Similar questions have been asked before, however they don't really answer what I'm actually looking for so I asked a new question instead.

TL;DR Why does the circuit below cause noise at the output?

I am learning about basic amplifier circuits, and after building a few CE and CCs, I decided to try and reinvent the wheel by myself in order to confirm my understanding. So, I started from the basics and thought about the following: I have an input signal which I take from my laptop's headphone jack. I set it to no more than 100mV peak-to-peak. I then build a common emitter amplifier circuit in the following way


simulate this circuit – Schematic created using CircuitLab

Notice how I have connected the input "in series" with the base, as opposed the usual configuration where you would put the input on the left side of R1 and R2, connect sleeve to ground and connect tip between R1 and R2 via a coupling capacitor.

Now, when I connect this in the usual configuration with a coupling capacitor, I get a nice clean amplified sine wave at Vout. If my calculations are correct, given the conditions above, I should get the same result with the above circuit as well - Vin sits on top of the DC bias and gets amplified. The only unknown variable here is my laptop's output impedance, but I assume it is low enough to not matter since it can drive headphones, earphones, etc.

Now, when I measure Vout with respect to ground, I get a sine wave that has a lot of high-frequency noise (small zig-zags) in it. If I plug an earphone, I hear a slight hiss and buzz with the sine wave - it is very similar to how guitar amps buzz when you short the input jack with your fingers, and I get the same sound when I do it on this circuit as well.

My question is, why does the noise show up? I suspect that it is interference from whatever is around me, including myself, however I do not understand how it gets picked up and why it doesn't get picked up in the "correct" configuration. If that suspicion is incorrect, please do inform me. Furthermore, is there a use-case for this kind of configuration, or is it unused because of the noise and/or other reasons?


Thanks to Andy aka's comment and Circuit fantasist's answer, I thought of what I think models this situation in the following way.

Let's say that we pick the voltage at the amplifier supply (9V)'s ground when it is not connected to the input ground as a reference and define it to be zero. Then, the output of the laptop can be written as

$$ \begin{aligned} \text{Tip} &= g(t) + s(t) \\ \text{Sleeve} &= g(t) \end{aligned} $$

where \$g(t)\$ is the difference between the supply's ground and the laptop's ground. Therefore the potential difference on the laptop's output is \$s(t)\$ which is the clean sinusoid. On the other hand, the voltage at the supply's terminals is 9V and 0V respectively.

Now, when we connect this in the configuration above, the potential at the transistor's base will be

$$ V = V_{DC} + s(t) + g(t) $$

where \$V_{DC}\$ is the bias voltage. If the amplifier has gain \$G\$ then the output becomes

$$ V_{Out} = G[s(t) + g(t)] $$

In other words - it amplifies the difference in potential between the amplifier's ground and the input's ground. On the other hand, if we connect the input in the usual configuration, the two grounds become equal, but the voltage is still summed and so the result is an output that only amplifies \$s(t)\$.

Finally, I also asked whether connecting the input in series ever makes sense. The answer to this, I think, is no, because in order to get a clean base voltage, you have to construct

$$ V = V_{DC} + [ s(t) + g(t) ] - g(t) $$

which in my knowledge would not be (practically) possible.

If the above is indeed correct, I will post it as an answer.


I think, is no, because in order to get a clean base voltage, you have to construct

$$ > V = V_{DC} + [ s(t) + g(t) ] - g(t) > $$ which in my knowledge would not be (practically) possible.

This is indeed possible - see Andy aka's answer.

I'm marking Andy's answer as accepted, because it gives the most important piece of information - what the noise source is and how to avoid it. However I think Circuit fantasist's answer also gives additional insight.

  • \$\begingroup\$ Did you try removing the power cord from your laptop? \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2020 at 16:56
  • \$\begingroup\$ @Andyaka I did now and the hum disappeared! This is interesting, but I still don't quite understand what is happening. \$\endgroup\$ Commented Apr 18, 2020 at 22:42
  • 1
    \$\begingroup\$ Sophisticated thoughts in EDIT... Let's consider the two cases when grounds are joined. In the case shown in the picture above, the output of the voltage divider (VR2) will be short connected; as a result, there woun't be bias voltage. If you swap the sleeve and tip, the transistor stage input will be short connected; now there woun't be any input voltage. Note the two grounds can be connected throught the mains neutral wire if both power supplies are grounded. \$\endgroup\$ Commented Apr 19, 2020 at 5:33
  • \$\begingroup\$ Did removing the power cord totally fix things or was there still some residual problem @Stealthmate ? \$\endgroup\$
    – Andy aka
    Commented Apr 19, 2020 at 8:18

3 Answers 3

Did you try removing the power cord from your laptop? – Andy aka

@Andyaka I did now and the hum disappeared! This is interesting, but I still don't quite understand what is happening. – Stealthmate

Try adding a capacitor here: -

enter image description here


The power supply to your Laptop will have seriously significant artefacts of mains hum on the DC output. Potentially many tens of volts p-p but, it is sourced from a circa 1nF capacitor inside the power supply so it isn't a shock-hazard. That capacitor is there (more than likely) to kill-off high frequency switching noise from appearing on the DC output but, the spin-off is that low frequency rectified AC mains hum appears. This normally isn't a problem but, when you use the jack output from the Laptop connected like you do, that hum will still be there and, the solution is to short it to ground via a capacitor as shown in red.

  • \$\begingroup\$ Removing the power cord fixed the hum, although I could still see slight noise on the oscilloscope. On the other hand, after adding the capacitor as you said, the noise disappeared completely. I suppose this capacitor acts as a "subtractor" of sorts - it removes the difference between input ground and supply ground, leaving only the clean signal. Thanks! \$\endgroup\$ Commented Apr 19, 2020 at 14:54
  • 1
    \$\begingroup\$ @Stealthmate Not a subtractor just an "over-powering" attenuator: If the source is 10 volts p-p and is fed through a 1 nF capacitor to a grounded 100 nF capacitor, the attenuation at the node where the two capacitors meet is close to 100:1 and the resultant interference becomes 0.1 V p-p. \$\endgroup\$
    – Andy aka
    Commented Apr 19, 2020 at 14:57
  • \$\begingroup\$ You mean there's essentially a path Vs - C1 - C2 - Gnd, where Vs is the noise, C1 is 1nF, C2 is 100nF and Gnd is the circuit ground, right? Calculating this indeed gives Vs/100 at the midpoint. \$\endgroup\$ Commented Apr 19, 2020 at 15:13
  • \$\begingroup\$ @Stealthmate yes, that's the path and that's the attenuation. \$\endgroup\$
    – Andy aka
    Commented Apr 19, 2020 at 15:14
  • \$\begingroup\$ @Stealthmate I probably would put a 1 kohm resistor across the sleeve and tip connections of the jack output in case there was an internal output capacitor in the laptop that might upset biasing by blocking DC. \$\endgroup\$
    – Andy aka
    Commented Apr 19, 2020 at 15:27

Your circuit solution is correct. Moreover, it is original and shows a good understanding of the bias idea.

Biasing simply means adding constant voltage to the input varying voltage; so it is implemented by a voltage summer. The possibly simplest summer is the series voltage summer.

Here, the R1-R2 voltage divider acts as a grounded bias voltage source that produces single-ended bias voltage VR2. The input voltage source (the laptop output) is floating and connected in series to the bias voltage source; so its voltage is added to the bias voltage as needed. It has to be galvanic and with relatively low output impedance.

There is no need of any blocking capacitors but note the input bias current will flow through the laptop output. Maybe this can cause some problems.

Since the input voltage source is floating, the two grounds must be separated.

  • \$\begingroup\$ Since the input voltage source is floating, the two grounds must be separated. <- This part is what I don't really understand fully I guess. However, after thinking about it some more, I think I developed a theory about it. Could you check my edit after this and see if it's correct? Thanks! \$\endgroup\$ Commented Apr 18, 2020 at 23:04

There are two unnecessary "noise" sources here.

One is the resistive (Johnson) noise source of R1 and R2 in parallel. In AC analysis terms, 9V and GND are both 0V (AC) putting the resistors in parallel. This will add a low level of white noise onto your output - this is not your problem, but it's unnecessary.

The other is not really noise but instability - coupling from the output side into either or both of:

  • the 9V supply
  • the high impedance node where R1 and R2 meet.

Both of these nodes should be decoupled with a nice fat capacitor to GND. 0.1uF will probably do, but for audio, a larger decoupling capacitor (10 to 100uF) is also recommended across the supply.

As a bonus, the second decoupling cap will also eliminate (at least, greatly attenuate) the Johnson noise from R1 and R2.


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