tryonziess said:
I don't know what an LME69810 is. I found an LME49810 that is the front end for a power amp.
I don't like integrated circuits, and I don't like feedback.
If you want to get started on something that works well and sounds good, I would build a Leach amp. That is the first project I ever built from scratch, and it is a very good design. When I built mine it never sounded that great until I removed the low-pass filter on the input. Then it sounded excellent. You could actually remove the feedback on this design if you wanted to.
Another way to go would be the JC3 (do a google search) that John Curl designed for Mark Levinson. This would be a lot more work as it's not a step-by-step construction article, and some of the parts are no longer available so you would have to figure out substitutes.
Charles, i have ordered the boards from Dr. Leach. I had been waiting for Delta Audio to make some more Leach Superamp boards. I am glad you mentioned Leach. I had wanted to hear someone say how good it SOUNDED and not just specs. I am beginning to take your attitude on non-discrete designs. Having built several chip amps with lm3886 i find they all run hotter than what i am used to. Must be all of the circuits having to be cooled by such a small contact area. I will also check on the jc amp. John Curls Parasound venture looks to be quite nice. Wish I could spend that kind of money on an amp. Wishful thinking. Though constructing is more rewarding. With regards Tad
Jim Hamley said:
Hi Jim,
Yes, this is a good app note that anyone using the ThermalTraks should read.
There is one thing in it that I wonder about, however, in Figure 2. Notice that they have a complementary Triple output stage. They have three pairs of ThermalTraks in the output stage so they have six ThermalTrak diodes to work with. Of course, with a Triple in the output stage, they need a bias spreader that provides about 6 Vbe of voltage drop. So they just string all six ThermalTrak diodes in series to form the bias spreader. Seems logical and efficient. BUT, I wonder if the arrangement they show will end up causing the output stage to be thermally over-compensated, since the pre-drivers, at least, are certainly not on the heatsink, yet their Vbe is being compensated effectively by a corresponding ThermalTrak diode on the heatsink and indeed on the output die. The situation is a little less clear for the driver transistors, since thay may or may not be mounted on the heat sink.
I would imagine that Charles Hansen might be able to comment on this based on his experience with them.
Cheers,
Bob
Bob Cordell said:
Yes, you should read it. But don't try to build your amp based on it. If the app notes had all the answers, then the engineers wouldn't have to do any work (and all the amps would be the same).
There are a lot of problems with the app note. They give comparative distortion numbers, but don't bother to even see if the bias current was the same. If you build this amp, you will quickly find that part-to-part variations will rear their ugly head. Et cetera, et cetera.
But that is the fun -- discovering the answers for yourself.
Bob Cordell said:
As noted above, using the ThermalTrak parts is not a straightforward proposition. For example, I understand that the McIntosh amp that uses these part came to a very different solution. Apparently they threw their hands up in the air and gave up, as they simply use the bias diode as a sensor that feeds a microprocessor. The micro is what sets the bias current. I can't blame them for taking this approach, but the last thing I would want to put in a high-performance amplifier is an internal RFI generator.
We figured out a different way that addresses the problems Bob alluded to in his post. I'll give you a hint:
What is the tempco of the MUR120 that is inside the ThermalTrak parts?
I'll give you another hint:
It's not -2.2. mV/K.
Have fun.
I am not sure that inside of an ThermalTrack BJT is a MUR120 diode.
I just downloaded from onsemi site two diffrent models, one for MUR diode and the other for Dnjl0302/0281
.MODEL Dmur120rl d
+IS=9.82553e-09 RS=0.0304418 N=1.57716 EG=0.849638
+XTI=0.5 BV=200 IBV=0.000002 CJO=5.37397e-11
+VJ=1.24876 M=0.515449 FC=0.5 TT=1.63843e-08
+KF=0 AF=1
.MODEL Dnjl0302dg_diode d
+IS=5.17122e-08 RS=0.0324668 N=1.78236 EG=1.12011
+XTI=0.5 BV=200 IBV=2e-06 CJO=5.59229e-11
+VJ=0.4 M=0.346657 FC=0.5 TT=2.59685e-08
+KF=0 AF=1
I just downloaded from onsemi site two diffrent models, one for MUR diode and the other for Dnjl0302/0281
.MODEL Dmur120rl d
+IS=9.82553e-09 RS=0.0304418 N=1.57716 EG=0.849638
+XTI=0.5 BV=200 IBV=0.000002 CJO=5.37397e-11
+VJ=1.24876 M=0.515449 FC=0.5 TT=1.63843e-08
+KF=0 AF=1
.MODEL Dnjl0302dg_diode d
+IS=5.17122e-08 RS=0.0324668 N=1.78236 EG=1.12011
+XTI=0.5 BV=200 IBV=2e-06 CJO=5.59229e-11
+VJ=0.4 M=0.346657 FC=0.5 TT=2.59685e-08
+KF=0 AF=1
Bob,
You make a good point. The complimentary predrivers Q7,8 and drivers Q9,10 are both operated class A (I assume.) So, I would think that, once they warm up to normal operating temperature, increased current drive through the NPN's Q7 and Q9 would create a Vbe shift that will essentially offset the corresponding reduced current and Vbe shift through the PNP's Q8 and Q10. If that is true, then the 6 Vbe diode string from the power transistors will overcompensate the 2 Vbe bias needed for the output stages.
One solution would be to resistor divider down the sum of the 6 Vbe's and use the diviided down voltage as the bias spreader voltage.
Of course, what really makes this all possible is Charles' statement that these devices all come fairly tightly matched from On Semi, thus assuring even load sharing. Buying a large quantity of devices and manually matching might be prohibitive for DIYers.
...Jim
You make a good point. The complimentary predrivers Q7,8 and drivers Q9,10 are both operated class A (I assume.) So, I would think that, once they warm up to normal operating temperature, increased current drive through the NPN's Q7 and Q9 would create a Vbe shift that will essentially offset the corresponding reduced current and Vbe shift through the PNP's Q8 and Q10. If that is true, then the 6 Vbe diode string from the power transistors will overcompensate the 2 Vbe bias needed for the output stages.
One solution would be to resistor divider down the sum of the 6 Vbe's and use the diviided down voltage as the bias spreader voltage.
Of course, what really makes this all possible is Charles' statement that these devices all come fairly tightly matched from On Semi, thus assuring even load sharing. Buying a large quantity of devices and manually matching might be prohibitive for DIYers.
...Jim
From this datasheet:
http://www.onsemi.com/pub/Collateral/MUR120-D.PDF
Figure 6. Typical Forward Voltage, at a current of 0.4 Amps, the diode has a forward drop of 0.9 Volts at 25 degrees C and a forward drop of 0.8 Volts at 100 degrees C. Therefore, at this current it changes 0.1 Volt over 75 degrees C, or -1.33 mV/degree C.
http://www.onsemi.com/pub/Collateral/MUR120-D.PDF
Figure 6. Typical Forward Voltage, at a current of 0.4 Amps, the diode has a forward drop of 0.9 Volts at 25 degrees C and a forward drop of 0.8 Volts at 100 degrees C. Therefore, at this current it changes 0.1 Volt over 75 degrees C, or -1.33 mV/degree C.
roender said:
Roender is correct. The thermal coefficient is -1.6 mV/K. I know the same way that Roender knows -- I've measured. It's simple to make an ice water bath with a known temperature of 0 degrees C. Then the easiest point to measure from is room temperature, typically around 20 or 25 degrees C. This give a big enough spread to make accurate measurements.
I don't think you could build an amp like the app note and have it work except by sheer luck. There is too much part-to-part variance.
roender said:
It is. This is straight from the mouth of the designer, and my measurements of the tempco of the MUR120 match the diode inside the ThermalTrak parts exactly. The models that ANY manufacturer supplies are rarely very accurate. Go to Andy_C's website for improved models for the OnSemi parts.
Charles Hansen said:
Hi Charles,
I am going to ask a very dumb question about the ThermalTrak transistors. I have read the spec sheet, and I have measured the real parts. I have from the beginning assumed that the ThermalTrak diode was fabricated right on the same die as the BJT power transistor, and that it was an identical-type transistor made by connecting the collector and base together, but with a smaller area. Is this true?
OR, does the package actually contain two die, both mounted on the same header?
I'm beginning to wonder.
The fact that there is no information on the spec sheet that talks about the matching and relative saturation current values of the BJT and diode makes me wonder if they are two separate die.
What's the answer?
Thanks,
Bob
Bob Cordell said:
No. They didn't take this approach because there would be too much capacitance between the transistor and the diode. This could cause all kinds of instability problems in an actual amplifier. Instead, they take a MUR120 (sans the epoxy body) and mount it to the copper lead frame near the transistor die. The resulting product has less than 2 pF of capacitance between the transistor and the diode.
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