It is important to be able to calculate the power characteristics of a transistor circuit. This includes both the input and output signal power, the power gain of the circuit, the efficiency and the power dissipation on the transistor.
How to do conversions between peak-to-peak (p-p) and rms sizes
Since we are talking about AC values, there is something first that we need to make clear: There are two ways to measure a AC sizes -for example voltage-, using an rms volt-meter or using an oscilloscope. If you measure the same signal with a voltmeter and an oscilloscope, you will not get the same results. That is because the voltmeter measures the rms-voltage while the oscilloscope measures the peak-to-peak voltage. So, depending on which measuring method you use, you may need to know how to convert between rms and p-p (peak-to-peak):
u_{rms} = 0.707 x u_{pp} / 2 => u_{rms} = 0.353 x u_{pp} u_{pp} = 2 x (1.414) x u_{rms} = 2.828 x u_{rms}
Now let's see how we can calculate the power of the input and the output signal. We will use the typical V x I formula with the rms values for voltage and current:
P = V_{rms} x I_{rms}
We can apply the Ohm's law (I = V / R) on the above formula to extract a more practical one:
P = V^{2}_{rms} / R
The above formula is valid if the voltage is measured with a volt-meter. If you are using an oscilloscope, then you need to convert the voltage from peak-to-peak into rms. The above formula can be directly converted to use peak-to-peak values like this:
P = V^{2}pp / 8 X R
Calculating the power gain
The formula to calculate the power gain is typical:
Ap = Pout / Pin
So we need to calculate the power on the output and on the input of the circuit. For both cases we will use one of the formulas for the power calculation from above. Usually, the last one with the peak-to-peak voltage is used. For the output power calculation, we only need to apply this formula to the load:
Pout = u^{2}_{L} / 8 x R_{L}
Remember again that we are talking about the voltage on the load, and since the load is coupled through a capacitor, it can only be an AC voltage. Regarding the input power calculation, we will apply the same formula, but this time on the base of the transistor:
Pin = u^{2}_{B} / 8 x R_{B}
Calculating the power gain using the hfe parameter
There is another simpler method to calculate the power gain through the voltage gain, using the h_{fe} parameter. The voltage gain can be easily calculated for all transistor connection types as we've seen in the previous pages. Let me briefely remind the formulas for the voltage gain calculation:
Common base: Av = R_{C} / r'e Common Emitter: Av = u_{C} / u_{B} = r_{C} / r'e Common Colector: Av = 0.99
Since Ap = Pout / Pin, Av = Vout / Vin, Ai = Iout / Iin and P = V x I, we can re-write the Ap formula like this:
Ap = Pout / Pin = Vout x Iout / Vin x Iin => Ap = Av x Ai
But as we know, the current gain is the h_{fe} parameter, so this leads us to the following formula:
Ap = Av x h_{fe}
Calculating the efficiency of the amplifier
This is again a very important value, especially for battery-powered applications which require low power consumption. The efficiency is calculated by dividing the output power with the DC power that is provided:
ÃŽÂ· = Pout / P_{S} x 100%
We need to calculate the DC power P_{S} first. To do so, we only need to multiply the supply voltage V_{CC} with the total DC current that flows through the circuit. Generally, this current is the quiescence current I_{CQ} plus the biasing current. For all sort of biasing methods EXCEPT the voltage divider bias, the biasing current equals to the base current, and since the base current is very small compared to the I_{CQ}, we can simply ommit it:
I_{S} = I_{CQ} + I_{B} => I_{S} Ã¢‰ˆ ICQ (Not valid for Voltage Divider Bias)
But if the amplifier is biased with VDB, then we need to calculate the total current that flows through the voltage divider:
I_{S} = I_{CQ} + I_{VD} => I_{S} = I_{CQ} + V_{CC} / (R_{B1} + R_{B2}) (Valid only for VDB)
And now we can calculate the DC power that the circuit will draw:
P_{S} = V_{CC} x I_{S}
Calculating the power dissipation on the transistor
Finally, a very important value that must be calculated is the power that the transistor is called to dissipate when no AC signal is applied. This value can be calculated simply by multiplying the DC quiescence voltage and current:
P_{D} = V_{CEQ} x I_{CQ}
You may now think that the power dissipation will increase if an AC signal is applied, but this is not true. If an AC signal is applied at the input, then part of the power that is being dissipated onto the transistor will be transferred onto the load (the amount depends on the efficiency of the circuit) and the power dissipation on the transistor will be decreased. Therefore, the above equation represents the maximum power dissipation.
The above diagram shows how the power dissipation on the transistor P_{D} decreases while the power on the load P_{L} increases.
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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?