know the emitter current and beta, we'll calculate base current like this: Ib = Ic/beta (since we don't know Ic yet, we need to calculate Ic from Ie, Ib and beta) Ie = Ib*beta + Ib 0.293 = Ib*45 + Ib 0.293 = 45Ib + Ib 0.293 = 46Ib Ib = 0.293/46 Ib = 6.37mA Now that we know the base current, we can calculate the collector current. Ic = Ie-Ib Ic = 0.293-.00637 Ic = 0.287 amps Now that we know the collector current, we know the current through the collector resistor (the transistors collector and collector resistor are in series which means they have the same current flow). If we want to know the voltage across the collector resistor, we can again use Ohm's Law. Vrc = Ic*Rc Vrc = 0.287*12 Vrc = 3.439 volts If we want to know the voltage across the transistor, we can subtract the voltage across the emitter and collector resistors from the power supply voltage. Vq = Vps-(Vrc+Vre) Vq = 12-(3.432+4.4) Vq = 4.168 volts We can calculate the collector voltage by subtracting the voltage drop across the collector resistor from the power supply voltage OR you can add the voltage across the emitter resistor to the voltage across the transistor. Either way, you'll come up with a collector voltage of 8.568 volts. With the information that we found above, we can calculate the power dissipation across the individual components. For the resistors, we can use either the current through the resistors or the voltage across the resistors. For the power dissipation in the emitter resistor: Pre = I˛*R Pre = 0.293 amps˛*15 ohms Pre = 1.29 watts You would need a resistor rated to dissipate 1.29 watts or more to prevent it's failure. For the power dissipation in the collector resistor: Prc = I˛*R Prc = 0.286 amps˛*12 ohms Prc = 0.98 watts For the power dissipation in the transistor, we will use the collector current and the voltage across the collector and resistor. For the power dissipation in the transistor: Pq = I*E Pq = 0.286*4.168 Pq = 1.19 watts Without a heat sink, this transistor would get very hot in a very short time. Even if the current flow through the transistor was below its rated maximum current at 25°c, it may still fail because its current rating is derated as its temperature increases (it can handle much less current when it's hot). NOTE: If the transistor is fully "on" (saturated), it will have virtually no difference of potential (voltage) between its collector and emitter. Current flows through it with virtually no resistance. This condition is known as saturation. I said 'virtually' no resistance because the transistor is not 100% efficient and will always have a small amount of resistance/loss. When operating in saturation, the transistor will dissipate relatively little heat because, while the current flow may be significant, the voltage drop is very small. Class A Amplification The following calculations are for a simple Class A amplifier. As you know, class A amplifiers are very inefficient. This example is no exception. The input impedance would be too low to be driven by a standard preamp but this example should give you an idea of what's required for this design. I used the values here so that I'd have values that were more easily understood. Most designs of this type are for small signal amplifiers, not power amplification. The small signal amplifiers have currents and power that are in the milli range. These are just a little higher.
Given:
As we said earlier, the gain is equal to the ratio Rc/Re. If we know that the collector resistor is 2 ohms (because we want a 2 ohm output impedance) and we want a gain of 10 we will use the formula: Av = Rc/Re or Re = Rc/Av Re = 2/10 Re = 0.2 ohms We now need to decide on the Q point or the operating voltage for the collector. So that the audio signal can swing equally both above and below the operating point, we want the Q point to be 1/2 of the power supply voltage. If we had a split supply (± voltage), the Q point would be 0 volts but we have a single ended supply. Now that we know the value of the emitter resistor, we can can calculate the current needed through the collector resistor to drop half of the supply voltage. For that we will go to Ohm's Law. V = I*R or I = V/R I = 17.5/2 I = 8.75 amps (yikes!) Since we know that the collector current is virtually equal to the emitter current (generally within 1%), we can use the collector current to calculate the voltage drop across the 0.2 ohm emitter resistor. V = I*R V = 8.75*0.2 V = 1.75 volts In the previous circuits we used a Vbe of .6 volts. When the current flow through the transistor is high, Vbe will increase. For this example, we will use Vbe=1.0 volts. You can get the Vbe vs. Ic curves from the transistor's datasheet. Now that we know that the emitter voltage is 1.75 volts and that Vbe is 1.0 volts, we know that the base voltage is 2.75 volts (1.75+1.0=2.75). For this amplifier we will use a voltage divider to bias the transistor. Since we know the DC current gain of the transistor, we can calculate the base current. I = Ie/Hfe I = 8.75/100 I = 0.0875 amps When using a voltage divider to bias a transistor, it's common to use resistor values that will pass about 12 times the current drawn by the base. This prevents the base current from significantly affecting the voltage divider's operation. Since we know the base current is 0.0875 amps, we will use a divider that passes 1.05 amps of current when connected to the 35 volt supply. To make this calculation, we will again use Ohm's Law. R = V/I R = 35/1.05 R = 33.3 ohms total We can now use the voltage divider rule to calculate the resistor that connects the base to ground. Rbase = Vbase/Ivdivider Rbase = 2.75/1.05 Rbase = 2.62 ohms Since we know that the total resistance is 33.3 ohms, we can calculate the upper bias resistor by subtracting the lower bias resistor from the total resistance. This would leave us with an upper bias resistor of 30.41 ohms (30 ohms would likely be close enough).
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