Hello electronics professionals and enthusiasts! Software engineer mingling with breadboards and discrete components in his spare time here.

I'm trying to build a simple BJT amplifier for my electric guitar that would drive a 1 watt 8 ohm speaker. I'm stuck after multiple attempts and redesigns and would appreciate if you could point me in the right direction.

1st failed attempt - common emitter

At first I tried to get away with just a one transistor common emitter amplifier, but it was not capable of driving the speaker. I figured that the challenge of this particular project is:

  • A high source impedance. The output signal from the guitar is around 300 mV in amplitude and the internal pickup resistance is 10K-20K ohm.
  • A low load impedance. Got a cheap 8 ohm speaker from AliExpress.

2nd failed attempt - common emitter + common collector

I tried to add a common collector stage after the common emitter so it can function as a voltage buffer and provide a low output impedance. This actually worked and I could hear the sound coming out of the speaker, but because the collector resistor in the common collector stage was only 100 ohm it got crazy hot in a matter of seconds (~100 mA.) This amplifier just can't be used for more than 10 seconds.

3rd and current attempt

I ended up changing the approach and using a common emitter stage for the signal amplification, followed by the push-pull (class AB) stage to drive the speaker. The output signal looks fine without the load, but it gets attenuated and heavily distorted as soon as I connect the speaker.

My understanding is that it's due to the massive output decoupling capacitor (1mF.) From my calculations, it has to be that big, otherwise the C+R load acts as a high-pass filter and cuts off the audible frequencies.

How could I work around this and balance the decoupling and output distortion?

Am I doing something conceptually wrong?

Schematics and measurements

Operation point:

operation point

Common emitter amplifier stage:

CE amp stage

Push-pull amplifier stage:

Push-Pull amp stage

Output signal at load:

Output signal at load

UPDATE - Iteration 1

Thanks everybody for you comments and especially @andy-aka for your detailed answer, corrections and proposals. The biggest take-away for me is the method Andy described in his response - start from the load and work your way backwards, all the way to the source.

Here's what I changed so far:

  • Re-calculated the capacitor values taking into account new resistors and the 20Hz cutoff frequency.
  • Rotated the wrongly connected PNP BJT in the push-pull stage (facepalm)
  • Added a buffer / impedance adapter upstream to the Common Emitter amp

I don't physically have any FET transistors, so in order to achieve the high input impedance for the buffer stage I tried forming a Darlington pair from 2 BJTs.

Now we have a terrible, but working amplifier - the total gain is 2.2. Here's how every amp stage is affecting the gain:

Stage Vpp Gain
IN 580 mV
BUFFER 475 mV 0.82
PRE-AMP 1.42 V 2.99
POWER AMP 1.27 V 0.89
OUT 1.27 V TOTAL 2.2

iteration 1 - operation point iteration 1 - transient

Now I see 2 main problems that I need to solve:

  1. The gain is quite modest - I'd like to understand if there's a way to increase it without introducing more amp stages. There is a suggestion from @Audioguru to use higher-gain transistors that would be able to push more current through the speaker, but I'm not sure how to calculate the optimal voltage & current needed to drive the 8Ohm speaker at its full potential. I'm also a bit afraid of getting in the hundreds-on-milliamps region for my first amp circuit on the breadboard.
  2. The current in the voltage divider that biases the Common Emitter stage is quite big. In particular, R3 dissipates 468mW of power, while my cheap AliExpress resistors are only rated 1/4W. Is there a way to reduce the biasing current without killing the gain, or should I just order some beefier resistors?
  • 2
    \$\begingroup\$ C1 and C3 are certainly too low in capacitance for 400Hz. \$\endgroup\$ Commented Nov 11, 2023 at 17:10
  • 3
    \$\begingroup\$ Q2 (PNP) is backwards...its collector should go to GND, emitter to Q1's emitter. This CLASS AB attempt is going in the right direction compared with previous attempts. \$\endgroup\$
    – glen_geek
    Commented Nov 11, 2023 at 17:22
  • 2
    \$\begingroup\$ The small output transistors have a maximum allowed output current of only 200mA. Then the maximum peak output voltage (plus or minus) is 200mA x 8 ohms= 1.6V which is 1.13V RMS. Then the maximum continuous unclipping output power is only 0.16W which is like a loud earphone. \$\endgroup\$
    – Audioguru
    Commented Nov 11, 2023 at 18:48
  • 1
    \$\begingroup\$ 8 Ω speaker at its full potential (there used to be cheap(ish) 18" drivers. (And giant enclosures, just after the extinction of dinosaurs.)) there's the potential of that 9 V battery, too: give everyone a hint about both. More important: What is the purpose of designing yet another discreet BJT guitar amplifier? \$\endgroup\$
    – greybeard
    Commented Nov 13, 2023 at 12:11
  • 1
    \$\begingroup\$ The purpose of this exercise is to get the intellectual satisfaction of building a working amp with the most low-level and basic components possible. I'll look in all the suggestions and try make tweaks here and there both in SPICE and on the breadboard. I see opamps as the "next step" that I should explore once I got these basics right. So I'll also consider that option for next project, but here I'm focusing on basic components. If that makes any sense :D \$\endgroup\$
    – Alex
    Commented Nov 14, 2023 at 20:42

3 Answers 3


I'm trying to build a simple BJT amplifier for my electric guitar that would drive a 1 Watt 8 Ohm speaker

Firstly, fix the output stage of your push-pull amplifier: -

enter image description here

Next, the most significant problem area is driving the speaker. So, focusing-on the most likely circuit to be successful (the push-pull amplifier) and, considering what both output transistors have to do in terms of supplying current to the 8 ohm load, base your biasing is not up to the mark.

For instance, the 2N3904 has to supply maybe up to 100 mA: -

enter image description here

You need the transistor to turn-on to a low \$V_{CE}\$ voltage in order to maximize the output voltage swing but, the \$h_{FE}\$ available at this sort of drive strength is only 30. This means that you need up to 3 or 4 mA into the transistor base terminal to achieve the output requirements.

This then means that the biasing resistors (R1 = R2 = 10 kΩ) need to be passing at least 10 mA in the quiescent condition. Some folk would say that they should pass maybe 30 mA but, stick with 10 mA for now and try that. So, to get 10 mA from a 9 volt battery supply requires a resistance (R1 + R2) of 900 Ω. Currently you have 20,000 Ω.

Thus, R1 and R2 need to be 450 Ω each but, because you need a forward bias of about 1.4 volts across the diodes, the resistor values for R1 and R2 need to be about 380 Ω.

As a result of this, C1 and C2 need to big bigger in value to ensure that you don't attenuate the bass frequencies too much. I'd say you need a cut-off frequency of about 70 Hz for a regular 6-string electric guitar.

If the battery voltage is expected to droop to (say) 7 volts, then the resistors need to be more like 280 Ω

Compare that with your original 10 kΩ values and I think you'll agree that this sort of change will produce serious knock-on effects on the class A preamp formed around Q3. That transistor has a collector resistor (R5) of 3.3 kΩ and, when loaded with the new low values of R1 and R2, gain will be decimated. In other words, R5 and R6 need to be much lower in value such as R5 = 150 Ω and R6 = 25 Ω.

Then, there is a knock-on effect on Q3's bias resistors; you might get away by lowering R3 to about 3 kΩ and R4 to about 470 Ω.

But, after all that is done, your input impedance will be appalling low for a guitar and your guitar sound will be as dull as dishwater.

So, then you need an extra input stage to act as a high-impedance to low-impedance buffer. For guitar amplifiers this is best achieved using a JFET.

  • \$\begingroup\$ I appreciate the time and effort you put into this detailed answer, thank you! I used your feedback to iterate on my circuit and updated the post. \$\endgroup\$
    – Alex
    Commented Nov 11, 2023 at 20:07
  • 1
    \$\begingroup\$ @Alex R3 and R4 can be ten times bigger in value. That will reduce C4 by ten. In future, stick with no more circuit amendments because answers are given to your original question and I for one don't want to go through several iterations of retries on my answer. Instead, complete this question by following the advice in this link then, begin a new question with links to this question for reference. \$\endgroup\$
    – Andy aka
    Commented Nov 11, 2023 at 20:20
  • 1
    \$\begingroup\$ Guilty as charged - first time posting here. Marking the post as resolved and will keep it in mind for the future. Thanks again! \$\endgroup\$
    – Alex
    Commented Nov 11, 2023 at 20:23
  • 2
    \$\begingroup\$ What we are trying to avoid is the perpetual chameleon question or Change to question invalidates my answer, what to do?. In other words, play the game and remember that new questions are what this site is all about. \$\endgroup\$
    – Andy aka
    Commented Nov 11, 2023 at 20:25
  • \$\begingroup\$ I repeat. The 2N3904 and 2N3906 output transistors are VERY overloaded and will soon burn out.. Use TO-220 size power transistors that will not be overloaded and can be biased by resistors with higher resistance. Then the preamp can have more gain. \$\endgroup\$
    – Audioguru
    Commented Nov 12, 2023 at 0:56

start from the load and work your way backwards, all the way to the source.

Agreed, this is the best method.

If you want 1W into 8Ω (for a sinewave) then some basic calculations show you will need peak values for voltage and current of:
Voltage: +/-4V across the load, and
Current: +/-0.5A into the load.

The +/- signs are there to remind us that we need both polarities, positive and negative, for a sinewave waveform (sound or music). Just note that, for a sinewave, 2W peak power is required to deliver 1W average power.

Since you have a 9V power supply (a battery?) that leaves only 0.5V remaining as "headroom" for the amplifier to work correctly in each polarity - and that is assuming that the power supply is split perfectly 50/50, ie: that both the biasing and the big dc-blocking capacitor (between the load and amplifier) are perfect, such there is effectively +4.5V and -4.5V available to drive the power into the load. From experience, I know that the biasing and dc-blocking will not be perfect, and you will end up with less than +/-4.5V to drive the load.

So achieving 1W with low-distortion from a 9V battery into an 8Ω load will be challenging, so let's set a more realistic target of 0.5W into an 8Ω load - you will probably prefer to hear music at 0.5W at low distortion, rather than 1W at high distortion. For this power level we will need peak values for voltage and current of:
Voltage: +/-2.83V across the load, and
Current: +/-0.35A into the load.

Design Procedure

Step 1: Power Dissipation in Output Stage
Assuming this power is being delivered by a push-pull output stage biased at the mid-point of the supply voltage (Vcc=9V), then each output transistor will suffer a power dissipation of about 0.25W average (0.5W peak). Looking at the datasheet for the transistors you used (2N3904, 2N3906), the thermal impedance from junction to ambient is 200C/W (no heatsink):

enter image description here

This means the internal temperature of the transistor (the "junction") will be at 50C above ambient (0.25W * 200C/W = 50C). For a BJT this is OK, so let's proceed.

Step 2: Voltage gain
Your signal source is a guitar pickup, having an output amplitude of ~300mV (let's assume that is peak, ie: 600mV peak-to-peak) with an internal impedance of 10kΩ. So as to not load our signal source too much, let's set our amplifier input impedance to be 10 times that of the source impedance, so 100kΩ (even this is too much of a load for real hi-fi, but for the purposes of this discussion let's proceed with this for now). The voltage gain we need will be:
Vout peak = 2.83V Vin peak = 0.30V Av (gain) = 2.83 / 0.3 = 9.4, so quite modest.

It is interesting to see the power gain we seek:
Input power = 300mV peak into 100kΩ = 0.9 micro-watts peak.
Output power = 1W peak into 8Ω.
Power gain ~ 1.11 million - yes, over one million!

Which brings me to the point I wanted to make: what we are really amplifying here is power rather than voltage.

Step 3: DC blocking capacitor
The lowest frequency for a guitar is about 70Hz. Let's calculate the dc-blocking capacitor required to have a cut-off frequency of 70Hz with a resistor of 8Ω:

enter image description here
The cutoff frequency is given by:-

enter image description here

...from which we compute C as 284uF, so select an electrolytic capacitor of any value larger than this, say, 330uF 16V. At full power and low frequency, there will be an AC voltage across the capacitor since its "impedance" is not zero (from basic circuit theory, any component carrying current will have a voltage across it). We can say there is a voltage drop across this capacitor, which means the amplifier must supply both the load voltage and the voltage drop across the capacitor. Let's get an estimate of what this voltage drop will be, we can use the formula for the impedance of a capacitor:-

enter image description here

Plugging in our values of 70Hz and 330uF we get:
Zc = j6.9Ω yes, that is similar to the load resistance.

To deliver 0.5W to the 8Ω load at low frequency, the output voltage of the amplifier must swing higher than the load voltage to compensate for the voltage drop of the blocking capacitor. The amplifier has to push the required current (0.35A peak) through the series connection of the blocking capacitor (330uF) and the load (8Ω); using basic circuit theory for a complex load we can calculate this as follows:
Zload = 8Ω + j6.9Ω,
which has magnitude = Sqrt (8^2 + 6.9^2) = 10.56Ω
To calculate the voltage across this load we convert the load current peak to its RMS value, then multiply by this impedance: V = 0.35A * 0.707 * 10.56Ω = 2.61V rms.
Now we convert this back to peak we get: 3.7V peak.

So, if we wanted to have a low distortion frequency response down to 70Hz, the output of the amplifier needs swing 0.9V higher than the voltage at the load in order to overcome the voltage drop of the blocking capacitor. So you can hopefully see that we are getting closer to our hard limit of 4.5V peak that is available from our 9V battery.

Step 4: Base Current Required to Drive the Output Push-Pull Stage
The load current is 350mA peak (for 0.5W average into 8Ω). Assuming the output transistors are as you have indicated (2N3904 & 2N3906) the datasheet advises that current gain falls sharply, for both types, as Ic increases over 20mA, refer charts below (note that the vertical scale (Hfe) is logarithmic).

enter image description here

enter image description here

The datasheet states that at Ic=100mA Hfe = 30 (refer below). The graph above tells us that Hfe drops from 0.3 (normalised) at Ic=100mA, to less than 0.1 (normalised) at Ic=200mA. So Hfe at 200mA will be one-third what it was at 100mA, ie: 10. And 300mA is off the chart - so it will be even lower again. This suggests that these transistors may not be suitable - we ideally want Hfe to remain over 50 for the entire load range up to full-rated power.

enter image description here

However, for the purposes of this discussion let's persist with these transistors, and let's assume that Hfe=10 at Ic=350mA, so at peak output current we expect the base current to be:
Ibase(peak) = 350mA/10 = 35mA.

To source this current from a load resistor in the collector of the voltage amplifier BJT, then the resistor value will be quite low, about: 1.0V / 35mA = 28Ω.
With such a low resistance it will be difficult to design a C-E single BJT voltage amplifier with sufficient gain.

There are at least two solutions to this problem. One solution is to have an active load (constant current source) as the load of the C-E stage. The other solution is to apply a bootstrap. The circuit below shows the bootstrap method, but even this required an extra transistor (Q3) to ensure sufficient base drive to Q1 without excessive loading of the voltage amplifier stage (VAS).

0.5W Amplifier with Boot-Strap

enter image description here

Q1 & Q3 form a "Sziklai pair" and together with Q2 forms the push-pull putput stage of the amplifier. D2 and D4 provide some bias for the output stage.

Q3 ensures there is sufficient base current to drive Q1. Q4 performs the same function for Q2 as Q3 does for Q1 (in addition to being the voltage amplifier). C6 provides the bootstrap function which dramatically reduces the swing of current in R34 as the output voltage swings, which greatly increases the voltage gain at signal frequencies of the voltage amplifier formed by Q4.

Q5 (emitter-follower) buffers the input signal, and together with the boot-strap of C5, R27 & R2, greatly increases the input impedance of the amplifier. DC bias of the output is set to ~4.5V by the bias network R32, R1, & R25, as well as R3 which provides the feedback at DC.

Overall voltage gain is set by R3, R5 & C3. At signal frequencies, the impedance of C3 goes to zero, so the voltage gain is set by the ratio: Av = R3 / R5 = 1k /120 = 8.33.

Update 2023-11-13: Improved 0.5W Amplifier with Boot-Strap:

enter image description here

Adding one more BJT allows significant improvement in performance. Referring to the circuit above, one BJT was added to make the output stage fully symmetrical, ie: Sziklai pairs for both the top and bottom locations. For this application (very low supply rail voltage) the Sziklai pairs are preferred over Darlington, due to one Vbe drop rather than two from VAS output to the load. Bias via D2 & D4 remains unchanged.

For Q5, the voltage amplifier, this change removes the burden of supplying heavy base current, so the entire VAS can be re-designed to focus purely on it's primary role: voltage amplification. To that end, two other changes of note:

  1. Separate DC and AC feedback paths. DC feedback is set by R8, C3, & R3, and set the DC bias (at [Q1_Vout]) = 4.5V ie: the midpoint of VCC. AC feedback is provided by R9 acting into R3 & C3, and is taken directly from the load after the DC-block capacitor, C1. This allows the amplifier to compensate for voltage drop across C1, which extends the low-frequency response, and allows for a smaller value for C1. However, there is a limit to what can be achieved here, of course, primarily due to the limited output voltage swing available from a 9V supply.
  2. Bias network (R32, R1 & R25) changed. This was necessary to permit the AC feedback, and is quite simple: R3 & R8 had to be larger value to permit high value for R9 (low loading effect on output of Q6). So bias voltage at base of Q6 had to be moved down toward 0V.

Minor Changes:

  1. Added high-frequency gain roll-off, refer C8.
  2. C5 is larger, due to smaller values of R2 & R27 that were necessary due to changes in bias.

0.5W Amplifier with Active Load

Update 2023-Nov-13, 2200hrs

Here is the schematic of a design using an active load in the voltage amplifier stage (VAS).

enter image description here

Changes compared with the previous schematic (boot-strap):

  1. Boot-strap capacitor (C6) removed. This has been replaced with a new BJT configured as a constant current source of about 1mA: Q7, R7, D5, D6 & R10.
  2. Bias changed to "ampflied diode": Q8, R5, R6. This allows fine control over bias voltage (change R5), which will be necessary in any practical implementation to control the quiescent bias current in Q3 & Q4.
  3. Input signal is passed through a band-pass filter, about 20Hz to 100kHz, formed by C6, R21, R20, R13, & C8. This is to limit the frequency range of the signal reaching the amplifier input at Q6.
  4. Bias current of Q6 set by R24=2k2 (was 1k). [Q1_Vout] bias voltage is set to 3 * Vbe above base voltage of Q6 (R24, R12, R11).
  5. V7 sets the bias voltage now - this is just for convenience, if this design was be built then R18 must be adjusted to get the same voltage as V7.
  6. AC feedback network changed for a mid-band voltage gain of 9 (refer the "strange" value of R9), and some high-frequency roll-off (R19, C9). This gain was chosen so that clipping was avoided with input of 300mV peak, but power is 450mW, a little below the target value of 500mW.
  7. C1 was increased from 330uF to 470uF. This was to extend the low end of the frequency response.

Waveforms are presented at three frequencies:
mid-band = 1kHz
Low = 75Hz
High = 10kHz
Output power is nominally 450mW into 8Ω at mid-band.

The average power dissipation in Q1 & Q2 is about 270mW in all cases, so these should run at an acceptable temperature without a heatsink.

Images below: Actual output voltage is shown compared with ideal output, compare [Vload] with [Vin * 9]. Also shown is [Q1_Vout - Vload]; this is the voltage across C1, which is significant at 75Hz. This demonstrates the benefit of the separate feedback paths, AC & DC, which compensate for this voltage drop.

F=1kHz. Pload = 453mW enter image description here

F=75Hz. Pload = 475mW
enter image description here

F=10kHz. Pload = 420mW
enter image description here

1W Amplifier using Bridged Twin Amplifiers

Update 2023-Nov-25, 0800hrs

Here is a design that uses two identical amplifiers arranged as a bridged pair. The main benefit is that there is plenty of voltage swing available, so it can easily deliver 1W into 8ohms from a single 9V battery. In fact, it can deliver up to 4W to the load provided the output stage (and the battery!) can handle the extra current.

The other major benefit: the big, bulky (and perhaps sonically disagreeable) DC blocking capacitor between the amplifier and the load is eliminated.

Each amplifier is identical, and has the standard topology: a diff-pair input stage followed by VAS stage using an active load, followed by the Sziklai output stage voltage-follower as presented before. The diff pair simplifies the feedback network and makes it easier to configure two amplifiers in a bridge arrangement. The active load in the VAS stage was mentioned previously.

Image below: The overall schematic, note:

  1. The connection of the output of Amplifier 1 to the inverting input of Amplifier 2.
  2. The connection of the input of Amplifier 2 to 0V via a low-value resistor, rather than to the input signal.

enter image description here

Image below: Detail of one amplifier. enter image description here

Image below: Frequency response, -3dB points at 20Hz and 120kHz: enter image description here

Image below: Waveforms for 1kHz @1W into the load. Note that each output transistor now dissipates 0.5W which may be too high (double the dissipation of the previous designs), so I suggest either putting transistors in parallel to share the load current (which will require resistors in the emitter and base circuits to force good current sharing), or use transistors of higher power rating, such as BD139, BD140.

enter image description here

  • \$\begingroup\$ Considering the voltage budget, I'd try to compare '23/11/13/22 constant current source to a current mirror, probably built from more transistors than two to reduce wasted current. How does it compare to the bootstrap variant? \$\endgroup\$
    – greybeard
    Commented Nov 13, 2023 at 15:50
  • \$\begingroup\$ @greybeard Yes, I was considering using a Wilson (either 3 or 4 transistor variants), but then decided the voltage budget was too constraining and the dc-blocking capacitor has to go. I'm now considering a full-bridge amplifier (+ve and -ve output nodes) that can access the full 9V and avoid the dc-blocking capacitor altogether. Cheers. \$\endgroup\$ Commented Nov 13, 2023 at 23:22
  • \$\begingroup\$ I was thinking of e.g one transistor for input, two for output of twice \$I_i\$, and possibly emitter resistors for about .1 V. An emitter follower from "input collector" to the connected bases wouldn't hurt, but I can't see it help, as the mirror would be used as a low output drop "constant" current source. \$\endgroup\$
    – greybeard
    Commented Nov 13, 2023 at 23:30
  • 1
    \$\begingroup\$ Whoa, that's plenty of food for thought and ideas to try in your response, thank you very much, Fabio! \$\endgroup\$
    – Alex
    Commented Nov 14, 2023 at 20:52
  • 1
    \$\begingroup\$ @FabioBarone this is my first analog electronics project and I enjoy the challenge of going as low-level as possible, to understand / being able to measure what every single basic component in the circuit does. Once I get the basics covered and a working version of the discrete circuit - I'm going to actually build it. And then I'll most certainly venture into the op-amp world too. I do realize ICs are the most practical way to go if you want a good, working solution, but I feel like I didn't suffer enough doing it the hard way yet. \$\endgroup\$
    – Alex
    Commented Nov 14, 2023 at 21:57

Here's a way that simulates ok, presenting "just enough" input impedance for guitar (really prefer > 500k, but this will be around 200k input impedance, which won't hurt the tone more than this amplifier itself will).

This includes feedback, which will help reduce the distortion somewhat. Overall the strategy was to get a gain of 10, which is enough to make this amplifier clip when you dig in. R5 and R6 are sized larger than other posts recommend just because I'm more conservative about heating these small transistors during idle state.

@Fabio Barone has what looks like a very good performing amp and a lot of good advice, but this question was too much fun to not pitch in another idea. As you can see, this version is just starting to clip at 500mW output.

Power amp schematic

sim output

  • \$\begingroup\$ Why seriously overload the little low current 2N3904 at the output? \$\endgroup\$
    – Audioguru
    Commented Nov 13, 2023 at 0:18
  • \$\begingroup\$ Too much fun is never enough! :-) How do paste smilies in here? \$\endgroup\$ Commented Nov 13, 2023 at 1:40
  • \$\begingroup\$ @Audioguru because the OP apparently has them in the parts bin. It's like using a minivan to tow a camper over the pass. You may prefer using a large pickup with a V-8, but you need to do it tomorrow and you have a minivan. The main difference is if you burn out a 3904 it costs <$0.01. The risk is higher with a minivan. \$\endgroup\$ Commented Nov 13, 2023 at 4:35
  • \$\begingroup\$ The 2N3904 was designed for a max current of only 200mA where its hFE is very low. In this very high current circuit, it does not work and simply burns out. Maybe 4 in parallel will work and survive. \$\endgroup\$
    – Audioguru
    Commented Nov 13, 2023 at 22:01
  • 1
    \$\begingroup\$ (@FabioBarone en.wikipedia on Emoticons (Unicode block)) \$\endgroup\$
    – greybeard
    Commented Nov 14, 2023 at 11:14

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