Are you suggesting a single bank would handle 8ch? Well, here's a stupid reason: they came w/the kit
And maybe another one, I would be flexible as to application once my present passion for spkr dsn was quenched. From the blissful world I live in, I could deliver to any HT I could imagine. It never occurred to me the parallel bridge thing was an issue, so why not be prepared? -bg
Don't know how much current (like an off-the-shelf commercial box would not). The initial use for this would be for various spkr designs and wind up some day in a HT.
For diodes, MUR860 w/8 continuous and 16 pk, I believe. But, I think that would be heat sunk and that is not the case for me. thx. -bg
RGrayton- I think you are on the right track, a simple dirty, ugly fix would be something like 0.1 ohm 5 W at each bridge input. I say "something like" because I don't know if 0.1 ohm is ok, or if you need 0.3 or 0.5 ohms.
Still I would recommend selling the diode boards and putting all those caps in parallel behind one large bridge. But your thinking may be acceptable to you; this is DIY we cannot tell you what to do and I certainly don't mean to criticize your effort.
Still I would recommend selling the diode boards and putting all those caps in parallel behind one large bridge. But your thinking may be acceptable to you; this is DIY we cannot tell you what to do and I certainly don't mean to criticize your effort.
Another solution would be to use one bridge per one or two amplifier channels. That has the same effect of lowering the current per bridge, but without having to worry about load sharing and derating. And you can use the PCBs you have.
If your kit is one that uses an LM series IC, the maximum continuous current would be 5,6 A per channel with 4 Ohm speakers. With 8 Ohm speakers you will hardly reach 4 A per channel.
If you replace the MUR860s with MUR1560s, assuming the PCB is designed for the increased amount of current, you could run four channels off one bridge.
So how do you determine the transformer, the fuses and the wire gauge you will need?
If your kit is one that uses an LM series IC, the maximum continuous current would be 5,6 A per channel with 4 Ohm speakers. With 8 Ohm speakers you will hardly reach 4 A per channel.
If you replace the MUR860s with MUR1560s, assuming the PCB is designed for the increased amount of current, you could run four channels off one bridge.
Would you mean I could keep 1/2 the boards I have? While that would ease my situation by 1/2 wouldn't it still have a load share? What effect has load share?
I am Russian. (No offence to any others out there.) Big usually works for us.
That sound right. One thing I did find interesting, maxing out a 4ohm load did not blow a 4A fuze. I would think it would have been at least 5A. There was 5.3A going to the load and whatever else the chip needs.
I will look up the MUR1460. pBlue, thanks again for sticking with me on this. This might be something I could work in over time. I could use the boards for some other amp, someday. I doubt I will ever need 5A instantaniously at the same time from more than one ch. -bg
Lets take a few numbers at random for a pair of parallel diodes.
Vf=650mV, I1=2A, I2=3A
Vf=700mV, I1=6A, I2=12A.
Now apply a forward voltage to the pair and draw 5A from them.
2A will flow through D1 and 3A will flow through D2. D1 dissipates 1.3W and D2 1.95W. D2 will run hotter and I2 will increase slightly reducing the load on D1. It might become 3.5A and 1.5A.
Increase the current draw to 18A, 12A will pass D2 dissipating 8.4W and heat very quickly.
6A will pass D1 dissipating 4.2W, increasing in temperature more slowly. Both will need heatsinks.
As these diodes heat up D2 current will increase maybe to 14A and D1 could decrease to 4A.
Now set up the parallel diodes and add a series resistor to each with a value of 0r01. try to find out what happens to current balance? Now step the resistor value to 1r0 and see the effect on diode current balance.
A spreadsheet and the diode datasheet should allow you to build a decent model.
Vf=650mV, I1=2A, I2=3A
Vf=700mV, I1=6A, I2=12A.
Now apply a forward voltage to the pair and draw 5A from them.
2A will flow through D1 and 3A will flow through D2. D1 dissipates 1.3W and D2 1.95W. D2 will run hotter and I2 will increase slightly reducing the load on D1. It might become 3.5A and 1.5A.
Increase the current draw to 18A, 12A will pass D2 dissipating 8.4W and heat very quickly.
6A will pass D1 dissipating 4.2W, increasing in temperature more slowly. Both will need heatsinks.
As these diodes heat up D2 current will increase maybe to 14A and D1 could decrease to 4A.
Now set up the parallel diodes and add a series resistor to each with a value of 0r01. try to find out what happens to current balance? Now step the resistor value to 1r0 and see the effect on diode current balance.
A spreadsheet and the diode datasheet should allow you to build a decent model.
You can probably keep 1/2 of the boards, yes.rgrayton said:
Each bridge provides the current for the amplifier channel(s) that is(are) connected to it.and that is it. You do not have to worry about tolerances from one diode to the other, because the current goes through one rectifier to the amp and back. It has no business with the other rectifiers.
If you use parallel diodes, the situation is different as Andrew explained above. The same current is split up before the rectifiers and added up again after them. And due to tolerances from one diode to the other the distribution is usually not the same for each diode. That is, why they must be derated, and why you should use resistors to level that out.
If you load fuses with a certain current, it takes some time for them to react. According to your application you have to choose the right characteristic, e. g. slow for a transformer or fast for electronics.rgrayton said:
You probably won't, but you never know, which channel has to deliver, so you should provide the possibility that all channels can do it.rgrayton said:
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