Can somebody please check my bias/heatsink calculations? - diyAudio
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Old 31st March 2005, 01:25 AM   #1
spence is offline spence  United States
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Default Can somebody please check my bias/heatsink calculations?

Hello all!

I have been reading this forum for quite some time but have not really posted anything of importance, because I had never tried a project before. So I finally did... here are the details:

Subject of Modification: B&K ST 125.2 Power Amp

Mods:
1) Increased power supply capatance from 15,000 uF per rail to 45,000 uF per rail. Added low value filtering caps in parallel. Result: much tighter bass, seems to have much more depth of soundstage

Ok... here is the question regarding the mod I have just completed, increasing the quiescent current in the output MOSFETs.

Output MOSFETs: Hitachi SK1058 / SJ162, Emitter resistors: 0.47R

1) I measured the initial voltage across emitters after amp warmed up (2 hours):

35.6 mV / 0.47R = 75.7 mA Bias per MOSFET.

2) I increased the bias using the trim pot is small segments (1 hour between each segment) until I reached a value which made the heatsink ~60C. This turned out to be:

60.8 mV / 0.47R = 129.4 mA of bias per MOSFET

3) Checked DC offset, was ~ 5 mV per channel.. .OK

4) Then I calculated the approximate junction temperature of the MOSFETs using this method, and assumptions:

a) Rail voltage = 64V (measured),

64V * 0.1294 A = 8.3 W Dissipated per MOSFET

b) Thermal resistance Junction to Case = 1.7 C/W
Thermal resistance Silicon Pad to Heatsink = 0.7 C/W
(this is an estimate, from Rod Elliots Articles)
So total thermal resistance heatsink to Junction = 2.4 C/W

c) At 8.3 W Dissipation,
Temp Increase from Heatsink to Junction =
2.4 C/W * 8.3 W = 19.92 C

d) Therefore if the heatsink reaches 60C, the Junction of the MOSFET is 60C + 19.92 = ~80C

So... from the MOSFET data sheet Power vs. Temp Derating graph, at 80C, max power dissip ~ 50W.... I guess this means that I'm ok.

Could you guys and gals (being the experts) please take a look at these calcs to make sure I did not miss anything important. Also, with the higher bias current, the amp sounds FANTASTIC, do you think that I can use these calcs to push the current even higher, or is 60C on the heatsink plenty?

One more thing, if anyone is interested I do have the schematic for this amp... very simple diff pair in class A driving class AB output stage, that's it. Sounds very good if you've never head a B&K.

Thanks in advance for the help!

Best regards, Spence
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Old 31st March 2005, 02:56 PM   #2
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Hi Spence,

your calculations look reasonable.
But please note.
The data sheet allows 50W losses at 80°C....
If your amp now produces an output signal:
Let's say 32V and 4 Ohms load is connected.
When the output slopes to 32V , then the output
current is 8 Amps. The voltage drop across the transistor
is rail minus output = 32V. At this moment the transistor will
have to disappate 32V x 8A = 256W. Of course just for a short moment
and the SOA diagrams usually allow high power for a short time.
But may be the margin mayless than what you would have expected.
Especially at low frequency signals....
How many output transistors does the amp have in parallel?

Another point which you should check:
Is the idle current stable, or does it increase when you heat
up the heat sink?


BTW:
Schematics are always welcome.

Bye
Markus
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Old 31st March 2005, 03:09 PM   #3
spence is offline spence  United States
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Markus,

Thanks for the reply... very helpful to make sure that my calculations are not completely off...

Good point about transients making the safety margin less than I may think... however, I measured the temperature of the heat sink at 60C after one hour of playing very loud music with very much bass... so it seems like the heatsink is able to absorb the additional dissipation of the transients without a problem. Does this sound right?

The amp has two pairs of SK1058 / SJ162 per channel

Also, the idle current is stable, in fact, it decreases slightly as the temp increases, I noticed this as I was slowly incrementing the bias current. Example: if I set the current at 100 mA, and left the temp to equalize for an hour, when I remeasured, it was about 95 mA.

I'm at work so I don't have the schematic... I'll attach it from home later.

Thanks, Spence
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Old 31st March 2005, 03:27 PM   #4
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Spence,

i would make an estimate of the heatsink thermal resistance(Rod.E. has one), and calculate for how much dissipation the B&K is laid out for.
That should give you a more accurate estimate on how much bias heat the amplifier can spare.
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Old 31st March 2005, 04:01 PM   #5
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Hi Steve!
Slight reduction of idle current with increasing temperature is perfectly fine.

To find out if you are in the SOA you would need to check the
in the data sheet if for some ms it is allowed to dissipate
128W (each of the two transistors at 128W make the 256W
as posted before) .
It is difficult to estimate the real silicone temp during pulses just
from the heat sink temp. For this you would need the dynamic thermal resistances of the transistor. Often they are not given in the data sheet. But the SOA diagram usually shows if you are in the allowed range.

Enjoy your amp!
Don't blow it by increasing the idle current to far.

Bye
Markus
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Old 31st March 2005, 05:01 PM   #6
spence is offline spence  United States
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Thanks so much to you guys for your help... I just have a few questions. I apreciate your willingness to clarify what may seem like obvious questions to you for this newbie.

ChocoHolic said:
"Let's say 32V and 4 Ohms load is connected.
When the output slopes to 32V , then the output
current is 8 Amps. The voltage drop across the transistor
is rail minus output = 32V. At this moment the transistor will
have to disappate 32V x 8A = 256W."

Why have you used 32V, if the rails are +/- 64V?

ChocoHolic said:
"To find out if you are in the SOA you would need to check the
in the data sheet if for some ms it is allowed to dissipate
128W (each of the two transistors at 128W make the 256W
as posted before)."

If there are 4 transistors per channel (2 X SK1088, 2 X SJ162), would the 256W be divided by the four transistors, not two?

ChocoHolic said:
"Enjoy your amp!
Don't blow it by increasing the idle current to far."

I'm trying hard not to blow it up!


Thanks!! Spence
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Old 31st March 2005, 05:55 PM   #7
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When the output is at the half of the DC rail, you have the
point of max. loss in the transistor.
Simply follow the sine wave at the output.
At every moment the losses in the transistor are roughly:
P_loss = I out x (DC_Rail - V_out )

4 transistors per channel (2 X SK1088, 2 X 2SJ162):
But only two of them are carrying the positive output current and
the other two are carrying the negative output current.
So SK 1088 and SJ162 are alternating in use.
The 256W do only split to two SK1088 or to two 2SJ162.
You will have roughly 128W in each transistor during short
periods of time.
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Old 31st March 2005, 06:40 PM   #8
spence is offline spence  United States
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Quote:
Originally posted by ChocoHolic
When the output is at the half of the DC rail, you have the
point of max. loss in the transistor.
Simply follow the sine wave at the output.
At every moment the losses in the transistor are roughly:
P_loss = I out x (DC_Rail - V_out )

4 transistors per channel (2 X SK1088, 2 X 2SJ162):
But only two of them are carrying the positive output current and
the other two are carrying the negative output current.
So SK 1088 and SJ162 are alternating in use.
The 256W do only split to two SK1088 or to two 2SJ162.
You will have roughly 128W in each transistor during short
periods of time.

Ahhh... I get it. Thanks!
So for my 8 ohm load: 32V / 8 Ohm = 4 Amps
4 amps * 32 V = 128W / 2 transistors = 64 W per transistor

At 60C heatsink and 1.0 C/W resistance of Sil Pad insulator, Case Temp = 70 C

From the transistor data sheet, max continuous dissipation at 70C ~ 60 - 65W (hard to tell exactly from the graph)

But since the 64W per transistor is transient. I'm guessing I will be well below this. So I think I'll keep the bias right where it is.

The amp sounds great with this higher bias setting, I'm glad that it appears safe from the calcs.

Thanks very much for the help.

Regards, Spence
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Old 31st March 2005, 07:05 PM   #9
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Old 31st March 2005, 08:11 PM   #10
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Quote:
Originally posted by ChocoHolic
When the output is at the half of the DC rail, you have the
point of max. loss in the transistor.
Simply follow the sine wave at the output.
At every moment the losses in the transistor are roughly:
P_loss = I out x (DC_Rail - V_out )
I think the situation can be even worse as a speaker is not purely resistive. At crossover freqs the phase shift between I&V may be quite high.
Nice to see you posting again Herr ChocoHolic

best regards
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