Dave Smith said a year or so ago that leaving it out makes a big improvement.
I have not tried, as my boards are now with Mark.
I think Uwe left them out.
So depends on how experienced you are and how much risk you want to take.
It is well know that I like shorting my outputs with relays to protect the speaker.
Relay is then not in signal path.
So some current limit is useful to save the FETs.
Bear in mind the Toshiba's are becoming rare commodity.
So you should have at least a few sets in reserve.
Patrick
I have not tried, as my boards are now with Mark.
I think Uwe left them out.
So depends on how experienced you are and how much risk you want to take.
It is well know that I like shorting my outputs with relays to protect the speaker.
Relay is then not in signal path.
So some current limit is useful to save the FETs.
Bear in mind the Toshiba's are becoming rare commodity.
So you should have at least a few sets in reserve.
Patrick
I left it out when I replaced the IRF boards with the Toshiba board. Did not do any A/B comparison, and managed not to destroy anything...until now.
My son shorted the output out the F5 with limiter, and anly the fuse blowed. Now the cable are screwed so tight that you need a wrench to unplug them.
D.
My son shorted the output out the F5 with limiter, and anly the fuse blowed. Now the cable are screwed so tight that you need a wrench to unplug them.
D.
In the grounded x, the amp is basically two separate SE F5's driving the same speaker with out of phase signals. If you only feed a SE signal to Vin+, and 0V to Vin-, one side of the amp will be idle.
By floating the X, bit sides of the amp would be operating even in the case of a sE signal, though we still recommend you strongly to use balanced signals as input.
Patrick
By floating the X, bit sides of the amp would be operating even in the case of a sE signal, though we still recommend you strongly to use balanced signals as input.
Patrick
My pleasure Patrick. It's been a lot of fun to get to where we are.
Hi Tsang,
Yes, no problem.
Grounded X is essential for biasing the amplifier after construction and is also useful if you want to use each board for two separate channels. In grounded X configuration the two halves of the circuit (left and right side in the schematic posted) are essentially two separate amplifiers on one board that are configurable as a single channel in bridged mono mode or as two seperate channels. Applications like multi-channel or bi-amp come to mind...
Floating X configuration connects the left and right halves to each other in the complementary configuration so that the amplifier becomes a true fully balanced amplifier from input to output. This allows the amplifier to accept either balanced inputs or single ended, RCA type inputs.
There will be more information on this in the build thread coming soon.
Dave
There are some good reasons to use grounded X.
For example as Dave has said :
If you want to use the amp as SE, and you have a two way speaker with active crossover, you can run each balanced PCB as 2 power amps for one channel.
All you have to do is ground the X, raise voltages to +/-24V, change bias to 1.3A per FET, and you are in business.
Patrick
For example as Dave has said :
If you want to use the amp as SE, and you have a two way speaker with active crossover, you can run each balanced PCB as 2 power amps for one channel.
All you have to do is ground the X, raise voltages to +/-24V, change bias to 1.3A per FET, and you are in business.
Patrick
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...which in turn, by means of adding a third output binding post tied to ground, allow you to use it as a 4 channel amp, for example for bi-amping purposes. Great flexibility!
For those who like analysing circuits, this is essentially one (bottom) half of a single ended F5, fed by current sources.
Except for the output stage which has a JFET phase Splitter to drive it as quasi Push-pull.
http://www.diyaudio.com/forums/solid-state/203431-all-fet-jlh-class.html#post2839388
Performance quite OK in Spice at least.
Patrick
PS
No, I have not built, and it probably would not be a new project.
Too many on the queue already.
Except for the output stage which has a JFET phase Splitter to drive it as quasi Push-pull.
http://www.diyaudio.com/forums/solid-state/203431-all-fet-jlh-class.html#post2839388
Performance quite OK in Spice at least.
Patrick
PS
No, I have not built, and it probably would not be a new project.
Too many on the queue already.
Patrick:
In your opinion, what's the better "use" of the voltage drop, the follower as you propose or cap multiplier?
The follower I proposed is essentially a cap multiplier using UHC MOSFETs.
You can always use a Darlington instead.
But I like FETs as they allow high impedance at the gate.
Whether follower is better or feedback regulator or CRCis a matter of taste.
My current reference system has CRCRC.
Happy New Year to all,
Patrick
You can always use a Darlington instead.
But I like FETs as they allow high impedance at the gate.
Whether follower is better or feedback regulator or CRCis a matter of taste.
My current reference system has CRCRC.
Happy New Year to all,
Patrick
Interesting. Which capacitor gets multiplied? I'm used to seeing a capacitor on the base of a high gain transistor, such that the "perceived" (I'm not sure what the proper word is) capacitance is a function of the beta of the device times the base capacitor value.
In this case using a MOSFET, what is the value of the resulting capacitance?
> I'm used to seeing a capacitor on the base of a high gain transistor, such that the "perceived" (I'm not sure what the proper word is) capacitance is a function of the beta of the device times the base capacitor value.
IMHO this is not the correct interpretation.
The BJT CM is nothing more than a emitter follower with its base locked at a filter voltage by an upstream RCRC filter. Whether you use a BJT (Darlington) or MOSFET is immaterial. Usually MOSFETs have a lower transconductance than BJT with the same current, hence higher output impedance. This is why people prefer BJTs. The MOSFETs I chose has a transconductance of 10S at 4A, which I consider sufficient.
So the input cap gets "multiplied not because of current gain of the pass device, but by the additional filtering action of the RCRC (and if you so wish R-C-R-ZD//C) which appears to act like a much larger bank of caps.
At least this is my understanding.
Happy New Year,
Patrick
IMHO this is not the correct interpretation.
The BJT CM is nothing more than a emitter follower with its base locked at a filter voltage by an upstream RCRC filter. Whether you use a BJT (Darlington) or MOSFET is immaterial. Usually MOSFETs have a lower transconductance than BJT with the same current, hence higher output impedance. This is why people prefer BJTs. The MOSFETs I chose has a transconductance of 10S at 4A, which I consider sufficient.
So the input cap gets "multiplied not because of current gain of the pass device, but by the additional filtering action of the RCRC (and if you so wish R-C-R-ZD//C) which appears to act like a much larger bank of caps.
At least this is my understanding.
Happy New Year,
Patrick
Patrick, thankyou for that simple and logical explanation of the capX, I have implemented them, but have been a little confused by the naming and they were a bit of a mystery to me, but thinking of them as multiplying the smoothing effect of the filter in the loop is much easier to understand. so 'in effect' the filtering is as of a larger bank of caps, but if you were to draw on them you wouldnt get the added storage of the extra capacitance. so it only amplifies the smoothing aspect, not the storage.
I do not recall publishing a regulator using 2SJ201 ??
But in any case, you can pretty much use any MOSFET in those positions.
Only two factors need to be consider :
1. The Vgs at operating current will determine the minimum voltage drop.
2. The transconductance will determine the output impedance.
Patrick
But in any case, you can pretty much use any MOSFET in those positions.
Only two factors need to be consider :
1. The Vgs at operating current will determine the minimum voltage drop.
2. The transconductance will determine the output impedance.
Patrick
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