Powering a 2-W Max 8-Ohm Speaker

Thread Starter

strikethedream

Joined Jan 23, 2024
4
Hi, I'm currently a student and need some help. As much as possible, I would like to be guided and don't get too straight answers. I really want to understand this concept and learn. I have some background in Q-point biasing (like reading the lectures) but I'm still confused about it. What I understand is, it's sort of to adjust the wave up or down to meet certain conditions for the transistor. Also, the load lines still make me scratch my head.

Context:
I had a previous circuit schematic, tried it with the breadboard, didn't work because I can't seem to make the output center at 0V even with a coupling capacitor. The values I was kinda sure was the CE, anything beyond was just trial and error (I know, quite desperate and dumb); tried all sorts of topologies like push-pull (AB), darlington, and CC. And now? Just decided to start from scratch and literally try to learn on the go through no more trial and errors. So bear with me.

I need to power on the 8-ohm speaker at a max of 2W; it must deliver 1.5W. So far, the input signal would be a sinusoid at 1Khz @ 1Vpp. I chose Vcc to be 12V because I researched online that it's most optimal for this power rating; also, we shouldn't go past that point.

Steps Did:
So far, I chose to start with a CE amplifier because of its stable configuration using a voltage divider bias. We are told we have around 10K of impedance at the line level load, hence I assumed that 10K-ohm would be Rc.

I started from Rc = 10K, and moved down from designing the amplifier to have a gain of -10 because of 1Vpp approximately giving 10Vpp. Anything above the Vpp would clip the output because I assume it is due to Vswing = Vcc - Vcesat = 12 - 0.2 = 11.8? I'm still wrapping my head around this but I noticed that the output wave shifts up as gain increases and will clip at a certain point.

I then referred from G = -R3/R4=>R4 = -10K/-10 = 1K. From a tutorial online, and from me understanding KVL assuming Ic = Ie with B >= 100, Icsat = 12/(R3 + R4) = 1.09mA. From the graph of Vce and Ic, this is the region of saturation and Icq would be half of it = 545uA.

Vceq is chosen to be 6V. I would also like to know is Vceq always half of Vcc or are there any exceptions?

I learned that the current through R2 would be >= 10Ib. To be honest, I'd like to delve more into this concept but all I search online is it's recommended. Anyways R2 = B * R4 / 10= (300*1K/10) = 30K. I chose beta to be 300 because in LTSpice, the NPN model I'm using has Bf of 300. As far as I know, this is the maximum beta so I just used that and anything beyond 100 is negligible, are there any specific design choices on choosing a different beta?

R1 = VccR2 /Vb - R2
For Vbe, I looked through the datasheet and I believe it's 0.6V. So, I just made it 0.65V just to be the middle between 0.7V.
Vb = 0.65 + Ve =0.65 + 545.45u(1K) = 1.2
R1 = 12(30K)/1.2 - 30K = 270K

For the coupling capacitors, I used 1/2(pi)(lowest freq)(resistance seen by capacitor). Two things: If the frequency is fixed, will the lowest frequency just be the set frequency that is gonna be used, in this case, 1K? Is the resistance seen just Rth?

Honestly, the values I gained were quite weird as they really jacked up my wave forms, so I decided with 0.47uF for coupling even though I have no idea why this would work. I am familiar with I = C dv/dt (and we already did Laplace, ugh) but I don't have the time to see why the waves act at certain capacitor values.

I thought of putting additional capacitors like a bypass one, but until I find out how not to mess up my waves using capacitors, I wouldn't put it right now.

The CE now looks like this.

Waveform looks pretty good, 10 gain, and did what an inverting amplifier would do.

Problem:
However, when I now connect the 8-ohm load to the coupling capacitor, the power and voltage cross it is virtually non-existent. I assume because the impedance is too low, thus there is virtually no voltage to be absorbed available. I figured to amplify current more so:

I'm thinking of cascading an emitter follower as I researched about it and it does amplify current, I just don't know how to start designing it given it's now connected to my voltage divider. Every time I try to connect a coupling capacitor, with my previous iterations of circuit schematics, it ends up diminishing the speaker power and voltage significantly.

If cascading an emitter follower is not enough, I think a push-pull AB configuration would do? Though I suppose a darlington with emitter follower suffices.

Could you help me out trying to design this? Thank you so much.
 

MrChips

Joined Oct 2, 2009
30,824
Think resistors in combination. What happens when a low value resistor is put in parallel with a high value resistor?

The driver circuit must have an impedance lower than the load.
 

Thread Starter

strikethedream

Joined Jan 23, 2024
4
Hi, usually the equivalent resistance would be the lowest resistor since current prefers to go through that way, so the voltage across would be quite negligible. I noticed this effect from Req = 1/(1/Ra + 1/Rb) so if Ra >> Rb, then Req = Rb. Also, what do you mean by the impedance of the driver circuit, is it Rc?

I'm trying to find out some way to make sure that even with the 8-ohm impedance, the voltage across should still be large, so I figured the current was the problem. But now since you mention about the driver circuit impedance, I'd like to know more about it, care to?
 

MrChips

Joined Oct 2, 2009
30,824
Rc must be lower than 8 ohms which is going to blow the transistor.
What has your prof taught you so far about impedance matching?

You need common collector Class B or Class AB complementary push-pull output stage.
 

Thread Starter

strikethedream

Joined Jan 23, 2024
4
For impedance matching, I was taught that Pmax = Vrms^2 / R, so I applied it where R = 8

To gain around 2W max, I need a Vrms of 4V from a Vpeak of 11.3V, 5.65V as amplitude, or at least I assume to be amplitude since it's divided by 2.

Let me try the common collector class B. As far as I know, Class B's have their outputs on by half the cycle, efficient however some say it can cause distortion though I don't know how much of an effect it is. May I ask why Class B, despite the points I mentioned, over Class A which the output is on for all cycles? Also, we still haven't discussed the differences between classes of amplifiers (for some reason) so I'm not that familiar on which components are different across classes.

Anyways, with Vrms of 4V, in the CC, I need current of about 4 =8Ie => 500mA. Or should I use the amplitude as V? I still don't know whether to use rms or amplitude itself.

Now, this is where I get confused, to what can I do with this information? Since I imagine adding a resistor in parallel with the load at the emitter would change the current equations. So I would have two missing values, Ie and the new Ire.
 

MrChips

Joined Oct 2, 2009
30,824
You use voltage or current to calculate power into the load R.
For DC calculations it does not matter if you use RMS or peak since they are the same.

Class A amplifier will waste as much power as what goes into the load.
Class B push-pull puts all the power into the load. Class AB eliminates cross-over distortion.

You need to focus on the 8-ohm load, before worrying about the voltage or current. Common collector (emitter follower) is a current driver with unity voltage gain.
 

Thread Starter

strikethedream

Joined Jan 23, 2024
4
Thank you for this information. So, I tried to design an emitter follower, with Ie as 500mA to Pmax of 2W.

I decided to try a voltage divider to bias Vb to remain 5V much like in the CE just without Rc. Knowing my Ie is 500mA, it's approximately Ic too, since beta max is 300.

I can solve for R1 and R2 because I know that Ib = Ic / B = 500m/300 = 1.67m

Using kcl at Vb, (-5+2)/R1 +5/R2 + 1.67m = 0

I'll choose R2 to be 1K, eventually R2 equates to 1K as well.

My equations were correct so Vb is still around 5 while I'm now getting a minimum of 1.5W at the load.

I noticed this iteration of my circuit is better than my previous ones. However, I faced this same problem as before, when I cascaded the EF, the output of Q1 and input of Q2 became "DC" where Vpp is so low. Although I am getting the power I needed, I still need a pure AC output to make the speaker vibrate.

I presume I add another stage on the amplifier that will lengthen the Vpp (or is this the equivalent of increasing gain)? Would a push-pull amplifier after EF do this? Or rather what kind of amplifier would preserve power all the while maximizing gain?

EDIT 7:42 PM: This is the circuit as of now
 
Last edited:

crutschow

Joined Mar 14, 2008
34,470
That circuit puts a large value of DC current through the speaker, which will likely zap the speaker.
As has been said, you can't practically use an emitter-follower to get that much power into an 8 ohm load.
Use a push-pull output.
If you don't want to do what we suggest, then you are just wasting everyone's time that's trying to help.
 
Last edited:

BobTPH

Joined Jun 5, 2013
9,003
A push pull stage has a gain of less than 1. What does that tell you about getting 1.5W into 8 Ohms from a 1V p-p input?
 

Audioguru again

Joined Oct 21, 2019
6,710
Please look at the datasheet of the low power 2N3904 output transistor you are using.
Why are you overloading the 2N3904 output transistor? Its absolute maximum output current is only 200mA and it works poorly above 100mA. The datasheet shows that its beta can be as low as 30 at Ic of 100mA.

For 2W in an 8 ohms speaker you need 4VRMS across it which is +5.6V peak (+700mA) and -5.6V peak (-700mA).
Select a transistor that works well at 700mA but not as an emitter follower. Use a push-pull pair of transistors.
DO Not put DC current in a speaker. Use a coupling capacitor.
 

Audioguru again

Joined Oct 21, 2019
6,710
A class-AB amplifier output uses a complementary (NPN and PNP) pair of emitter-follower transistors biased so that they conduct a small idle current to avoid crossover distortion. Their biasing is usually produced with a similar pair of diodes or a voltage-setting transistor so that the idle current of the output transistors automatically follows temperature changes.

My circuit uses a bootstrap capacitor to increase the maximum output voltage swing and negative feedback to reduce distortion.
 

Attachments

MrAl

Joined Jun 17, 2014
11,496
Hi, I'm currently a student and need some help. As much as possible, I would like to be guided and don't get too straight answers. I really want to understand this concept and learn. I have some background in Q-point biasing (like reading the lectures) but I'm still confused about it. What I understand is, it's sort of to adjust the wave up or down to meet certain conditions for the transistor. Also, the load lines still make me scratch my head.

Context:
I had a previous circuit schematic, tried it with the breadboard, didn't work because I can't seem to make the output center at 0V even with a coupling capacitor. The values I was kinda sure was the CE, anything beyond was just trial and error (I know, quite desperate and dumb); tried all sorts of topologies like push-pull (AB), darlington, and CC. And now? Just decided to start from scratch and literally try to learn on the go through no more trial and errors. So bear with me.

I need to power on the 8-ohm speaker at a max of 2W; it must deliver 1.5W. So far, the input signal would be a sinusoid at 1Khz @ 1Vpp. I chose Vcc to be 12V because I researched online that it's most optimal for this power rating; also, we shouldn't go past that point.

Steps Did:
So far, I chose to start with a CE amplifier because of its stable configuration using a voltage divider bias. We are told we have around 10K of impedance at the line level load, hence I assumed that 10K-ohm would be Rc.

I started from Rc = 10K, and moved down from designing the amplifier to have a gain of -10 because of 1Vpp approximately giving 10Vpp. Anything above the Vpp would clip the output because I assume it is due to Vswing = Vcc - Vcesat = 12 - 0.2 = 11.8? I'm still wrapping my head around this but I noticed that the output wave shifts up as gain increases and will clip at a certain point.

I then referred from G = -R3/R4=>R4 = -10K/-10 = 1K. From a tutorial online, and from me understanding KVL assuming Ic = Ie with B >= 100, Icsat = 12/(R3 + R4) = 1.09mA. From the graph of Vce and Ic, this is the region of saturation and Icq would be half of it = 545uA.

Vceq is chosen to be 6V. I would also like to know is Vceq always half of Vcc or are there any exceptions?

I learned that the current through R2 would be >= 10Ib. To be honest, I'd like to delve more into this concept but all I search online is it's recommended. Anyways R2 = B * R4 / 10= (300*1K/10) = 30K. I chose beta to be 300 because in LTSpice, the NPN model I'm using has Bf of 300. As far as I know, this is the maximum beta so I just used that and anything beyond 100 is negligible, are there any specific design choices on choosing a different beta?

R1 = VccR2 /Vb - R2
For Vbe, I looked through the datasheet and I believe it's 0.6V. So, I just made it 0.65V just to be the middle between 0.7V.
Vb = 0.65 + Ve =0.65 + 545.45u(1K) = 1.2
R1 = 12(30K)/1.2 - 30K = 270K

For the coupling capacitors, I used 1/2(pi)(lowest freq)(resistance seen by capacitor). Two things: If the frequency is fixed, will the lowest frequency just be the set frequency that is gonna be used, in this case, 1K? Is the resistance seen just Rth?

Honestly, the values I gained were quite weird as they really jacked up my wave forms, so I decided with 0.47uF for coupling even though I have no idea why this would work. I am familiar with I = C dv/dt (and we already did Laplace, ugh) but I don't have the time to see why the waves act at certain capacitor values.

I thought of putting additional capacitors like a bypass one, but until I find out how not to mess up my waves using capacitors, I wouldn't put it right now.

The CE now looks like this.

Waveform looks pretty good, 10 gain, and did what an inverting amplifier would do.

Problem:
However, when I now connect the 8-ohm load to the coupling capacitor, the power and voltage cross it is virtually non-existent. I assume because the impedance is too low, thus there is virtually no voltage to be absorbed available. I figured to amplify current more so:

I'm thinking of cascading an emitter follower as I researched about it and it does amplify current, I just don't know how to start designing it given it's now connected to my voltage divider. Every time I try to connect a coupling capacitor, with my previous iterations of circuit schematics, it ends up diminishing the speaker power and voltage significantly.

If cascading an emitter follower is not enough, I think a push-pull AB configuration would do? Though I suppose a darlington with emitter follower suffices.

Could you help me out trying to design this? Thank you so much.
Hello,

You should upload your schematic here rather than link to it because external links often go bad and then readers cannot see the drawings.
I am uploading it here now.

One of the problems right off is the 10k resistor. You can not drive 8 Ohms with a 10k resistor.
When the transistor turns on, it puts the 10k resistor in series with the 8 Ohm output load. That means if you have a 12v power supply, the maximum voltage across the 8 Ohm load will be (from the voltage divider formula):
v8=12*8/10008=0.00959 which is approximately 10mv.
That of course is nothing even close to the require output voltage swing.
You need to lower the collector resistor value if you want more output voltage, but there are other considerations too, such as can the transistor take the current and the resulting power dissipation. You have to look into that.
 

Attachments

MrAl

Joined Jun 17, 2014
11,496
Why are there many new people here trying to heat their home with an antique class-A audio power amplifier?
Probably looking for simplicity over efficiency. Long time ago I made a headphone amplifier like that but that was very low power so not as big a deal.
 
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