SSLV1.1 builds & fairy tales

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merlin el mago

merlin el mago

Member
Joined 2009
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Salas said:
The guide writes 1K8 resistor also. Is that resistor 1K8? Anyway, add 1K instead of the Zener jumper, should give you enough conversion to reach 24V with ease.

P.S. I see you edited. The examples in the guides state they assume a certain in circuit IDSS. If a FET chances a bit weaker or stronger it takes compensation in the values.
Click to expand...

Negative rail 1K805
Positive rail 1K797

Negative rail trimmer 5K42
Positive rail trimmer 5K13

I suspect JFETs I will follow your advice adding 1K resistor instead diode zener jumper
 
Salas

Salas

diyAudio Chief Moderator
Joined 2002
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You must always calculate the dissipation and the temperature rise at least roughly as in the guide. In this case of your 384mA @ 24V with ~90mA dummy consumption you will get 4.5C/W times 7.2W =32.4C above ambient in the output Mosfet sink. High for long term reliability. Those PCBs are designed with best layout in mind and enough dissipation. So long side or vertically parallel heatsinks, or thick metal floor are best. Small individual ones are for less than 200mA up to 15V. Get a 3 stage M3 thread taps set and make nice blind threaded holes on thick metal long sinks or make see through holes to thinner surfaces and arrange PSUs nicely and securely on good ventilation places when you will finalize constructions.
 
Salas

Salas

diyAudio Chief Moderator
Joined 2002
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The loss value introduced by the 1 Ohm resistor on the MKT there is always known and stable regardless of ambient temperature, aging. Using a 4.7u to 10u electrolytic, loss would range a lot between brands and types and be much more susceptible to alteration than with an MKT+resistor. The output local termination impacts OLG bandwidth, phase margin, and its a sensitive matter in this design if to perform as intended.
 
N

necplusultra

Member
Joined 2011
thanks Salas,

that dummy load comes out to 15W 10r (unless my formula or calculations are wrong)

tried one test with single 5w 680r
and another with single 10w 2k2r


the irfs get pretty warm. the 10r also gets hot but not as much as the irf. sinks may be a bit small.

thanks again, learning a great deal from your high quality products.
all works now.
 
Salas

Salas

diyAudio Chief Moderator
Joined 2002
Paid Member
By using 10 Ohm across 18V, your dummy was asking for 1.8A. You had shorted the output effectively when you had much less CCS of course. On a positive note, your build had just passed a hard short test. :up: What is it for? OPAMPS circuit? What will be the real load current demands?
 
N

necplusultra

Member
Joined 2011
This was very good learning for me.

was measuring the current and effectively got about 28 mA (18v / 680r) and 8 mA (18v / 2200r)

So if I understand correctly my 10R R301 is limitting the setup to 200mA
My max dummy load would then be 90r (18v / 90r = 200mA)

Since I was using a dummy load of 15W 10r, it was overpowering the 10r r301.
What is the formula to calculate the r301?

PS: Don't really have anything planned for the build. just a testing and learning rig. It'll probably end up in some kind of dac build later.


thanks again!
 
Salas

Salas

diyAudio Chief Moderator
Joined 2002
Paid Member
You did not actually overpower the 10R setting, you just devoured all its output. The CCS acts as a current limiter along other useful things it does here. Thus, this reg can survive an output short as you experienced. You drove all its available CCS current to ground. Nothing was left for the parallel voltage circuit to operate, so you could not see the 2 red leds light up, and was unable to fix a voltage.
Your dummy load consumption should never match your CCS current. Because it will consume it all, leaving zero for the reg's own needs. 25mA is about the shortest stable operating distance to CCS you should reach. 100mA is a decent distance for base performance. Your dummy load should match the consumption of your real projected load. If for learning purposes and no real app in mind yet, a good balanced experiment in your case would be to assign 80mA at 18V output by using a 220R 5W dummy. A 10 Ohm R301 could give anything from 120mA-200mA depending on various characteristics of the associated LEDS and Mosfet. The formula is you measure voltage across the setting resistor and then you divide by its value. I.e. R301Volts/R301Value=CCS current. The bigger the part of the CCS current the dummy uses, the less heat you will notice on the output Mosfet, and more heat on the dummy itself. Did you read the guide thoroughly by the way?
 
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