# How to determine wattages on an applications note schematic?

Specifically, I refer to Fig 18 on Page 14 of this datasheet for dual supplies.

Really nice, really "simple" to use 2x40W power amp I want to try out. Unfortunately, I'm a newbie and their schematic doesn't list any wattages or voltages. I assume with a max +/- 35V supply, if I get 50V caps I should be safe; but what about the resistors?

I imagine that the 4.7ohm ones would need to be large, but how large? 5W? How do I determine that? What about the others?

I'm not quite so foolish I want to just throw 1/4Ws everywhere and see what blows, but I'm also not skilled enough to be able to fully figure out how much amperage is flowing where.

I've tried finding similar problems but maybe I'm not good at research, I turned up empty. Any help or direction is really appreciated since I also don't want to make an amp out of 10W resistors. Not yet anyway.

• So let me try this - R8 lies between one of the speaker's in/out lines. Doesn't that mean it can experience 2x35 = 70V max in worst case? If that's true, then 70/18k leads to ~ 0.27W of power, so a 1/2W would do for that. (I may be confused as I'm also really inexperienced especially with feedback)
– MJXS
Commented Jun 28, 2014 at 19:41
• Additionally the resistors in the mute/standby section are all ~10k order and hooked to Vs, thus 35V^2/10k = 0.1225 so I'd be okay with 1/4W.
– MJXS
Commented Jun 28, 2014 at 19:44
• Thanks everyone - I learned a lot of what you've mentioned in my undergrad but never had anyone show me how to practically apply it.
– MJXS
Commented Jun 28, 2014 at 20:09

Average (only slightly pathological) worst-case (20W average power out into 8 ohms at 20kHz continuous sine wave) power dissipation in the 4.7 ohm resistors will be around 120mW.

Edit: In case you're curious how I got this whilst avoiding any math mistakes, well it's Saturday, and I'm lazy, and calculations look a bit too much like what I do for paid work, so I just simulated it in SPICE. Of course I could do it manually or in Matlab etc. too, but this was easy.

The W markers are like probes that measure the instantaneous power dissipated in the components. The 17.888 peak voltage was picked to give 20W out (and verified by the marker):

VAMPL = $\sqrt{2} \times \sqrt{20W \times 8\Omega} = \sqrt{320}$ V.

Real dissipation playing any sensible signal source for listening will be a lot less.

You can use 1/4W for everything resistor-wise. 35V rating is okay for the capacitors since you should never get close to that for the supply (and they usually have a surge rating that allows you to slightly exceed the rating), but I'd use 50V and 105°C for some margin.

As a check - look at the parts on the sample PCB layout. If the resistors are not physically large, they're not high-power.

That doesn't mean that if you did something really perverted like feeding a 100kHz sine wave into the input (and cranking it up so you got 20W out despite the roll-off of the amplifier) you couldn't muck up those resistors by making them dissipate several watts. If you're nervous, 1W resistors are cheap and not that huge.

If you want to read more about the series RC on the outputs, they're called Zobel networks or Boucherot cells.

• Thank you VERY much, I've heard the "zobel" term but had no idea what it was
– MJXS
Commented Jun 28, 2014 at 20:07
• What's a good, preferably free program to use for simulation? I have PSPICE but it's so limited, or are models easy to find?
– MJXS
Commented Jun 28, 2014 at 21:55
• LTSpice, but it's a bit harder to use. Free. Commented Jun 28, 2014 at 22:19

The $R_7$-$C_8$ series impedance can be easily calculated: $$Z=R_7+Z_{C8}=R_7+\frac{1}{sC_8}=\frac{R_7C_8\cdot s + 1}{C_8\cdot s}$$ that means that when frequency is zero the impedance approaches $\infty$, i.e. $C_8$ is open, while as frequency increases the impedance lowers, and when frequency approaches $\infty$ you have $Z\rightarrow R_7$. Since $R_7=4.7\Omega$ and $V_{cc}=-V_{dd}=35V$ the maximum voltage across $Z$ would be less than $35V$. In the worst case, i.e. when $Z=R_7=4.7\Omega$ you have: $$P_Z=\frac{V_{CC}^2}{R_7}=\frac{1225}{4.7}W=260W$$ Wait, what?
Well, the point is that frequency is way beyond $\infty$, actually we can assume $f_{MAX}=20$kHz, that's audio signal for you. Since $s=j\cdot 2\pi\cdot f$ we just need to calculate: $$Z_{MIN}=Z(f=f_{MAX})= \dots = (4.7 - j\cdot 80)\Omega$$ Of course we need the absolute value: $$|Z_{MIN}|=\sqrt{4.7^2 + 80^2}=80\Omega\rightarrow I_Z=\frac{V_{CC}}{|Z_{MIN}|}=\dots=437mA$$ And finally: $$P_{MAX}=R_7\cdot I_Z^2 \approx 900mW$$

Just use something a little bigger and you are good to go, 1W is good enough.

Please note that 35V is the absolute maximum rating. You are better powering the amp with a lower voltage, check table 4, page 6: to get 40W @ 8$\Omega$ you just need $\pm26V$, that leads to $P_{MAX}\approx500mW$ meaning that $1W$ resistors are perfectly safe for you.

About the capacitors, all power caps need to withstand full $V_{CC}$, i.e. $C_{4,5,6,7}$. The same goes for $C_{8,9}$. $C_3$ must support a narrower swing, check table 5 and fig 17, page 13. Finally, $C_{1,2}$ can be "smaller" too since they only see line levels, that's some 2V. You can just buy a bunch of 35V caps and use them.

• Holy moly, thank you kindly. How do I tell what the other resistors should be? Are they just handling signal-level values?
– MJXS
Commented Jun 28, 2014 at 19:37
• yes, other resistors can be 1/4W Commented Jun 28, 2014 at 19:40
• @MJXS please note that I did a huge mistake, I am correcting it. Commented Jun 28, 2014 at 19:44
• Uh oh. Vladimir, you muffed the last line. Pmax represents total power assuming the capacitor impedance is a resistor. The power across the 4.7 ohm resistor is then PMAX x (4.7 / 80)^2, or .05, since the voltage across the resistor is (4.7 / 80) x VCC. Commented Jun 28, 2014 at 19:45
• As I remarked, yours is a wildly conservative number. I can pretty much guarantee that if you drive a 40 watt speaker (or even a 100 watt speaker) with +/- 26 volts at 20 kHz, the tweeter is toast. Commented Jun 28, 2014 at 20:07

Start with the 18k / 560 combo. Total power dissipated is V x V / R, or 25 x 25 / 18,560. That's .034 watts, so both can be 1/8 watt (25 volts is used by finding the peak voltage for a sine wave to produce 40 watts into 8 ohms. That's sqrt (2 x P x R)). For the 4.7 ohm units, consider that they are in series with .1 uF caps, and the impedance of a cap is 1 /( 2 x pi x f x C). For a C of .0000001 and a maximum frequency of, let's say 10 kHz, that's about 160 ohms. Voltage across the resistor (disregarding phase shifts) will be 25 x (4.7 / 164.7) or .7 volts. Power on the resistor is then V x V / R, or (.7 x .7 / 4.7) or .1 watts. So a 1/4 watt unit should do you.

The reason you can use a 10 KHz frequency is that, for real music, harmonic levels are way below the lower frequencies, and you don't need to deal with the entire 20 kHz.

• Well that's interesting, I'm redoing the math with 10kHz but our values are way different... Commented Jun 28, 2014 at 19:40
• Ok I'm just stupid. Commented Jun 28, 2014 at 19:43
• Why is the max voltage across the 18k 40V? Is the IN- hooked to ground through the 560? (Sorry I can't fully follow the current path, I don't understand the differential outputs)
– MJXS
Commented Jun 28, 2014 at 19:47
• Ack! Sorry, brain fart. I'll edit. Good catch. Commented Jun 28, 2014 at 19:49
• Haha, no worries, just means I actually know what's going on, thank you kindly for your help and time.
– MJXS
Commented Jun 28, 2014 at 19:52