# Audio Power Amplifer

Discussion in 'Homework Help' started by JasonL, Jun 19, 2013.

1. ### JasonL Thread Starter Active Member

Jul 1, 2011
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This isn't a homework problem, but I guess it is similar to one.
I'm looking at the circuit from http://www.electroboom.com/, the latest post. I also included the circuit in the pdf attached.

I want to know how this circuit works by calculating values in the circuit. This circuit is a power amplifier where Vin is an audio signal from a mp3 player. My work is found in the same attachment.

I'm not sure if I calculated the modes of operations correctly for the transistors but I think they look about right.

Besides the question if my work is correct, here are my other questions
1. How do I calculate the current going through the NPN_CE and PNP_EC
junction when they're not in cutoff?
2. How do I calculate the current or the power going into the speaker
assuming its something like an 8ohm speaker?
3. Is Vbe_on for transistors always 0.7V? I'm not sure how to read the
parameters on the datasheets. I didn't see 0.7V anywhere. I just assumed
it was 0.7V since the blog from http://www.electroboom.com/ said so. I'm
having a lot of trouble understanding datasheets.

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2. ### WBahn Moderator

Mar 31, 2012
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I don't have time to look at the circuit or address the other two, but I'll through out some quick points on this one.

No, the Vbe is not always 0.7V. This is a very, very useful approximation that is close enough for most circuit design and analysis purposes. Some people use 0.6V or 0.65V for small-signal transistors and 0.5V or 0.55V for power transistors. And, of course, we are talking about silicon NPN and PNP BJT transistors, here.

But all of these are approximations. In reality, as the base current increases so does the base-emitter voltage. If you plot the base current (y-axis) as a function of Vbe (x-axis) you would find that it starts out very, very small until you get up to somewhere in the vicinity of 0.5V to 0.7V. At that point you reach a "knee" in the curve and the current starts increasing dramatically as Vbe continues to increase.

In reality, this behavior and the existence of the "knee" is an artifact of plotting the current on a linear axis. If you were to plot it on a logarithmic axis you would find something close to a straight line. In fact, near room temperature, the base current increases by a factor of ten each time you increase the Vbe by about 60mV. So as you go from, say, 0.5V to 0.74V you would see an increase in current of about a factor of 10,000, say from 10μA to 100mA.

Note that there is nothing fundamental about this knee being in the vicinity of 0.7V. You can make a transistor that would exhibit the knee at much lower, or higher, Vbe voltages. You just have to scale the junction areas accordingly -- but you have to increase the junction area by a factor of ten to get that 60mV drop in Vbe (assuming you operate the junction near room temperature).

You see this in high power, high current diodes (which have a nearly identical exponential relationship between current and voltage) where you can purchse diodes that will conduct 1000A while having Vd of under 50mV. There is strong motivation for doing this when you figure that this diode still dissipates 50W, but a diode with a voltage drop of 0.7V would dissipate 700W.

There isn't as much of a driving motivation in BJT transistors, so making them in the 0.5V to 0.8V range for the typical current levels they are expected to see is an acceptable compromise. Basically, you balance off a number of design issues such as current density, electromigration, power density, and what comes typically comes out of the mix are junction areas that park you in that "normal" range.

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3. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
Thanks Bahn! This explanation especially helped a lot. I looked at the datasheet for the PNP KSB596YTU. There was a graph Vbe vs Ic. The knee was at around 0.6V. The graph wasn't Vbe vs Ib like you were describing but I can see how it would still look about the same if the gain between Ib and Ic was constant. I don't see a graph for Vbe vs Ib for the NPN D44H11TU datasheet though. My guess would be that figure 2 would be the graph that approximates Vbe_on because that's the threshold voltage before Ic receives any noticeable gain?

4. ### WBahn Moderator

Mar 31, 2012
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Usually you will see Ic vs Vbe because the graph is limited to operation in the linear region. I was allowing for operation in saturation, as well. The data sheets focus on different parameters when the device is in saturation.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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For which Vin voltage?
I will try to show you how we can determine approximate value of a max power that this circuit can deliver to the load (8ohm speaker).
As you can see on the diagram the base current for transistors is deliver by op amp output.
So with op amp we have two main limitation. First limitation is maximum positive or negative output voltage that op amp can produce on his output. So for example the positive saturation voltage for TS912 and RL = 100Ω, Vcc = 5V and is equal to Voh_max = 3.7V.
http://www.st.com/web/en/resource/technical/document/datasheet/CD00000501.pdf
So the positive saturation voltage is equal to 1.3V.
And if we assume this value, the V_load voltage will be equal to
VL = 6V - 1.3V - VbeQ1 ≈ 4V and NPN transistor emitter current is equal to Ie = 4V/8Ω = 0.5A.
And if we read from NPN transistor data sheet http://www.fairchildsemi.com/ds/D4/D44H11.pdf
Hfe ≈ 130 the base current will be equal to
Ib = 0.5A/131 = 3.8mA

So next we need to check if op amp is capable to provide such a base current (second limitation). But as we can see in data sheet the Output short-circuit current Isc = 65mA typical and 45mA min. So we don't need to worry about that.

Because our op amp will see ours load not just like a 8Ω resister but (Hfe+1)*RL ≈ 1KΩ (4.7V/3.8mA). And because of that, our op amp saturation voltage will be smaller then 1.3V. Probably smaller than 0.45V for 0.6KΩ load.

Now we can finally determine approximate value of a maximum power that we can to the load (8ohm speaker).

Pmax = VL_max^2/(2RL) = (6V - 0.45V - 0.7V)/(2*8Ω) = 4.85V^2/16Ω = 1.47W

Last edited: Jun 21, 2013
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6. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Can you give an example of a diode that will conduct 1000A while having Vd of under 50 mV; manufacturer and part no. or hyper link? I would be very interested in such a diode.

7. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0

I'm not sure how do you get RL and Voh_max = 3.7V. I think the saturation voltage was from doing Vcc - Voh_max, but I don't understand what's going on here. Was Vcc = 5V because it was ~80% of the supply voltage to the op amp?

I understood your explanation here, thanks a lot! So, Vo = 6V - 1.3V
and Ve = Vo -VbeQ1 where the current through the speaker is Ve/8Ω. I just don't know where the 1.3V comes from.
I know the current gain through Ie is (Hfe+1) but I don't know where that equation you posted comes from. Just looking at it, I don't think the units match because the left side seems to be in Ω and the right side is in Ω^2

Thanks again for the help! I appreciate it a lot.

8. ### WBahn Moderator

Mar 31, 2012
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A quick search of Digi-Key shows this as the best performer (in this regard) that they stock:

Microsemi LX2400ILG Vf_max=65mV @ 10A

I don't remember which site I was looking at when I ran across the monster diodes and I had to take four or five looks before I was able to convince myself that I was seeing what I was seeing. I saw a selection page that went to over 10,000A but the one I found a picture of was in the 600A range and it looked like an oversized hockey puck that was over six inches in diameter and something on the order of a half-inch thick. A single diode sold for just over \$100.

I've spent a few minutes seeing if I could find them again and haven't been able to. Mitsubishi Electric carries several lines that are definitely in the same vein, but are physically much smaller (about 2" in diameter) and with forward voltage drops at 1000A that are in the typical 500mV to 750mV range.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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I use TS912 data sheet to find positive saturation voltage for the op amp.
But data sheet don't show Voh for 6V so I use data for 5V Vcc.

OK, sorry for the confusion. The resistance seen by the op amp is equal to
R = (Hfe + 1)*RL or we can use Ohms law to find this resistance.
R = Voh/Ioh = 4.7V/3.8mA

Any more question or maybe you still have some doubts.

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10. ### JasonL Thread Starter Active Member

Jul 1, 2011
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Thanks, a bit more clear now.

I know now that you used the table from the datasheet to approximate the saturation voltage, but how do you arrive to the conclusion that it must be smaller than 1.3V and then approximate it to 0.45V for a 600Ω load?

This is a silly question but, how do you get P = V/2RL? I was always familar with the equation P=IV or P= (V^2)/R

11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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393
Well, I'm disappointed. I was hoping you had found a "miracle" junction diode technology. The LX2400ILG is not a diode. It's a power FET with supporting circuitry built into a single package for solar panel bypass:

We've been looking for such a low Vd diode. The very thin Schottky diodes shown in that same reference are the best true diodes available for solar panel bypass use. They need to be thin for lamination into the panel.

I don't think there is such a thing as an actual junction diode with Vd as low as 50 mV.

12. ### WBahn Moderator

Mar 31, 2012
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Always, always, ALWAYS check UNITS!!!

13. ### WBahn Moderator

Mar 31, 2012
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Boy, now I REALLY wish I could find that site again, just to find out what the fine print might say about what I was really looking at.

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Because our load resistance is larger then 100Ω.
Load resistance seen by the op amp output is close to (Hfe + 1)*8Ω = (131) *8Ω ≈ 1KΩ
So if load resistance is larger then 100Ω. The saturation voltage also will be smaller than 1.3V. Ans this is also shown in datasheet.

As you know the real power or active power for AC signal is equal to

$P =V_{rms} *I_{rms}$

or if we use Ohms law

$P = \frac{V_{rms}^2}{R}$

And we all know that

$V_{rms} = \frac{V_{peak}}{\sqrt{2}}$

And finally

$P = \frac{V_{rms}^2}{R}= \frac{\frac{V_{peak}}{\sqrt{2}}* \frac{V_{peak}}{\sqrt{2}}}{R}=\frac{\frac{Vp^2}{2}}{R} =\frac{{Vp}^2}{2R}$

So to find real power we need to use this equation

P = Vp^2/(2R) = 4.85V^2/16Ω = 1.47W

Last edited: Jun 21, 2013
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