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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:
The base and emitter voltages are always within ~0.6 volts of each other (as long as the base/emitter junction is forward biased).
The emitter voltage is LESS than the base voltage.
The voltage on the collector is virtually a mirror image of the emitter voltage. When a transistor is used as an amplifier and the output is taken from the collector, the output is inverted (180° out of phase) when compared to the input signal.
As the voltage on the base increases and voltages on the collector and emitter pull away from their respective power supply values, the current through the resistors increases (indicated by the density of the yellow arrows).
The collector and emitter resistors are the same values. Changing the ratio of the collector resistor to the emitter resistor will change the rate that the voltage on the collector changes with a change in base voltage. Later we will see how the ratio is changed to increase or decrease the voltage gain on the collector.
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 = I²*R
Pre = 0.293 amps²*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 = I²*R
Prc = 0.286 amps²*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:
2 Ohms output impedance. The output impedance is determined by the collector resistor. This is the first thing we need to choose.
The gain is 10. The output voltage on the collector is 10 times the input voltage.
The transistor's Hfe is 100.
This is not a fully functioning amplifier. It's just the final stage of what could be a real amp. I'm just using it to show the calculations.
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).
All of the calculation would get you close but the actual values would likely have to be tweaked a little. To get the q point exactly right, you'd have to adjust the value of the lower bias resistor up or down. One thing you should also realize is that many of the values that we used for the transistor (such as beta and Vbe) will change with temperature. Since this amplifier would run very hot, you'd have to wait until it reached operating temperature to set the final values.
This amplifier would take its output from the collector. This means that the output signal is floating at ~17.5 volts DC. To drive a speaker with it, you'd either have to block the DC with a capacitor or would have to use 2 of them to drive a speaker in bridged mode.
This amplifier has no feedback. This is simply a single stage of what would need to be a multi-stage amp to be remotely useful.
A transistor with higher gain (1000 instead of 100) would make the circuit much easier to drive. It would increase the input impedance by a factor of 10.
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