Can you check and see what your input voltage (wall) is. A lot of transformers are spec'd at 115 VAC whereas many houses have 120 volts at the wall. You are going to use this transformer on a regulated power supply I presume, so an extra few volts on the input side is likely to be of no consequence. It is the DC output of the power supply that is more important.
Small transformers have very poor regulation, the off-load voltage will be as much as 20% higher.
http://www.amveco.com/pdf/Amveco_Catalog.pdf#page=25
http://www.amveco.com/pdf/Amveco_Catalog.pdf#page=25
the no load secondary voltage will be =
mains voltage / rated input voltage * rated secondary voltage * transformer regulation.
eg mains is 121Vac.
rated voltage is 115:18
regulation is 15%
no load output voltage = 21.78Vac.
Now look at the range of mains voltage that your supplier can send to you.
In the UK, this is between 216Vac and 254Vac.
Your design MUST take account of these variations.
mains voltage / rated input voltage * rated secondary voltage * transformer regulation.
eg mains is 121Vac.
rated voltage is 115:18
regulation is 15%
no load output voltage = 21.78Vac.
Now look at the range of mains voltage that your supplier can send to you.
In the UK, this is between 216Vac and 254Vac.
Your design MUST take account of these variations.
it's actually 2 * 4780 ..i understand that internally each is 2*LM3886...I wanted to try this extremely simple 100W amp. you would agree paralelling a 4780 is relatively easy compared to a discrete. I have another design in progress using a LME..series driver chip..but i wanted to try out paralelled chip amp first 
With a 2*18 V or 36 V CT transformer you can expect ~22 W output power into 8 Ohm and ~42 W into 4 Ohm, no matter, if you use a single channel or two parallel channels.
For 100 W into 4 Ohm you need two parallel channels and a 2*26 V or 52 V CT transformer.
You can get 100 W into 8 Ohm with two bridged channels and a 2*20 V or 40 V CT transformer.
If you use the two channels of an LM4780 parallel or bridged for that amount of output power, you will need an incredibly big heatsink or, better still, fans on top of a still big heatsink.
For 100 W into 4 Ohm you need two parallel channels and a 2*26 V or 52 V CT transformer.
You can get 100 W into 8 Ohm with two bridged channels and a 2*20 V or 40 V CT transformer.
If you use the two channels of an LM4780 parallel or bridged for that amount of output power, you will need an incredibly big heatsink or, better still, fans on top of a still big heatsink.
actually the national suggested heatsink for PA100 is not that big....I have two of them made to order, it's quite cheap here in India ...around 30 euros. Regarding the transformer I was more concerned about the capacity ( VA ). National data sheet suggests around +- 32 volts for ~100W into 8 ohm.
National specified the heatsink for two LM3886 not one LM4780. Two LM3886 can dissipate 250 W at 25 °C. One LM4780 can dissipate 155 W at 25 °C. Both can disspate 0 W at 150 °C. The power dissipation for 100 W output into 4 Ohm will be around 60 W, hence two LM3886 can run at ~120 °C, one LM4780 can run at ~100 °C. That means a heatsink for the LM4780 must provide a 20 °C lower temperature than for two LM3886. Translated into thermal resistance the heatsink for one LM4780 must be 3 K/W smaller than for two LM3886. That will be difficult, because the specified heatsink has around 1,5 K/W.sasmit said:
National did not specify the ambient temperature for those heatsinks, so it is probably 25 °C. What is the highest temperature, where you live?
It should have more VA than the amplifier has W.sasmit said:
Yes. And you will not get ±32 V from a 36 V center tapped transformer. Not even unloaded, and the ±32 V must be present at nominal output.sasmit said:
sasmit said:
Your heatsink is far too small even for 4 3886 chipamps.sasmit said:
And Pacific is right, the 4780 needs even more cooling.
It's all there in National's datasheet and application notes, but they bury the real situation, by always starting from Tc=25degC.
In Bangalore your starting Tc could be anywhere from 35degC to 60degC. You must de-rate everthing that is affected by elevated temperatures.
It's far simpler going discrete than going parallel.
I should read, who posts what. The 36 V CT was somebody else's.
The 2*25 V transformer should bring you into the 90 W range with 4 Ohm. Heat dissipation could become an issue nevertheless. If your heatsink has 20 fins, as usual for that size, you can expect ~0,9 K/W for Pd = 50 W. With that dissipation the LM4780 can operate up to ~110 °C. The total thermal resistance should be less than ~1,66 K/W. Subtract 0,8 K/W for the LM4780, 0,2 K/W for thermal grease and some more for an isolation washer. The worst case heatsink should have less than 0,66 K/W even without safety margin.
The 2*25 V transformer should bring you into the 90 W range with 4 Ohm. Heat dissipation could become an issue nevertheless. If your heatsink has 20 fins, as usual for that size, you can expect ~0,9 K/W for Pd = 50 W. With that dissipation the LM4780 can operate up to ~110 °C. The total thermal resistance should be less than ~1,66 K/W. Subtract 0,8 K/W for the LM4780, 0,2 K/W for thermal grease and some more for an isolation washer. The worst case heatsink should have less than 0,66 K/W even without safety margin.
They will not be big enough for all listening situations. If you want that, you will need either bigger heatsinks or an additional fan for each heatsink.
Music is not a worst case signal, temperatures are not always high and you will probably not always be listening at full throttle. So with the heatsinks you have you will be able to listen at high levels, when it is cold or to listen on hot days at moderate levels.
Music is not a worst case signal, temperatures are not always high and you will probably not always be listening at full throttle. So with the heatsinks you have you will be able to listen at high levels, when it is cold or to listen on hot days at moderate levels.
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