# Transistor Audio Amplfier - Clipping Problem

Discussion in 'The Projects Forum' started by j.ep, Apr 16, 2014.

1. ### j.ep Thread Starter New Member

Apr 16, 2014
1
0
Hello there,

I'm designing a pretty basic transistor amplifier circuit to play music out of an 8 ohm speaker. It will be three stages, but I'm having problems on the first stage with clipping.

When the input signal goes above around 60 mV the output starts to clip (it's closer to 200 mV when simulated, but this is expected I guess).

I set Vc to close to 4.5 V to allow maximum voltage swing, but as soon as the output starts clipping Vc starts to get bigger.

I'm wondering if I've biased it wrong? Or if the transistor is saturating and that's why it's clipping? (Though would this be a problem of biasing?) Also I calculated a value of about 4 V for Vce but the actual measured value was 5 V so I wonder if this could have affected things?

I've attached a picture of the circuit and the input signal just before it clips, showing where it sits. Any help would be much appreciated, cheers.

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2. ### PRS Well-Known Member

Aug 24, 2008
989
35
I checked your circuit and get the following bias values:
Vb=1.06 volts (from divider)
Ve=1.06-.6=.46v
Ie=.46/RE=2mA
Vc=2.5v from 9-Rc*Ie
Vce=Vc-Ve=2v
max pos swing is 9-Vc=6.5v
max p-p swing=4v
Av=3000/220=13.6
max output; 100mV*13.6=1.36v peak to peak
The 100mV maximum input is due to harmonic distortion for larger signals

Now how do you feed stage two 1.36 volts without getting distortion? The answer is in using feedback. But you'd make life more simple if you just used an op amp followed by a power amp that can supply 8 ohms.

3. ### Rolf Zetterberg Member

Sep 20, 2008
14
2
To get symmetrical clipping of the output,don't just think about half power supply voltage.That's a simplification.In real life you have to take in consideration both the voltage loss across the emitter resistor and the voltage loss across the saturated transistor.
These two voltages must be added to the 'half supply',which means that it should be raised by some 0.66V+0.09V=0.75V.
So the theoretical 'half supply' would in this case be 4.5V+0.75V=5.25V.

Now that's in theory.Since,in real life,values of all components may differ a symmetrical clipping is best achieved by checking with an oscilloscope and adjusting one of the bias resistors until satisfactory.
As an example of real life values,the BC547B base-emitter voltage can differ from 0.58V to 0.7V at 2mA collector current according to the data sheet.

4. ### Veracohr Well-Known Member

Jan 3, 2011
540
75
You're saying that scope image is suppose to be showing the base voltage? If so, something isn't right. The biasing should center the base voltage at 1.06V, but the image shows 69mV peak-peak centered at 109mV. One possibility is that you actually have a 820Ω resistor for R2, rather than 8.2kΩ. Make sure the bias is set up correctly as in your schematic, check the resistor values.

Last edited: Apr 17, 2014