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DESIGNING TRANSISTOR AMPLIFIERS
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| Home > Tutorials > Designing Transistor Amplifiers |
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Modified:22:39, 22 October 2013
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| INTRODUCTION
This is a guide on how to design a transistor audio amplifier using a DC supply in 7 easy steps.
The circuit we will use is shown right. This tutorial will show you how to calculate the correct bias resistor values for R1 and R2.
The gain of the amplifier is set by Rc / Re. In this example, we have set a gain of:
Gain = Rc / Re = 1 x103 / 22 = 45 (approx)
so if the input signal is 10mV, then the output signal will be 450mV.
Q1 is any high gain low power transistor. In this case we will use a BC549 transistor which has a gain (hFE) of 100-200. The transistor gain should be at least twice that of the amplifier gain. We will take hFE at an average of 150
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CIRCUIT DIAGRAM
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| STEP 1 - FIND VRc - Voltage across Rc |
We want the output to sit halfway between 0v and 9v so the voltage across Rc will be half that of the supply, so
VRc = +V / 2 = 9 / 2 = 4.5v
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| STEP 2 - FIND Ic - Current flowing through Rc |
There will be current flowing through Rc since there is a potential difference across it. We can use Ohms Law (I = V/R) to work this out.
VRc = 4.5v
Rc = 1x103
Ic = VRc / Rc = 4.5v / 1x103 = 4.5 x10-3 A
We will take Ie as the same as Ic by ignoring the contribution from Ib which will be less than 1% of the total, and eases calculation.
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| STEP 3 - FIND VRe - Voltage across Re |
There will be voltage generated across Re since there is current Ie flowing through it. We use Ohms Law (V = IR) to work this out.
Ie = 4.5x10-3 A
Re = 22
VRe = Ie x Re = 4.5x10-3 x 22 = 0.099 V
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| STEP 4 - FIND VR2 - Required bias voltage |
| This is easy to do since the voltage at the base will always be 0.7v higher than the emitter when the transistor is conducting.
So VR2 = VRe + 0.7v
= 0.099 + 0.7 = 0.799v = 0.8v
This voltage is the required bias voltage to keep the output at half that of the supply.
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| STEP 5 - FIND VR1 |
Find this value while we are at it to make calculating the bias resistances easier later on. VR1 is simply the supply voltage minus VR2 (Kirchoffs Voltage Law)
+V = 9v
VR2 = 0.8v
VR1 = +V - VR2 = 9 - 0.8v = 8.2 V
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| STEP 6 - FIND Ib - Required Base Current |
This we need to use Ib = Ic / hFE, the formula for NPN bipolar transistors.
Ic = 4.5x10-3 A
hFE = 150
IB = Ic / hFE = 4.5x10-3 / 150 = 30x10-6 A
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| STEP 7 - FIND R1 |
| Finally we can calculate the resistor values to bias the transistor correctly using Ohms Law.
We will be using a potential divider to do this. In order for the voltage to remain stable in the divider, we need to allow roughly 10x the amount of current needed for Ib, in the divider.
VR1 = 8.2 v
IR1 = 10 x 30x10-6 = 300x10-6 A
R1 = VR1 / IR1 = 8.2v / 300 x10-6 A = 27333 or 27K
For R2, the current will be slightly less, since Ib flows into the transistor.
VR2 = 0.8 v
IR2 = 300x10-6 - 30x10-6 = 270x10-6 A
R2 = VR2 / IR2 = 0.8v / 270 x10-6 A = 2962 or 3K.
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