As said, this setup achieves maximum voltage amplification ratio. Let's see how we can design such an amplifier. This amplifier will drive a high impedance load (10 Kohms). The DC biasing voltage is 12 Volts and the input signal is 10 mVolts p-p. The internal resistor of the signal generator is not significant so we leave it out of our calculations for now. The frequency is about 1KHz. The transistor is a BC338 with a typical h_{FE}=250 (DC current gain). We will use the same value for hfe as well (AC Current gain) . First, let's design the circuit schematic:
We can start the design by estimating the RC resistor. We will select one collector resistor with an educative guess, taking into account the load resistor R_{L}. The higher the collector resistor, the higher the voltage amplification of the amplifier. The point is that we cannot have large collector resistor if the load resistor is low. Since the load resistor is high, we can start by selecting a medium to high collector resistor, like for example 1000 Ohms. As said this is an educative guess, so experience helps to choose. Let's draw the DC equivalent and do the DC analysis:
We want the Q point to be in the middle of the load line, therefore we set the V_{CE} to V_{CC} / 2, which is 6 volts:
Now we can use the h_{FE} parameter to calculate the base current. This is where most errors take place, since the h_{FE} parameter is not very stable nor easy to measure. The typical h_{FE} of the BC338 is 250:
I_{B} = I_{C} / h_{FE} = 6 / 250 => I_{B} = 24 uA
The base current must be 24 uA for the transistor to operate in the middle of the load line. We set the base current with the base resistor:
It is quite large value, but that is normal for this amplifier. At more extreme example you may have results in the scale of Meg-Ohms. I will use a 470 Kohms resistor as a closer value. Since we've used different resistor from the calculated value, it is good practice to re-calculate the I_{C} value for this resistor following the opposite way:
I_{B} = V_{CC} / R_{B} = 12 / 470 = 25,5 uA
I_{C} = I_{B} x h_{FE} = 25,5 x 250 = 6,3 mA
The difference is not very big, but if you want to be academically correct then you have to do this recalculation. Now we can proceed with the AC analysis of the circuit to estimate the characteristics of the amplifier. To do so, we make the AC equivalent:
It is important to calculate the internal emitter resistance of the transistor and the base impedance before we start:
The first characteristic that we can easily calculate is the input impedance of the amplifier. Since the base resistor r'e is parallel to the base impedance Zin(base), we use the Ohm's law to calculate the amplifier's input impedance:
Zin = R_{B} // Zin = (R_{B} x Zin(base)) / (R_{B} + Zin(base)) = (470 K x 990)/( 470 K + 990) = 987,9 Ohm
This value can be used to calculate the AC voltage drop from the AC source on the internal generator resistor, but in our case we said that this resistor is zero, so there is theoretically no voltage drop. The next value we can calculate is the AC current in the base:
i_{B} = u_{B} / Zin(base) = 10 mV / 990 = 10,1 uA
Using the hfe parameter, we can calculate the ac current on the collector:
i_{C} = Î² x i_{B} = 250 x 10,1 = 2,5 mA
Now that we know the collector current, we can use the ohm's law to calculate the voltage drop that this current will generate onto the output resistor R_{C} // R_{L}. But first we calculate the equivalent resistor Rext:
Rext = R_{C} // R_{L} = (R_{C} x R_{L}) / (R_{C} + R_{L}) = 909,1 Ohms
u_{Rc} = i_{C} x Rext = 2,5 x 909,1 = 2272,7 mV
Av = u_{Rc} / u_{B} = 2272,7 / 10 = 227
So this amplifier will amplify the input voltage by 227 times. It is obvious that the voltage amplification is very big. One thing to determine is if this amplifier works as a small signal amplifier. Let's remember the rule once more:
The AC peak to peak current of the emitter must be smaller than 10% of the DC current of the emitter.
The AC current is 2,5 mA and the DC current is 6,3 mA. For the amplifier to operate as small signal, the AC current should be smaller than 10% of the 6.3 mA, this means that is should be smaller than 630 uA. The circuit does not operate as a small signal amplifier and this means that there may be some sort of distortion at the output. Let's make this circuit on a breadboard and see the results:
The yellow channel of the oscilloscope is connected to the base of the transistor with the yellow wire to measure the input voltage. The green channel comes from the green wire which measures the output:
The base-emitter characteristic of the transistor
First, let's examine the output amplitude. We calculated that the output had to be 2.2 Volts. With the oscilloscope we measure the output and it is 1.95 Volts. Although the numbers are not identical, we can still consider that the circuit operates normally, because the error is very small.
Before we calculated that this amplifier does not operate as small signal amplifier. Nevertheless, the output does not seem to be distorted whatsoever. The waveform is a nice smooth sine wave with no clipped or rounded edges. How comes that? Well, the answer is in the base-emitter characteristic. It is obvious that any distortion would start if the collector current was higher than 50mA in any temperature. Since our collector current is limited to 6 mA, the transistor works in the linear area of the characteristic and that is why we do not see any distortion.
Power Gain Calculation
As we saw in the Power Calculation and Efficiency page, we can calculate the power gain of this amplifier simply by using the voltage gain Av and the h_{fe} parameter:
Ap = Av x h_{fe} = 227 x 250 => Ap = 56750
So the power of the input signal is amplified by 56750 times.
Efficiency Calculation
This amplifier works as an A-class amplifier, which means that the efficiency will not be very good. To calculate the efficiency we need first to calculate the DC power that the transistor draws (P_{S}):
P_{S} = V_{CC} x I_{S}
As we saw in the previous pages, the DC current I_{S} for this sort of transistor bias is equal to the I_{CQ} current. Therefore:
P_{S} = V_{CC} x I_{CQ} = 12 V x 6.3 mA => P_{S} = 75.6 mWatt
Now we must calculate the output power that is provided to the load by the circuit:
Pout = u^{2}_{out_p-p} / 8 x R_{L} = 5,15 / 8 x 10000 = 0.064 mWatts
So, the efficiency is:
Î· = Pout / P_{S} x 100% = 0.064 / 75.6 x 100% => Î· = 0.084 %
Transistor Power Dissipation Calculation
Finally, we can calculate the power that will be dissipated on the transistor. The maximum dissipation happens when no signal is applied, so we will calculate this value using the DC voltage and current quiescence values:
P_{D} = V_{CEQ} x I_{CQ} = 6 x 0.0063 => P_{D} = 37.8 mWatts
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?