Transistor Circuit Essentials
Choosing the right bias
After selecting the proper connection, the one that is most suitable for your application, you must select a biasing method. Biasing in general means to establish predetermined voltages and currents at specific points of a circuit, so that the circuit components will operate normally. For transistors, biasing means to set the proper voltage and current of the transistor base, thus setting the operating point, also known as quiescence point (Q). We will discuss in details the quiescence point within the next chapters. For now, you need to know that this point will determine how the transistor will operate (amplifier or switch). A correctly placed Q offers maximum amplification without signal distortion or clipping.
The most efficient and commonly used biasing method for transistor amplifiers, it the voltage divider bias (VDB). We will analyze this method in detail, but first we need to discuss the other biasing methods. In this chapter, we will use a Common Emitter NPN transistor amplifier to analyze the various biasing methods, but each method can be used for other connections as well.
This is the most rarely used biasing method with transistor amplifiers, but it is widely used when the transistor operates as a switch. The base current IB is controlled by the base resistor RB. From the second law of Kirchhoff, we have:
VCC = VB + VBE
VB is calculated using the Ohm's law:
VB = IB x RB
So, by selecting the proper base resistor RB, we can define the required base voltage VB and base current IB. Now we can calculate the collector current using the appropriate hybrid parameter. Since this is a common emitter circuit, we use the hfe:
IC = IB x hfe
The problem with this method is that the collector current is very sensitive is slight current gain changes. Suppose for example that this is a silicon transistor and operates as a B-class amplifier with current gain 300, RB=80 Kohms, RC=200 Ohms and VCC = 10 volts:
The output of this circuit is taken from the collector resistor RC:
VRC = IC x RC = 6.7 Volts
Now suppose that the temperature rises. As we've discussed in earlier pages, this will increase the current gain. An increment by 15% is a realistic and rather small value. From 300 it will climb up to 345. This means that the collector current will become 38.7mA, and the output voltage will also become 7.7 Volts! A whole volt higher than before. That is why this biasing method is not used for transistor amplifiers.
On the other hand, due to the fact that this method is very simple and cost-effective, it is widely used when the transistor operates as a load switch, for example as a relay or LED driver. That is because the Q point operates from cut-off to hard saturation, and even large current gain changes have little or no effect at the output.
Emitter feedback bias (Fixed bias with emitter resistor)
This is the first method that was historically used to fix the problem of the unstable current gain discussed previously. In a transistor circuit with fixed bias, a resistor was added at the emitter. This method never worked as it should, so it is not used anymore. This is how it was supposed to work. If the collector current is increased due to a temperature increment, the emitter current is also increased, thus the current through RE is also increased. The voltage drop across RE is increased (emitter voltage) which eventually increases the base voltage. Finally, this base voltage increment has as a result the decrement of the voltage across the base resistor RB, which eventually decreases the current of the base IB. The idea is that this base current decrement decreases also the collector current!
This sounds amazing since a change of the output of the circuit has an effect on the input. This effect is called "feedback" and more specifically it is a "negative feedback", since an output increment causes an input decrement. Here is how the new collector current is calculated:
IC = (VCC - VBE) / (RE + (RB / hfe))
Let's see how the previous circuit (Fixed bias) would react if we add a 100 Ohms RE feedback resistor.
IC = (10-0.7) / (100 + (80000 / 300)) = 9.3 / 366.6 = 25.3 mA
We assume again that the current gain is increased by 15%:
So, a 15% current gain increment caused a 15.1% output current increment. By adding a 100 Ohms feedback resistor at the emitter, a 15% current gain increment caused a 10.6% output current increment. The increment is 4.5% less which means that this method works somehow, but still the shifting of the Q-point is too large to be acceptable.
Collector feedback bias (Collector to base bias)
The next method that the researchers used to stabilize the Q point is the collector feedback bias. According to this method, the base resistor is not connected at the power supply, instead it is connected at the collector of the transistor. If the current gain is increased due to temperature increment, the current through the collector is increased as well, and this decreases the voltage on the collector VC. But the base resistor is connected at this point, so less current will go through the resistor in the base. Less current through the base means less current through the collector.
Again, there is a negative feedback in this circuit. But how much is it? Lets do some math. The collector current is now calculated by the following formula:
IC = (VCC - VBE) / (RC + (RB / hfe))
To see the change, we will apply this formula in our first example (fixed bias):
IC = (10 - 0.7) / (100 + (80000 / 300)) = 9.3 / 366.6 = 25.3 mA
When the current gain is increased by 15%:
IC = (10 - 0.7) / (100 + (80000 / 345)) = 9.3 / 331.8 = 28 mA
The effectiveness of this method compared to the emitter resistor feedback bias shown before is exactly the same. The difference is that, RC is usually much larger than RE, which results in higher stability. Nevertheless, quiescence point Q cannot be considered stable.
Collector feedback bias (Collector to base bias)
It did not take long before someone tried to utilize both the previous methods to work together to achieve better results. And indeed, the stabilization is much better than each one separately. The formula to calculate the collector current is the following:
IC = (VCC - VBE) / (RC + RE + (RB / hfe))
Let's apply this formula to our previous examples
IC = 9.3 / (100 + 100 + (80000 / 300)) => IC = 9.3 / 466.6 = 19.9 mA
With a 15% current gain increment:
IC = (VCC - VBE) / (RC + RE + (RB / hfe)) = 9.3 / (100 + 100 + (80000 / 345)) = 21.3 mA
So, a 15% current gain increment causes a 7% output current increment. Although it is better than the previous circuits, still the Q point is not stable enough. Add to this that hfe is extremely sensitive to temperature changes and the transistor generates a lot of heat when it operates as a power amplifier. So we need a much better stabilization technique.
Would you please tell that why the mentioned curvature happens? I have this problem in my TFT (Thin Flim transistor) and in the low voltage of D-S, D-S current does not like a diode curve and it has a curvature like you mentioned.
@Giorgos Lazaridis Your Diagrams for the current flow for a NPN connection are wrong. BIG MISTAKE Fix it please http://pcbheaven.com/wikipages/images/trans_theory_1317761009.png & http://pcbheaven.com/wikipages/images/trans_theory_1317761285.png
@Mint Electronics Sorry for 2 reasons: first for the loooong delay (i thought that i had posted the answer immediately), and sorry for not explaining this in the article. I will re-read the whole theory when i finish it and fix some issues like that. I though that it was not so important to explain it, but maybe i will put some spoilers with the proof. Anyway:
Vcc = Vb Vbe Ve => Vcc - Vbe = Ve Vb
But we can approximate that Ve = Ic x Re (since Ic almost = Ie)
Also, Vb = Ib x Rb => Vb = (Ic x Rb) / hfe
From the above:
Vcc - Vbe = (Ic x Re) [(Ic x Rb) / hfe] => (We get Ic in common)
Vcc - Vbe = Ic x [Re (Rb / hfe)] => (divide both sides with term)
(Vcc - Vbe) / [Re (Rb / hfe)] = Ic
The voltage divider current will always be bigger than the base current, since it is composed by the voltage divider current PLUS the base current.
@Mint Electronics Although i try to keep the math as simple as possible, the way you want me to re-arrange it would be more like a math tutorial rather than a transistor tutorial. I keep it simple but not that simple, it would be tiring and confusing for those who want to learn transistors.
As for the "arrow", it is not an arrow, it is the Greek letter %u03B2 (Beta) which probably you cannot see due to your browser's encoding used. It is good to know that there are people who cannot follow this encoding. This %u03B2 letter is the same as the hfe. In formulas we use %u03B2 rather than hfe for short. I think i have to find something else to show this....
In the meanwhile try to change your encoding and the correct letter will reveal.
The use holes to explain any part of a transistor function is confusing, a positive charge does not move because of the mass of the positively charged nucleus.[ie; Protons, so called holes.]The flow of electrons is convincingly demonstrated by the cathode ray tube and other experiments carried out a hundred years ago. Electron deficiency and excess better explain the attraction or repulsion which is used for a transistor to function
@almalo you're right, the typo is obvious. I used the minus sign used to show the reverse current directin, as an algebric sign. I had to use ABS numbers for the comparison. Common collector has the maximum current amplification. hfc>hfe>hfb
I write -hfc = IE / IB but i should write instead |-hfc| = IE / IB. The result with this change is:
hfe = hfc - 1
I tripple check this editorial because i do not want to make such mistakes, sometimes i fail to locate them though. Thank you for noticing.
Can you recommend a book that goes into the history of transistor biasing and the development of other basic circuits? I am interested in how this developed. From hindsight it seems so clear and I wonder how difficult it really was.
The resistance of a transistor varies as its temperature changes, am I right?
I have a circuit which has a +5V voltage regulator, a transistor is used to amplify the signal of a PIEZO sounder and it is placed nearby the voltage regulator because of the routing of tracks. However, due to the quite-high current, the regulator heats up when it is in operation, heatsink is not setup yet, not only itself, the tracks under the board are also heated up.
Due to the case stated above, the transistor is heated up later on, it changes the frequency to the PIEZO sounder (ie unable to keep the original frequency) because of the change of its internal resistance. As I need to add the heatsink on the regulator, what things should I do in order to keep the frequency (that is to keep the internal resistance of the unit) of the amplifying circuit?