# CE Amplifier problem

I am new to learning about electricity and I was trying to make a common emitter amplifier.

I did some research and I tried to make the circuit. The circuit amplified the signal but the signal was not clear and was cutting out. The signal I am trying to amplify is audio from an iPad headphone jack. The circuit diagram is below.

Does anyone know why this is happening and how I can fix this? Thanks.

simulate this circuit – Schematic created using CircuitLab

(Vcc = 9 V)

• Because the capacitor bypass R3, the gain is R3/re which is very high and may cause clipping. You can remove C3 or split R4 into 2 resistors and bypass one. Dec 16, 2023 at 20:36
• R3/Re. what is re? Dec 16, 2023 at 20:39
• @MarcusMarchese your schematic contains a resistor labeled "Re" Dec 16, 2023 at 21:48
• @MarcusMarchese I get where the input is coming from -- a headphone jack designed to drive a headphone. But what are you listening to? Is this some speaker you have? Or what, exactly? Details matter. Dec 17, 2023 at 1:13
• It is a youtube video from the Ipad that I plug the cable into and I am trying to amplify that signal. I have the amplifier connected to a 1 watt speaker Dec 17, 2023 at 1:49

# The beginning

I added some elements (input source, load and power supply) necessary to bring your circuit to life in CircuitLab...

simulate this circuit – Schematic created using CircuitLab

... and I saw it working.

After you expressed your desire to understand the ideas in this famous circuit (which rarely happens here), I decided to write this amazing story on Christmas Eve as an alternative to boring textbook explanations. This is how I want to show that handling circuits is not only a craft but can also be an art.

# How we understand circuits

It would be great if we could get ready knowledge directly from some very reputable sources. If so, I would now answer you with only a few paragraphs of concise laconic text taken from such sources and repeatedly relayed. However, Nature has decided that we should do it in a more difficult way - slowly and painfully, making mistakes and learning from our experience.

To really understand a circuit, you need to know not only what and how it was done, but also why it was done. The best way to do it practically is to start from the very beginning, building the circuit step-by-step and exploring it at each step. Here is a possible scenario for making such an amplifier.

# What an amplification is

There is no such thing as power amplification in the literal sense of the word; rather it is power regulation (control). This is not proven, it is a postulate. In such an "amplifier", a low power source controls a high power source, and this creates the illusion of power amplification. Let's see the possible ways in which we can make an amplifier following this recipe.

# Building an amplifier

## Controlled source

If we could control the (output) voltage of the power source by another smaller (input) voltage, it would be the simplest amplifier. In the simulation below, I (like the input voltage) have imitated it by sweeping the supply voltage Vs from 0 to 10 V. Run DC Sweep Simulation to see it.

simulate this circuit

Voltage-supplied load: If we could control the load resistance, it would be an even more exotic "amplifier" (so, for example, by switching the heaters, the power dissipation in electric stoves is regulated). I have imitated it by sweeping the load resistance RL from 100 Ω to 1 kΩ (run DC Sweep Simulation to see it).

simulate this circuit

However, we see that the relationship between current and resistance is non-linear.

Current-supplied load: Of course, we can solve this problem if we supply the variable resistor with a current source.

simulate this circuit

## Controlled resistor

But in most cases we can neither control the source nor the load and then we are forced to insert in series an additional regulating element. In the simplest case it can be a variable resistor Rvar. Its role is to dissipate excess power (voltage VRvar) from the maximum (Vs) to the minimum (0 V). Thus, the load voltage VL changes from the minimum (0 V) to the maximum (Vs).

Indeed, this "discarding technique" is a rather silly way of regulation ("amplification"), but all analog electronics is based on it (in pulse electronics, they do it smarter).

Floating load: If Rvar is grounded, the load RL will be "floating" which creates problems when cascading the amplifier stages (the input of the next stage will be grounded).

simulate this circuit

The other trouble is that the load voltage VL depends non-linearly on the varying resistance Rvar.

Grounded load: But we easily guess to apply a little trick - instead of the voltage across the load, take its complement to the supply voltage. This means to replace the floating load with a constant resistor R, and to connect the grounded load in parallel with the variable resistor. Note that it has to have high resistance (in the schematic below, the voltmeter Vout serves as such a "voltage load").

simulate this circuit

The graph is vertically flipped (the "amplifier stage" is inverting).

## Controlled current source

So far, we have manually controlled the resistance of the regulating element to mimic an amplifier. In a real amplifier, however, this must be done by the input voltage. So we need voltage-controlled resistors.

For some reason beyond my understanding, electronic regulating elements (transistors, tubes, etc.) behave like non-linear current-stabilizing "resistors", and their IV (output) curve is an almost horizontal line what is the curve of a current source. That is why they model transistors as current sources even though they do not produce but consume power. Let's take such an "ideal transistor" (VCCS) from the CircuitLab library to make a real voltage amplifier.

simulate this circuit

As you can see, its transfer characteristic is linear.

## Unbiased transistor

Now it remains only to replace the "ideal" transistor with a real one (Q) to get a real amplifier. I have adjusted the input voltage Vin applied to the base-emitter junction so that to set the output voltage Vout equal to half of the supply voltage. You can change it carefully (because the current vigorously changes) by opening the Vin parameters window and setting the desired value.

simulate this circuit

When sweeping Vin, we see something unpleasant - the amplifier characteristic is non-linear. What is even more unpleasant is that it does not start from 0 V, but somewhere from 600 mV.

## Biased transistor

Input "lifting": So we have to somehow jump over the dead zone of several hundred mV at the beginning. Practically this means adding a bias voltage to the input voltage. The simplest way to do it in the conceptual schematic below, is by connecting a constant voltage source Vbias1 in series with the input source.

simulate this circuit

As a result, when sweeping Vin from -10 mV to 10 mV, it appears "lifted" with 641.5 mV at the base.

Output "lowering": The next thing we need to do is to drop the output (collector) voltage by inserting a 5 V bias voltage source Vbias2 between the output and the load.

As a result, when sweeping Vin from -10 mV to 10 mV, the collector voltage appears "lowered" with 5 V at the load and wiggles around ground.

## Negative feedback

The time has come to introduce the greatest (and not only circuit) concept called by the prosaic name negative feedback (in the past, it was called by the more figurative name "emitter degeneration").

Activated: To implement it, we only need to insert a resistor Re between the emitter and ground. Thus, we force the transistor to copy the input voltage across this resistor, and the voltage VRc across the collector resistor Rc is proportional to VRe and accordingly, to Vin. As a result, VRc/Vin = Rc/Re.

simulate this circuit

The gain decreases but becomes stable; linearity is improved.

Deactivated: The trick above was that the emitter voltage was not constant but was "moving" along with the input voltage. If we fix it, the "magic" of negative feedback will disappear, and the circuit will revert back to schematic 8. We can test it by measuring VRe and connecting a constant voltage source of the same voltage in parallel with Re.

simulate this circuit

Gain increases and linearity worsens.

So the conclusion is: if we want a low but stable gain, we disconnect Ve; if we want high gain, we connect it. We will use this trick in the AC amplifier below.

## AC amplifier

Circuit diagram: In real circuits it is not convenient to use (floating) voltage sources; we can replace them with capacitors. We just need to provide circuits for charging/discharging them - C1 from the voltage divider R1-R2 through the input source (1 kHz sine), and C2 from the output through the load RL.

simulate this circuit

Operation: The behavior of the circuit already depends on the frequency of the input signal - when it is low, it is represented by schematic 9; when the frequency is high - by schematic 10. Thus we provide stability for the DC mode and amplification for the AC mode.

The previous stages were DC amplifiers. Their behaviour did not depend on time, and therefore we examined them with DC Sweep Simulation.

This stage is an AC amplifier whose behavior depends on time (frequency). So we have to explore it through the Time-Domain Simulation.

# Why not both voltage and power amplification?

From your comments I see that "you want to connect a 1 watt speaker". This means a low-resistance (e.g., 8 Ω) load that will consume significant current. At the same time, you also want significant voltage gain. Unfortunately, it turns out that both are difficult to achieve at the same time, and they connect (cascade) two amplifier stages one after the other - a voltage amplifier followed by a power amplifier (buffer). Why?

Current-driven load: The problem is that your load (no matter floating or grounded) is connected to the collector of the transistor which acts as a current source creating a voltage drop across the load. So to get significant voltage gain, the load must be high resistance but it is not. It turns out that this stage alone won't do it.

Voltage-driven load: If you connect your load to the emitter of the transistor (emitter follower), there will be enough current but no voltage gain because this stage just copies the input voltage.

• @Marcus Marchese, I roughly animated your schematic. I want to ask you if you understand it (what each element is for). If not, are you willing to understand it or just interested in making it work? Dec 16, 2023 at 21:02
• I dont understand the graph or what is attached to the output. I am intrested in understanding. I have been trying to build a CE amplifier with a voltage divider method of biasing for a couple months and can't figure out how to. Dec 16, 2023 at 23:57
• @Circuitfantasist I bet the load Rl does not match a 1W speaker. Dec 17, 2023 at 11:27
• @Circuitfantasist that would be amazing. Thank you so much. Dec 17, 2023 at 17:59
• Thank you so much for the explanation I have been busy with school but I will defiantly be able to read it Thursday night. This is a very big help and I am very thankful Dec 19, 2023 at 23:04

This is because of very high gain caused by Rc/re ratio. Removing Ce simplifies gain equation to Rc/Re, stabilizing gain by making it dependent only on external resistors rather than internal transistor resistance.

• Welcome to SE EE! Good answer... By the way, what is "re" and what "internal transistor resistance"? Are they something real? Dec 16, 2023 at 21:23
• When I remove Ce the sound output is very quiet Dec 17, 2023 at 0:22
• @Marcus M. With Ce the gain is G=-gmRc=-Rc/re (re is an abbreviation for 1/gm - it is NOT a real resistance!). Withot Ce you have signal feeedback and the gain is much smaller: A=-gmRc/(1+gmRe).
– LvW
Dec 17, 2023 at 10:27

Apart from the problem of overdriving the amplifier you also have the problem of overloading the amplifier.

In a comment you state that you are connecting the amplifier to a 1 W speaker. Generally a speaker like this will have an impedance somewhere between 4$$\\Omega\$$ and 32$$\\Omega\$$. Your amplifier will not be able to drive this efficiently as it has a much higher output impedance, roughly $$\R_C\$$, 1200$$\\Omega\$$ (in simulation I get 1.13k).

So how much voltage does that allow the speaker to get? The load forms a voltage divider with the output impedance, imagine your 9 V supply connected to 1200$$\\Omega\$$ in series with say 16$$\\Omega\$$. The voltage across the 16$$\\Omega\$$ would be: $$9V\times \frac{16\Omega}{1200\Omega+16\Omega} = 118mV$$ which gives you less than a milliwatt of power.

You can see this in the AC load line, in the first image the load is 100k and the AC load line (green) extends above and below the Q point quite a bit.

In the second image with a 16$$\\Omega\$$ load the AC load line barely gets above the Q point, there's no room for the voltage to swing very far.

So even if you solve the overdrive problem by reducing the gain (removing emitter bypass cap) or attenuating the input signal you're still not going to get a useful output into a typical speaker. A low impedance speaker needs to be driven by an amp with a low impedance output, typically emitter followers (common collector) in a push-pull arrangement. A common emitter amplifier such as this is meant to provide voltage gain into a high impedance load.

• Thanks so much for this. Is there a website or a book or a youtube video that explains how to come up with the correct values for the resistors and capacitors that you know of. Because I have been trying to learn how to build an amplifier for months and can't figure it out. Dec 17, 2023 at 3:40
• @MarcusMarchese There are lots of books and websites and videos explaining it, shouldn't be hard to find some. You have to understand that there are different types of amplifiers though, for audio there are voltage amplifiers, such as the one you're asking about, and there are power amplifiers. The voltage amplifier would be used to amplify a low level signal to get enough voltage to drive a power amplifier which then drives the speaker. Google Class A/B amplifier circuit, that should get you headed in the right direction. Dec 17, 2023 at 4:03
• Thank you I will defiantly do more research Dec 17, 2023 at 17:55