In order for a coupling or bypassing capacitor to operate effectively, it must have the right size. As said before, the capacitor acts like a resistor in AC current. The resistance of a capacitor is called "impedance". Unlike resistors, capacitors do not have a fixed impedance. Instead, the impedance is determined by the frequency of the AC signal. The equation to calculate the impedance is the following:
X_{C} = 1 / (2*Ï*F*C)
C is the capacitance in Farads, and F is the AC signal frequency in Hertz. So, a 10uF capacitor connected in series with a 1 KHz signal generator, will present an impedance of:
X_{C} = 1 / (2 x 3.14 x 1 x 103 x 10 x 10^{-6}) = 15.9 Ohms
So, what is the proper value for a coupling or a bypassing capacitor? There are two factors that must be taken into account: The frequency and the circuits total resistance. Let's talk about the frequency first. If the amplifier operates as a signal amplifier of a fixed circuit, then the answer is straight-forward: The frequency is the signal's frequency. But if the amplifier operates in a wide frequency range, then we must choose the worst case scenario.
Let's see an example of an audio amplifier. Audio amplifiers typically operate from 20Hz up to 20Kz. To choose the right frequency for our calculation, we must first think what we want the bypassing or coupling capacitor to do: We want this capacitor to act as a short-circuit in AC currents. In other words, we want the capacitor to present the lowest impedance possible in AC current. Since the impedance X_{C} is reverse-proportional to the frequency F, the lowest impedance is presented at the highest frequency. Thus, the worst case scenario (highest impedance) is presented at the lowest frequency. So, in our calculations we will use the lowest frequency that the capacitor will operate at. In an audio amplifier for example, the lowest frequency is 20Hz.
The second factor is the total resistance of the circuit. Let's talk first for a coupling capacitor. In this case, we are talking about the total resistance of the circuit in series with the capacitor. Here is a simplified example of a circuit with a coupling capacitor:
In this circuit, the total resistance is the sum of the internal resistance of the generator Rg, plus the internal resistance of the transistor ri.
In a circuit with a bypassing capacitor, the total resistance is the total resistance of the circuit parallel to the bypassing capacitor. Here is a simplified example of a circuit with a bypassing capacitor.
In this circuit, the total resistance is the R_{E}.
So, now we know how to choose the worst case scenario in terms of frequency, and how to calculate the total circuit resistance according to the capacitor type (coupling or bypassing). The optimum capacitor value that we choose must be 10 times smaller than the total circuit resistance, calculated for the worst case scenario.
To calculate the capacitor, we solve the impedance equation for C:
C = 1 / (2 * Ï * F * X_{C})
Let's see an example. In the simplified example of a circuit with a coupling capacitor shown above, Rg is 50 Ohms. r_{i} is 2.2 KOhms and the frequency is 20Hz to 20KHz. To ^{calculate the optimum capacitor value, we must first calculate the total resistance of the circuit in series with the capacitor:}
R_{total} = 50 + 2200 = 2250 Ohms
The worst case scenario is 20Hz (lowest frequency), so the capacitor must present a resistance of less than 225,0 Ohms (R_{total}/10) at 20Hz frequency:
C = 1 / (2 * 3.14 * 20 * 225) = 35.3uF
So, the capacitor must have at least 35uF capacitance. This makes sure that the capacitor will have less than 1% effect on the total resistance of the circuit. And since the calculated value does not exist as a standard capacitor value, we choose the next bigger value - in our case that is 47uF.
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?