A deep breath, a steady hand, and a bit of luck.
EDIT: Here, this is what the above can get me on a good day......
http://www.diyaudio.com/forums/showthread.php?p=1860855
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Impressive stuff.
Incidentally, as luck would have it, I've just remembered that many moons ago I orded an MSOP10 to DIL adapater - & even more lucky, I've found it in my messy workshop...therefore thank you for your kind offer, but there's now no need.
here's to a steady hand over the next hour or two!
Incidentally, as luck would have it, I've just remembered that many moons ago I orded an MSOP10 to DIL adapater - & even more lucky, I've found it in my messy workshop...therefore thank you for your kind offer, but there's now no need.
here's to a steady hand over the next hour or two!
Ok, I managed to get it soldered onto a DIL8 adapter & wired it up - it works.....
This chip shows a little bit less distortion than the NCS2211 (no kink, just a little bit of straightening of the rising/falling edges of the sine wave - nothing horrific), but it really does beg the question.... How on earth is this thing meant to dissipate 2W - I put about 1.5Vpp into it (which yields about 3Vpp on each leg of the BTL)...the thing was *way* to hot to touch.
I can't see how such a tiny little thing could sustain any length of heat that I'm observing?
Puzzled.
An externally hosted image should be here but it was not working when we last tested it.
(apologies that it's a bit blurry - since the thing only measure 3mm long - about 1/8" - to magnify it, I was using a 50mm focal length lens reversed which gives an impossibly small depth of field!))
This chip shows a little bit less distortion than the NCS2211 (no kink, just a little bit of straightening of the rising/falling edges of the sine wave - nothing horrific), but it really does beg the question.... How on earth is this thing meant to dissipate 2W - I put about 1.5Vpp into it (which yields about 3Vpp on each leg of the BTL)...the thing was *way* to hot to touch.
I can't see how such a tiny little thing could sustain any length of heat that I'm observing?
Puzzled.
Well, the exposed pad on the bottom of the IC really needs to be attached to something.
The only way I see to do that using an adapter would be to mount the IC upside down. You could then clamp or epoxy a small heatsink to it.
Perhaps it would perform better on a proper PCB with close decoupling.
The only way I see to do that using an adapter would be to mount the IC upside down. You could then clamp or epoxy a small heatsink to it.
Perhaps it would perform better on a proper PCB with close decoupling.
the pwr rating for these pwr pad chips is based on the exposed pad on the bottom being attached to a given area of Cu foil on the pcb - often 2" sq so the small size of the pkg isn't the space savings it might seem - you still have to have the pcb area or other heat sink
I've speculated that a thru hole a little smaller than the pad could allow hand soldering without too much loss of themal performance but they really spec these for solder paste/reflow assembly on multilayer pcb and lots of small thermal conducting vias
I've speculated that a thru hole a little smaller than the pad could allow hand soldering without too much loss of themal performance but they really spec these for solder paste/reflow assembly on multilayer pcb and lots of small thermal conducting vias
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Just to give some feedback here - the signal 'kink' as observed across the coil was down to the output current draw creating too much ripple on the VCC rail, which in turn was feeding back into the preamp (the preamp shares the same VCC rail as the output amp) ...this ripple was therefore being merged into the incoming signal - & then back into the power amp - pretty Fugly. If I put a heap of capacitance on there, the kink straightens out - but the physical dimensions of the capacitor are way too big to fit into a small guutar cavity! It probably needs an inductor on the VCC rail ....but I'm getting out of my comfort zone!
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Hi Elvee,
Apologies for not responding to your post - don't know how/why it slipped me by ...that's quite an impressive little circuit - is it of your own design? (is there an PCB layout available anywhere for it?)
wg_ski, re your quote here....
I'm not understanding the bit I've bolded....could you add a bit more meat (I'm keen to learn!)
Also, I'm puzzled why you say that Audio Amp chips whill only yield 1Vp-p? I'm scoping 2.5V peak to peak on a TDA7052A @5V supply?
Incidentally, could someone clear this up for me. The TDA7052A is a BTL chip, so If I'm scoping a sine wave that's 2.5V p-p on the +ve leg ....what voltage appears across the coil (it's not a trick question .....but the +ve leg, -ve leg is confusing me) ...is it 5V (ie the difference between the +ve leg & -ve leg) or just 2.5V?
Can anyone confirm what the power wrt a scoped 2.5Vp-p (+ve leg of a BTL chip) into a 4.4hm coil? (I've factored in the inductive reactance @330Hz within that resistance figure)
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Hi,
an amp running on a single supply of 5Vdc, or a dual polarity supply of +-2.5Vdc, cannot put out a signal voltage exceeding the supply rails.
Into a very high load impedance, the maximum output voltage can approach the rail voltage quite closely. Some amps are described as rail to rail. They make a better job of approaching the supply rail voltages at moderate output currents, but are never perfect. There is always a loss between supply rail voltage and maximum output voltage with any amplifier. the higher the output current the worse this loss becomes.
Look at some graphs from the datasheets of some rail to rail and "ordinary" opamps.
You will see the "loss" plotted vs load resistance.
1Vrms ~=2.8Vpp. An ordinary output stage amplifier can usually manage to develop this output signal from a 5Vdc supply if the output current is kept low (=high load resistance).
At high output currents, say into a 600r load, the maximum output is likely to be <<1Vrms.
an amp running on a single supply of 5Vdc, or a dual polarity supply of +-2.5Vdc, cannot put out a signal voltage exceeding the supply rails.
Into a very high load impedance, the maximum output voltage can approach the rail voltage quite closely. Some amps are described as rail to rail. They make a better job of approaching the supply rail voltages at moderate output currents, but are never perfect. There is always a loss between supply rail voltage and maximum output voltage with any amplifier. the higher the output current the worse this loss becomes.
Look at some graphs from the datasheets of some rail to rail and "ordinary" opamps.
You will see the "loss" plotted vs load resistance.
1Vrms ~=2.8Vpp. An ordinary output stage amplifier can usually manage to develop this output signal from a 5Vdc supply if the output current is kept low (=high load resistance).
At high output currents, say into a 600r load, the maximum output is likely to be <<1Vrms.
Thanks Andrew - when you talk of opamps above, do you include audio amp ICs too?
For example, with a TDA7052A (BTL chip), with a 5V supply rail, scoping the TDA's +ve leg output, I'm seeing a 2.5V sine wave into a 4 Ohm DC resistance coil before clipping occurs....that suggests it is pretty much getting to rail even with an impedance of approx 5 Ohms? (the 5 ohms here being the impedance @330Hz)
On a slightly different tack, have I got my facts/figures right here.....
Scoped signal Peak to Peak (+ve leg of TDA7052A) = 2.5V wrt earth
Voltage across coil = 5Vp-p
Therefore RMS Signal Voltage across coil = 1.77V RMS
Frequency = 330Hz
Coil inductance = 1.6mH
Total Impedance @330Hz = 5.2 Ohms
Current through coil = 0.34A
Power = 0.60W
If that Power figure is right - it opens up a lot more audio output ICs as being contenders (previously I thought I needed much higher output ICs)
The bit I'm unclearest about is that bolded line! If I'm scoping the +ve leg of BTL chip (TDA7052A) wrt Earth & seeing 2.5V onscreen ....do I therefore take it that the signal across the coil is actually double that - therefore 5V? (ie differential voltage output of the TDA7052A)
Edit: Actually, I've just found a good explanation from some chap at TI...
http://archive.chipcenter.com/analog/c029.htm
What I'm not clear about though, is if the TDA7052A is biased at 1/2 the supply rail (therefore for a 5V rail, it's biased at 2.5V) ...the maximum (theoretical) swing each output leg is capable of is 2.5V....how does that map out wrt calculating RMS voltage swings as seen across the load?
For example, with a TDA7052A (BTL chip), with a 5V supply rail, scoping the TDA's +ve leg output, I'm seeing a 2.5V sine wave into a 4 Ohm DC resistance coil before clipping occurs....that suggests it is pretty much getting to rail even with an impedance of approx 5 Ohms? (the 5 ohms here being the impedance @330Hz)
On a slightly different tack, have I got my facts/figures right here.....
Scoped signal Peak to Peak (+ve leg of TDA7052A) = 2.5V wrt earth
Voltage across coil = 5Vp-p
Therefore RMS Signal Voltage across coil = 1.77V RMS
Frequency = 330Hz
Coil inductance = 1.6mH
Total Impedance @330Hz = 5.2 Ohms
Current through coil = 0.34A
Power = 0.60W
If that Power figure is right - it opens up a lot more audio output ICs as being contenders (previously I thought I needed much higher output ICs)
The bit I'm unclearest about is that bolded line! If I'm scoping the +ve leg of BTL chip (TDA7052A) wrt Earth & seeing 2.5V onscreen ....do I therefore take it that the signal across the coil is actually double that - therefore 5V? (ie differential voltage output of the TDA7052A)
Edit: Actually, I've just found a good explanation from some chap at TI...
http://archive.chipcenter.com/analog/c029.htm
What I'm not clear about though, is if the TDA7052A is biased at 1/2 the supply rail (therefore for a 5V rail, it's biased at 2.5V) ...the maximum (theoretical) swing each output leg is capable of is 2.5V....how does that map out wrt calculating RMS voltage swings as seen across the load?
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Which ties up with my earlier calculation...
but I'm still not just getting this BTL melarkey, so I'll try & post the question as best I can....
wrt BTL output swing, what would be the max voltage swing for a BTL amplifier with a rail of 5V as seen across the load.
LIke I say, for a 5V rail, scoping the +ve leg (reference to ground) I'm seeing 2.5V before clipping ....but this differential aspect is not clear to me - is it a simple doubling of what I'm seeing voltage wise - therefore a 5V swing across the load?
What's puzzled me is this extract (source = http://archive.chipcenter.com/analog/c029.htm )...
"If the speakers are 4 W and the supply voltage is 5 V the maximum output power from a SE and BTL configuration is:
PSE = ((4.5V/2.83)2)/4-W PBTL = ((9V/2.83)2)/4-W
PSE = 0.63 W PBTL = 2.53 W"
The 9V seen in that calculation is 4.5V doubled (& the 4.5V figure is apparently the max 'real world' signal before clipping when a 5V supply is used), which suggests that even though I'm seeing just 2.5V on the +ve leg, using that above calculation as my lead, then somehow magically the voltage level is 4 times that?!!!!
but I'm still not just getting this BTL melarkey, so I'll try & post the question as best I can....
wrt BTL output swing, what would be the max voltage swing for a BTL amplifier with a rail of 5V as seen across the load.
LIke I say, for a 5V rail, scoping the +ve leg (reference to ground) I'm seeing 2.5V before clipping ....but this differential aspect is not clear to me - is it a simple doubling of what I'm seeing voltage wise - therefore a 5V swing across the load?
What's puzzled me is this extract (source = http://archive.chipcenter.com/analog/c029.htm )...
"If the speakers are 4 W and the supply voltage is 5 V the maximum output power from a SE and BTL configuration is:
PSE = ((4.5V/2.83)2)/4-W PBTL = ((9V/2.83)2)/4-W
PSE = 0.63 W PBTL = 2.53 W"
The 9V seen in that calculation is 4.5V doubled (& the 4.5V figure is apparently the max 'real world' signal before clipping when a 5V supply is used), which suggests that even though I'm seeing just 2.5V on the +ve leg, using that above calculation as my lead, then somehow magically the voltage level is 4 times that?!!!!
a bridged amplifier is TWO amplifiers.
Start out by considering ONE amplifier.
Understand what it can do and what it's limitations are.
Then series connect the outputs of the two bridge connected amplifiers and double the voltage across the load.
If one amp can drive a 4r0 load to 1Vrms, then a bridged pair can drive an 8r0 load to 2Vrms.
What the bridged pair CANNOT do is drive a 4r0 load to 2Vrms, if it's limit is 4r0 and 1Vrms.
A 4ohm speaker load draws upto three times as much current as a 4r0 resistor.
Start out by considering ONE amplifier.
Understand what it can do and what it's limitations are.
Then series connect the outputs of the two bridge connected amplifiers and double the voltage across the load.
If one amp can drive a 4r0 load to 1Vrms, then a bridged pair can drive an 8r0 load to 2Vrms.
What the bridged pair CANNOT do is drive a 4r0 load to 2Vrms, if it's limit is 4r0 and 1Vrms.
A 4ohm speaker load draws upto three times as much current as a 4r0 resistor.
Bear with me here - I'm sure the penny will drop soon!
I guess what's troubling me, is that we're talking about 2 Amplifiers - & both biased at 2.5V (ie for a rail of 5V).
Now, wrt the signal - as the non-inverting amp slews up towards the extent of its travel (ie towards +5V, from a starting bias point of 2.5V), the other inverting amp will be slewing down (towards 0V - but also from a starting bias point of 2.5V)...the end result is that there's a 5V peak to peak swing across the load - so where do we get the doubling of the output voltage from?!!
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I guess I'm just going to have to log this one under "things I can't quite grasp" - with both amps in the BTL config biased at 2.5V & swinging 2.5V either side 180 degrees out of phase, the most I can ever see being presented across the load is 5Vp-p! (which in my books is about the same level that a single ended amp can achieve for a 5V rail!). Now I realise I'm wrong, but I understand why I'm wrong!
no a single amp fed from a single supply is more likely to give ~1Vrms maximum into a low load impedanace.
A bridged pair should give ~2Vrms (~5Vpp) into a doubled load impedance.
Yes it is.
Sorry, it has just been simulated and breadboarded...
You could extract a little more performances by using superbeta transistors, from Zetex, or whatever it's called now.
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