Transistor Circuit Essentials
Voltage Divider Biasing
The most effective method to bias the base of a transistor amplifier is using a voltage divider. In the next chapter, we will analyze each transistor connection in detail and we will be using always this biasing method. Therefore, let's take some time to explain this method thoroughly.
The idea is that the voltage divider maintains a very stable voltage at the base of the transistor, and since the base current is many times smaller than the current through the divider, the base voltage remains practically unchanged. The resistor Re provides the negative feedback as explained before (Emitter Feedback Bias). Due to the fact that the base voltage remains unchanged, the negative feedback works very effectively and any unwanted increment of the current gain produces an -almost- equal negative feedback. The collector and emitter currents change just a few, and the Q point remains practically stable. Now, let's see in detail how this works...
Voltage Divider Bias Equations
We start with the assumption that the base current (I_{B}) is many times smaller than the current through the voltage divider (I_{VD}). A ratio of 20 is a good approach. This means that the base current must be at least 20 times smaller than the voltage divider current. This condition allows us to exclude I_{B} from our calculations, with an error less than 5%. Now we can safely calculate the base voltage as follows:
V_{B} = I_{VD} x R_{2}
Or using the classic voltage divider equation:
V_{B} = (V_{CC} x R_{2}) / (R_{1} + R_{2})
The current that flows through the voltage divider is:
I_{VD} = V_{CC} / (R_{1} + R_{2})
From the base voltage we can calculate the emitter voltage and the collector-emitter voltage drop as follows:
V_{E} = V_{B} - V_{BE}
V_{CE} = V_{C} - V_{E}
The emitter current is calculated using the Ohm's law:
I_{E} = V_{E} / R_{E}
And since the collector current is practically equal to the emitter current, we can calculate all the transistor currents and voltages:
V_{RC} = I_{C} x R_{C}
V_{C} = V_{CC} - V_{RC} = V_{CC} - I_{C} x R_{C}
V_{CE} = V_{CC} - I_{C} x R_{C} - I_{E} x R_{E} = V_{CC} - I_{C} (R_{C} + R_{E})
As you see, we can calculate everything we need without using any hybrid parameter. This is an amazing and unexpected result. Two transistors with different current gains can operate as amplifiers with exactly the same amplification, only because they are biased with a voltage divider.
Moreover, since V_{BE} is many times smaller than V_{B} and V_{B} remains unchanged all the time, the emitter voltage V_{E} remains also unchanged, hence maintaining a stable emitter current.
Firm and Stiff Voltage Divider
Previously, we made the assumption that the voltage divider current I_{VD} is many times bigger than the base current I_{B}, about 20 times as big. This is a good approach for an error less than 5%. This is not always possible thought. If the base current is high, the resistor values for the voltage divider must become very small, and this leads to numerous problems.
In such cases, we design the voltage divider with a ratio of 10 instead of 20. This approach has an error less than 10% when the IB_{} is excluded from the calculations, which is still acceptable. The voltage divider that satisfies this condition is named Firm Voltage divider:
I_{VD} > 10 x I_{B} => R_{VD} < 0.1 x Î²_{dc} x R_{E} (do not forget that Î² = h_{fe})
On the other hand, the application may require a very good Q stability with an error less than 1%. A ratio of 100 can be used to calculate the resistors, if this is possible:
I_{VD} > 100 x I_{B} => R_{VD} < 0.01 x Î²_{dc} x R_{E}
The voltage divider that satisfies this condition is named Stiff Voltage divider, and has an error of less than 1%.
Condition Confirmation
Suppose that the designer wants to design a transistor amplifier with stiff voltage divider bias. He designed a circuit that has (along others) emitter current I_{E}=1mA. The voltage divider he designed is calculated according to the stiff VDB condition, which means that the base current must be 100 times smaller than the voltage divider current. According to this calculation, the maximum base current cannot be greater than 40uA. The question now is: Does this circuit works efficiently for the whole h_{fe} range?
The fact that I_{B} and h_{fe} are excluded from the calculations, does not mean that these values do not affect the operation. They still have a small affect, but this is very small (1 to 10%). What we have to confirm now is that this affect will always remain small, even at the worst case scenario.
But which is the "worst case scenario"? Well, simply: The worst case scenario is when the transistor operates with minimum current amplification. When this happens, the base current becomes maximum to supply the required emitter current. Suppose that the transistor that our designer used, has an h_{fe} with range from 30 to 300. We have to confirm that the base current will remain under the calculated value (40uA) and still it will be able to provide full emitter current (1mA), even at the lowest h_{fe} (30):
So, the base current for the worst case scenario (33uA) is still less than the calculated base current (40uA), therefore we can say that this voltage divider remains stiff.
What each part does
Designing a transistor amplifier with VDB (Voltage Divider Bias) is not that hard, but sometimes it takes time to select the proper part values to begin with. And many times the designer has to change some parts to change the amplifier parameters. Here is a quick reference for the designer to know what each part controls:
R1 - This resistor should be used to control the current through the voltage divider
R2 - This resistor controls the base voltage V_{B}
R_{E} - This resistor controls the emitter current I_{E}
R_{C} - The collector resistor can control the V_{CE} voltage
to get more information about <a href="http://911electronic.com/tunnel-diode-characteristic-symbol-definition/">tunnel diode</a> click hotlink. I found this site yesterday and i think there is a lot information about diodes.
Hi,
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.
At 9 January 2014, 14:04:31 user Nash wrote: [reply @ Nash]
@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:
Question 1:
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
Question 2:
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.
Hey!
This is a good tutorial, however I am a bit lost in the math by the way you write it.
Would it be possible to rearrange it and have the formula go down the page e.g.:
5*(2 2)
=5*(4)
=5*4
=20
Also can you please explain to me the arrow that you use in the formulas, what does it mean? e.g.: IE = %u03B2 x IB => IB = IE / %u03B2 = 1mA / 30 => IB = 33uA
@john I know what you mean, but i do not show the conventional current flow but the electron flow (which is the reverse). I think i have to explain this somehow in a short paragraph.
At 14 April 2012, 9:19:47 user john wrote: [reply @ john]
I think the current flow in your explanation is for PNP transistor, because for NPN transistor current flow going in to basis not going out.
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.
At 16 January 2012, 11:07:08 user almalo wrote: [reply @ almalo]
Sorry I'am a bit confused:
First you wrote: hfc>hfe>hfb.
2nd: hfe = hfc + 1
So what is the truth?
@Russ i took a quick search around but i could not find one with more details about the biasing history.
At 20 October 2011, 1:01:32 user Russ wrote: [reply @ Russ]
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.
Thanks
At 13 June 2011, 15:40:12 user Fung wrote: [reply @ Fung]
The L7805 1A voltage regulator, there are 2 situations when a switch is open and close.
When open, only an LED and a 74HC00 IC are working, the total current may be about 30mA.
When closed, at least 5 ICs will be in operation, with an 7-segment dual display, the total current may increase to about 150mA. Then the regulator heats up in a short time.
The transistor is also heat up because of the voltage regulator.
Will the problem be solved by moving it away from the voltage regulator?
@Fung yet is varies. but the situation you describe does not sound normal to me. which voltage regulator you use, and how much current this is supposed to provide?
At 21 April 2011, 7:46:10 user Fung wrote: [reply @ Fung]
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?