Firstly, just a bigger toroidal will not solve that for you. It will only help to keep the voltage up as the load increase. You would also need to make sure that your heat management is up for it. If you can keep the heat on the junctions of the outputs and drivers in a sensible range you may increase Iq. With one pair of drivers (MJE15032/33) you should be able to drive at the maximum 4 output pairs. As soon as you want to drive more output pairs I would suggest an addition pair of drivers. I prefer 3 output pairs per pair of drivers as a sensible configuration. I am using the MJL1302/3281 BJTs for my outputs.
To get I am running at 640 ma Iq. The heatsink is massive (out of a 75hp motor variable frequency drive) and I also have a 5 inch fan helping out. The reason I need bigger toroidals is that mine is only 600va at 29-0-29. So rectified and loaded only 35v or so. The boards are to have 50v for the output stage and 65 for the front end. I have 36V and 45v for the front end.
So I need more volts and amperage capacity.
Local shop has big isolation totoisldals for cheap. I am 120v in so will just remove
So I need more volts and amperage capacity.
Local shop has big isolation totoisldals for cheap. I am 120v in so will just remove
That depends on the supply-voltage. The SOA of the 5200 limits the max current at 60 Vce to ca. 2 Amps continue. For a Sanken 2sc2922 this is a little over 3 Amps @ 60 Vce
So, in case of the 5200 and sufficient cooling, about 1A quiescent current per device.
I just picked up two very big toroidals. They are for active sub's I think. They have no center tap and are 1to1 ratio so 120v in and 120v out. I want to connect the primaries in series and deliver half the voltage too each primary and resulting secondary. Each secondary is rated at 120v and 8.3amps out.
Will I create a 16.6a mp 60-0-60 or a house fire?
Thanks.
Will I create a 16.6a mp 60-0-60 or a house fire?
Thanks.
No, you will create an 8.3 A transformer at half voltage. You say 1:1, no center tap, so I'm guessing there are 2 secondaries at 0-120. If this is the case you will have 0-60 out.
You may be able to pull more than 8.3A on peaks, but for continuous duty the copper and iron losses limit you to the original VA rating.
Chances are your 15A breaker will not trip unless you go crazy testing high power sine waves. In home use even for subs the continuous output is rarely above a few watts. This assumes you're using a soft start, since the instantaneous peak will be quite high.
You may be able to pull more than 8.3A on peaks, but for continuous duty the copper and iron losses limit you to the original VA rating.
Chances are your 15A breaker will not trip unless you go crazy testing high power sine waves. In home use even for subs the continuous output is rarely above a few watts. This assumes you're using a soft start, since the instantaneous peak will be quite high.
Last edited:
You don't need to centre tap it. You may leave the secondaries separate. This means you will have 2 separate 60V secondaries. You will then connect the 2 secondaries to 2 bridge rectifiers, one for each secondary. Then you will connect the + DC out of the first bridge to the - DC out of the second bridge to form the PSU 0V or PSU ground.
Very good. Thank you henryve. Now I also need a separate sourse for my inputsstage that will be 15v higher than my output. If I ran three toroidals in series on the primary's and used the third dissimilar toroidal for the imput stage would I get 1/3 voltage from each of the the output stage toroidala?
Sorry. That last thought wouldn't work as the inrush current for three big toroidals would be too much I think. I removed 33% of the secondary windings then respaced and center tapped a 1kva toroidal. It was 120 in 120 out and. Now its 120 in 40-0-40 exactly out. Assuming its still good for 8.33 amps as the wire size has not changed.
Got Them
Thanks for the quick response. These boards look beautiful. Can anyone share the latest BOM and schematics. I have been reading this thread, but wanted to order the parts and read while I am waiting for the components.
Drivers, who needs lots of them anyway?
If you keep the voltage rails the same, and the peak output current the same, the load on the drivers does not increase with the number of output devices, under any normal circumstances.
The current load on the drivers (which are basically simply a small output stage) is basically the current load of the output stage divided by the current gain of the output stage...why is this important? I hear you ask...
Adding more output devices decreases the current load on each output device and for almost all values we care about this will increase the overall current gain of the output stage...which is to say adding output pairs decreases the load on the drivers.
So, why does a KSA100 have two driver pairs?...it has 40% higher rails and it's designed to drive twice as much current. To do that it has twice as many output pairs, since each of these is current loaded as it was in it's smaller sibling, and the drivers are already dealing with 40% more power, two pairs were needed. Plus from all cost perspectives more drivers are cheaper than more outputs...
So if you are worried about the number of drivers, just make sure you have as many as dan designed in the first place then add more output transistors...and make sure you have the correct size of emitter resistor, or you will likely cause the drivers to leave Class A, at which point you wasted a lot of money on hardware that is not being used the way you intended...Smaller emitter resistors are never going to be the correct answer in this amp. Twice as many outputs should have emitter resistors that are twice the value...
HTH
Stuart
If you keep the voltage rails the same, and the peak output current the same, the load on the drivers does not increase with the number of output devices, under any normal circumstances.
The current load on the drivers (which are basically simply a small output stage) is basically the current load of the output stage divided by the current gain of the output stage...why is this important? I hear you ask...
Adding more output devices decreases the current load on each output device and for almost all values we care about this will increase the overall current gain of the output stage...which is to say adding output pairs decreases the load on the drivers.
So, why does a KSA100 have two driver pairs?...it has 40% higher rails and it's designed to drive twice as much current. To do that it has twice as many output pairs, since each of these is current loaded as it was in it's smaller sibling, and the drivers are already dealing with 40% more power, two pairs were needed. Plus from all cost perspectives more drivers are cheaper than more outputs...
So if you are worried about the number of drivers, just make sure you have as many as dan designed in the first place then add more output transistors...and make sure you have the correct size of emitter resistor, or you will likely cause the drivers to leave Class A, at which point you wasted a lot of money on hardware that is not being used the way you intended...Smaller emitter resistors are never going to be the correct answer in this amp. Twice as many outputs should have emitter resistors that are twice the value...
HTH
Stuart
Last edited:
Hi, I have not posted for a while but now have re-started building the KSA 50. After reading Stuarts post I wonder if he can answer some questions for me, so that I am 100% clear in my own mind.
I am using Jan's boards with rails of +and- of 42volts under load. I intend to use 4 output pairs per Chanel (I have 2 fan cooled heatsinks the same as used on the KSA100 )
I take it from Stuarts post that the one pair of drivers, MJE 15032 and 15033 will be sufficient to drive the extra output devices, is this correct?
Will the driver em miter resistors need changing in value, I am using 25ohm?
Will any other resistors need changing due to using 42V instead of 37V which I think the board was designed for?
I will use 1ohm emiter resistors on the output devicesis that correct?
Thanks for your help
Alan
I am using Jan's boards with rails of +and- of 42volts under load. I intend to use 4 output pairs per Chanel (I have 2 fan cooled heatsinks the same as used on the KSA100 )
I take it from Stuarts post that the one pair of drivers, MJE 15032 and 15033 will be sufficient to drive the extra output devices, is this correct?
Will the driver em miter resistors need changing in value, I am using 25ohm?
Will any other resistors need changing due to using 42V instead of 37V which I think the board was designed for?
I will use 1ohm emiter resistors on the output devicesis that correct?
Thanks for your help
Alan
Quick answers:
1) With good heatsinks a single pair of 15032/2 are sufficient for all sane loads.
2) If you keep a single pair of drivers, the original emitter resistor value is good. It limits the driver current to ~3A, which AFAICT should be enough for all purposes.
Longer answer:
To do a perfect job of choosing the various transistors drivers / outputs requires a clear understanding of the goals you are trying to achieve followed by a few iterations of a few calculations.
I think it is important is to think about the output load, which will, given the rail voltage decide the output current, and then later you have to decide how much of the power that is delivered into that load should be 'class A'. Nelson Pass wrote a great paper called (IIRC) "Leaving class A", it's worth a read.
With 42v and theoretically perfect output stage you have an absolute maximum power output (v ^ 2 / load) of 220w into 8r (ohms). 440w into 4r, 880 into 2r etc.
Peak current into 8r will be ~5A, we assume this is split equally among the output transistors. We probably need to design in the "party margin", when there are 6 pairs of speakers all connected at random to the output and a drunken idiot is controlling the volume ;-) So lets say we want a safety factor of 4x, we need the entire output stage (drivers, resistors, outputs, resistors, wiring, terminal posts etc) to be able to deliver 20A for longer than the upstream protection device (you did remember protection right?)...
If the output stage is delivering 20A, the driver stage will be delivering ~1A (20A divided by the current gain (hfe) of the output transistors)...choice of output transistors makes a big difference here, a gain of 20 means the driver is dumping 1A into the output stage, while a gain of 100 would give 200ma. More output transistors will increase the gain of the output stage, and therefore draw less current from the drivers.
In either case it is within the reach of a single 15032/33 pair, assuming you have good heat-sinking. At 1A peak output the drivers will be dissipating ~10w each.
Ideally you'd know the average temperature of the transistors as they are used in your case, and you'd de-rate them appropriately. Failing that it's probably wise to assume half the quoted power capacity.
Now I'm going to shut up unless people want to hear more.
HTH
Stuart
1) With good heatsinks a single pair of 15032/2 are sufficient for all sane loads.
2) If you keep a single pair of drivers, the original emitter resistor value is good. It limits the driver current to ~3A, which AFAICT should be enough for all purposes.
Longer answer:
To do a perfect job of choosing the various transistors drivers / outputs requires a clear understanding of the goals you are trying to achieve followed by a few iterations of a few calculations.
I think it is important is to think about the output load, which will, given the rail voltage decide the output current, and then later you have to decide how much of the power that is delivered into that load should be 'class A'. Nelson Pass wrote a great paper called (IIRC) "Leaving class A", it's worth a read.
With 42v and theoretically perfect output stage you have an absolute maximum power output (v ^ 2 / load) of 220w into 8r (ohms). 440w into 4r, 880 into 2r etc.
Peak current into 8r will be ~5A, we assume this is split equally among the output transistors. We probably need to design in the "party margin", when there are 6 pairs of speakers all connected at random to the output and a drunken idiot is controlling the volume ;-) So lets say we want a safety factor of 4x, we need the entire output stage (drivers, resistors, outputs, resistors, wiring, terminal posts etc) to be able to deliver 20A for longer than the upstream protection device (you did remember protection right?)...
If the output stage is delivering 20A, the driver stage will be delivering ~1A (20A divided by the current gain (hfe) of the output transistors)...choice of output transistors makes a big difference here, a gain of 20 means the driver is dumping 1A into the output stage, while a gain of 100 would give 200ma. More output transistors will increase the gain of the output stage, and therefore draw less current from the drivers.
In either case it is within the reach of a single 15032/33 pair, assuming you have good heat-sinking. At 1A peak output the drivers will be dissipating ~10w each.
Ideally you'd know the average temperature of the transistors as they are used in your case, and you'd de-rate them appropriately. Failing that it's probably wise to assume half the quoted power capacity.
Now I'm going to shut up unless people want to hear more.
HTH
Stuart
thx.