Emitter resistor at output stage transistor.

[Mooly]

Lets take your circuit.

You're running -/+85 volt rails and lets say you are putting -/+70 volts across an 8 ohm load + the Zobel network (wrong side of inductor ?) and the 1uF cap across the 8 ohm load.

All that gives a peak current in each 0.1 ohm of around 11.25 amps over half a cycle (its a push pull output stage).

Convert peak to RMS and a full 360 degree sine wave across 0.1 ohms gives a dissipation of (8*8)*0.1 which is 6.4 watts dissipation.

Your only seeing half that in each resistor because of the push pull nature of the output stage, so current only flows for 180 degrees. So the RMS value is around 5.7A giving 3.2 watts dissipation.

 

[sajti]

Your calculations show that the four emitter resistors are overrated at 0.22 Ohm, 5W

Dissipation wise overrated. But small resistor will not able to handle the peak current, and can act as a fuse... So 5W is safe side.

So, using two 0.22 Ohm, 5W resistors in parallel will be more than sufficient

If You want to oversecure, yes.

Sajti

 

[edbarx]

My calculation erroneously omitted the fact that the 0.1 Ohm resistor is loaded only for 50% of every cycle. However, this error did not warrant the type of rude, manipulative and arrogant reply I received. The calculation is justified logically but with an easy to make error of omission. A simple reminder that that power had to be divided by two because of sharing between two resistors should have been enough.

For those who may wish to verify the model is a Wharfedale S1500. For some reason it failed, otherwise I would not have bothered to make such a drastic modification.

I am qualified to teach mathematics and physics up to A-Level. So, I should have a reasonable feeling for mathematics, and should also know how to apply Physics for calculations involving DC conditions. I can also use complex numbers for AC conditions, but these calculations have the tendency of becoming too complicated for those taking 'the from first principles' path.

For a 2 Ohm load, and, assuming an average power of 650W as per specifications:

Peak Load Current:
P = I^2*R/2
==> I = sqrt(2P/R)
==> I = sqrt(2*650/2)
==> I = 25.50A

Using this peak current to calculate the necessary power rating of the 0.1 Ohm resistor:
P = I^2*R/2
==> P = 25.50^2*0.1/2
==> P = 32.5W

Noting that this power is alternatively shared between two resistors:
P = 32.5/2
==> P = 16.25W

Using this maximum rating for a wirewound resistor makes the latter's temperature rise to values that are better avoided.

Let us explore the rating of the emitter resistor to find the over-rating factor for our calculation:
Current per Re resistor, Ire = 25.50/4
==> Ire = 6.375A peak
For a 0.22 Ohm resistance, the power is:
P = I^2*R/2
==> P = 6.375^2*0.22/2
==> P = 4.47W

Power Rating Factor, PRF = 5W/4.47W
==> PRF = 1.12 = 112%

For a 16.25W, this becomes:
16.25*1.12 = 18.2W

Therefore, two 10W resistors should be in agreement with the manufacturer's over-rating factor.

 

[sajti]

With 2ohm load, and +/-85V, Your amplifier will definitely go smoke. You need double amount of output devices at least, even MJL21193/94 instead of the weak Toshibas...

If it survive, the 0.1 ohm resistor will dissipate 6,5W with continous sinusoidal output.

Sajti

 

[edbarx]

With 2ohm load, and +/-85V, Your amplifier will definitely go smoke. You need double amount of output devices at least, even MJL21193/94 instead of the weak Toshibas...

If it survive, the 0.1 ohm resistor will dissipate 6,5W with continous sinusoidal output.

Sajti

Could you please, post the complete calculation quoting any formulae used and defining variables in full?

The 6.5W value is about half of what I am getting and I did two different calculations. I also took into account having two identical resistors sharing the power.

 

[sajti]

Could you please, post the complete calculation quoting any formulae used and defining variables in full?

650W/2ohms means 25,5A peak, which is equal 8.11A DC across a single 0,1ohm resistor. This gives the dissipation 6.5W.

Sajti

 

[edbarx]

650W/2ohms means 25,5A peak, which is equal 8.11A DC across a single 0,1ohm resistor. This gives the dissipation 6.5W.

Sajti

You are dividing the current by two before calculating the power which is I^2*R. The total current of 25.5A flows in both resistors for half the time, this means, to calculate the power dissipated one has to square 25.5A first, then divide the result by two. A square is not a linear relationship, implying the order of calculation steps makes a difference as in this case.

Let me demonstrate it symbolically:
Let the tatal current be: I
Let the resistance be: R

Case 1: dividing the current before the calculation.

P = (I/2)^2*R = I^2*R/4

Case 2: diving the current after the calculation:

P = I^2*R
Sharing this power we get:
p = I^2*R/2

The problem with your calculation is the fact that the 25.5A are flowing in both resistor but with time sharing. Your calculation is valid for the case of current sharing not time sharing.

 

[sajti]

I calculated the dissipation of ONE resistor. My way is to calculate the average current flowing across the resistor, and use it for the power calculation.

This is the maximum I can do for you. If you don't believe me, do it as you want. I just simply don't understand, why did you asked, if finally refuse to accept our help...

sajti

 

[edbarx]

Using 2/pi as a factor to find the DC equivalent current or voltage of an AC sinusoidal signal is incorrect. 2/pi is the factor obtained by finding the average of a sine wave as the following derivation shows:

Let our signal be:
a = Q*Sin(2*pi*f*t)

The area under the t-axis is given by integerating:
da = Q*Sin(2*pi*f*t) dt

area = Integral of Q*Sin(2*pi*f*t) dt evaluated between 0 and 1/(2f)
==> area = { Q/(2*pi*f) }*{ -Cos(2*pi*f*t) } evaluated between 0 and 1/(2f)
==> area = { -Q/(2*pi*f) }*[ (-1) - 1 ]
==> area = { -Q/(2*pi*f) }*( -2 )
==> area = Q/(pi*f)

Since we calculated the area for half a cycle the average is as follows:
Mean Q = { Q/(pi*f) }/(1/2f)
==> Mean Q = { Q/(pi*f) }*(2f/1)
==> Mean Q = Q*(2/pi)

As you can see, the factor 2/pi is an average for Q which can be a current or a voltage. For power dissipation in a resistor this value does not apply as power is proportional to the SQUARE of voltage or current at any instant. For the calculation of power dissipation, the RMS value is what makes sense.


Calculation of Power Energy dissipation in a resistor from first principles:

Power is defined as P = VI
For a resistor this becomes:
P = VI
==> P = (I/R)*I
==> P = I^2*R

Expressing this in differential form:
dE = I^2*R dt ( dE is an increment of Energy consumed. This holds as I^2*R is power and dt is a time increment)

For a sinusoidal current this becomes:
dE = { Im*Sin(2*pi*f*t) }^2*R dt ( Im is peak current )
==> E = Integral of {Im*Sin(2*pi*f*t)}^2*R dt evaluated between 0 and 1/(2f)
==> E = Integral of {(1/2)*(Im)^2* [ 1 - Cos(2*2*pi*f*t) ]*R dt evaluated between 0 and 1/(2f)
==> E = {(1/2)*R*(Im)^2*[ t - ((1/(4*pi*f))*Sin(4*pi*f*t) ] evaluated between 0 and 1/(2f)
==> E = (R*(Im)^2/2) * [ ((1/2f) - 0) - (0 - 0) ]
==> E = (R*(Im)^2/2) * (1/(2f))
==> Prms = (R*(Im)^2/2)*(1/(2f))/(1/(2f))
==> Prms = R*(Im)^2/2

Writing this in the form of r*i^2:
Prms = R*(Im/sqrt(2))^2

Thus, the first factor is 2/pi and the second is 1/sqrt(2). The values of these are 0.63662 and 0.70711 respectively.

 

[PRR]

You don't need pie to figure the emitter resistors, if you know the load power.

Say 100 Watts in 8 Ohms. Say a single 0.33r resistor in the amplifier. Obviously the dissipation in the 0.33r is 0.33/8 of 100 Watts, or 3.75 Watts.

Since there are two 0.33r resistors conducting alternately, your factor of 2 comes in, 1.88 Watts.

If you are going to over-drive to Square waves, for perfectly stiff supply, you double the Sine values.

You must add the idle power. Say 100mA. This amounts to 0.003,3 Watts.