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Since I'm a beginner to electronics, I'm trying to build the following circuit in order to see the way a common emitter amplifier works. I'm using the transistor of type BC547, as indicated in the picture.

CE amplifier

My problem is that the voltage \$U_{out}\$ is lower (by a factor of app. 5) than \$U_{in}\$. This is a picture of my circuit on a breadboard. The red circles indicate the points at which I measured the voltages.

My circuit

Can anybody see what I'm doing wrong? Please tell me if I have to provide more information about the circuit or something else.

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  • \$\begingroup\$ @SpehroPefhany I'm using a 9V DC source because that's what the first picture tells me. The pot is used to adjust the voltage between basis and emitter. If I cannot use this circuit as a DC amplifier, what would be an alternative? \$\endgroup\$ – Elko Apr 14 '17 at 18:17
  • \$\begingroup\$ Doggone those blue-background resistors...is R1 (47k) yel,violet,blk,brn? Or yel,violet,blk,red? (call me colour-blind). \$\endgroup\$ – glen_geek Apr 14 '17 at 18:21
  • \$\begingroup\$ What are you using for a signal source? This is an AC amplifier and the input should be capacitively coupled- otherwise the source will interfere with the DC bias. Can you use the 100K pot to adjust the collector voltage to an appropriate value (such as 2-3V) with the source attached (but amplitude near 0)?? \$\endgroup\$ – Spehro Pefhany Apr 14 '17 at 18:21
  • \$\begingroup\$ If you want to amplify DC you can use an op-amp (eg. LM324) as the easiest way, but you will not be able to reach the power supply rails let alone exceed them. You can amplify 0.1V to 5V but not 9V to 450V. For the latter you would need a switching power supply. \$\endgroup\$ – Spehro Pefhany Apr 14 '17 at 18:23
  • \$\begingroup\$ BTW if you actually applied 9V to the base, with emitter grounded, then the transistor is probably toast. \$\endgroup\$ – Spehro Pefhany Apr 14 '17 at 18:24
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I've provided your circuit in the top schematic below:

schematic

simulate this circuit – Schematic created using CircuitLab

With a simplified, equivalent circuit at the bottom.

Note that I've provided a new parameter, \$X\$, which tells us what percent of the potentiometer has been dialed in. The added \$R_7\$ there is also dependent on \$X\$ and acts in series to \$R_1\$. At the mid-point, where \$X=.5\$, the value of \$R_7\$ is at its maximum value of \$25\:\textrm{k}\Omega\$.

The collector voltage (\$V_O\$) vs % of potentiometer setting looks like this:

enter image description here

As you can see, you only get about half the potentiometer range (a little less than that) to sweep the collector voltage from about half of your power supply rail down to close to zero. Keep turning after that and not much will happen. That's in the case with \$\beta=150\$, so your case may have a value that is perhaps twice as high and therefore you will have much less range of movement for the potentiometer before reaching a point where changing it further doesn't change the collector voltage much, anymore.

The value you will measure at \$V_I\$ will look like this over that range:

enter image description here

As I just mentioned, the above charts aren't meant to be accurate for your situation, as I have no idea what BJT parameter values to apply. They are only approximate pictures. Each BJT is different and I decided to use \$\beta\approx 150\$ for the above charts and to neglect some details (such as what happens to \$V_I\$ when the transistor goes into saturation.) But it should give you an idea about the reality here.

You can easily see that \$V_O\$ can be less than \$V_I\$ towards the right side of these graphs. This is the area where saturation is taking place.

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Remove R3 (it's limiting the maximum output voltage to half the supply voltage).

This is an inverting amplifier, so the output voltage will go down as the input voltage goes up. There should be plenty of voltage gain, but only over a narrow range of around 0.6-0.7V on the Base. Above this voltage the transistor will be 'saturated' when it runs out of Collector voltage as the output nears 0V.

If you want to amplify an AC signal then couple it to the Base via a capacitor. You can then set the potentiometer to provide the DC 'bias' voltage required to get the Collector voltage to about half the supply voltage. The signal will then make the Base voltage rise and fall around the bias point, and the Collector voltage will follow it (amplified and inverted).

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Here is a good DC amplifier

schematic

simulate this circuit – Schematic created using CircuitLab

For more gain, reduce R5 to (100 - 2*26 ohms) = 47 ohms; expect gain of 100X.

I edited the schem, to provide offset adjustment: R6, R7, R8 allow cancelling mismatched of Vbe, up to 18 milliVolts. Short Vin to GND, and change R8 to get 0.0000000 volts between +vout and -vout.

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