The simplistic Salas low voltage shunt regulator - Page 554 - diyAudio
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Old 1st November 2012, 01:51 PM   #5531
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The idea behind the Reflektor (correct me if I am wrong) is to reflect the current change in the reference leg (with the LED), in the regulator leg (where the FET is) in order to counteract a change in output voltage. So the mechanism relies on current change in the reference leg. This current change is greater when the dynamic impedance (dV/dI) in this leg is lower. That is the main reason I suppose why the LED is there. Since Vgs of the IRF640 is approximately between 4 and 4.5V, wouldn’t it improve the regulation of the FET when for exaple two red LEDS in series were used (together approx. 3.2V and well below 4V) instead of one green LED? That would lower the dynamic impedance in that leg (you need less series resistance) and thus cause more current change, which could improve the regulation. Is it a good idea to boost current change anyway or am I missing important design considerations?

Peter
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Old 1st November 2012, 05:31 PM   #5532
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I realize now that the capacitor across the LED and resistor does that job, but that is very frequency dependant. It would be more elegant to have a more frequency independent solution.

Peter
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Old 1st November 2012, 05:49 PM   #5533
Salas is offline Salas  Greece
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More LEDS will push tempco more negative also. You need some appreciable and well judged for gate capacitance drive load impedance to the output side for some gain. Hence the 1K. There lies the gain VS frequency bottleneck.
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Old 2nd November 2012, 06:52 AM   #5534
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Quote:
Originally Posted by Salas View Post
For that purpose OK but to generally sink it, plus the dioded 4401s dissipation roof, small current stuff. How hot is it now for how much current you spare through it?
I was worried about same points before making.
In my case constant current is set on 115mA, and the current of target board is 10mA. MOSFET is running without heat sink. Touching finger, temperature of 4401s and MOSFET is a little bit warm.If more current is needed, they must be changed to another one which is more power dissipation.
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Old 2nd November 2012, 06:58 AM   #5535
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Quote:
Originally Posted by pietjers View Post
Sabre DACS can sound fantastic with the right power supplies. The sound is really as good as power supplies go. With this DAC you (even) get rewarded for every subtle change you make in your power supply.

Peter
Thanks pietjaers.
I have same opinion with you. Sabre DAC is strongly affected by quality of power supply.

Good luck.
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Old 7th November 2012, 07:45 AM   #5536
alibear is offline alibear  United Kingdom
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Hi all, I want to supply the input, VAS and pre-driver stages of a class A amplifier with + and - 37V using a Version 1 regulator. The DC voltage I have after rectafacation and filtering is + and - 51V. How do I calculate the value of R1 if I require 250mA ( or any other current ). Also at a later stage I may use this supply for the drivers where more current is needed if so would it be OK to select R1 for this higher current even if initially I do not require it? I appreciate that the shunt transistor will dissapate more heat .
Thanks for your help
Alan
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Old 7th November 2012, 08:33 AM   #5537
Salas is offline Salas  Greece
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If you mean a post 1 like arrangement then Iccs=Vbe/R1. Using 100mA or more spare helps for lower output impedance. Its more about how much thermal burden your sinking and your consumption scheme may allow.
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Old 7th November 2012, 10:49 AM   #5538
AndrewT is offline AndrewT  Scotland
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your +-51Vdc is likely to rise to ~+-53Vdc worst case mains voltage.
This combined with 100mA for the CCS will dissipate a worst case 5.3W in each CCS mosFET if the output gets shorted.
Decide whether you want the mosFET to survive this worst case scenario. Then select the heatsink appropriately. If you go for this "survivability" and the heatsink is isolated from the chassis, then the FET can be bolted/clamped direct to the sink for lowest Rth c-s.
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Old 7th November 2012, 12:08 PM   #5539
alibear is offline alibear  United Kingdom
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Thanks for the replies. Attached is the circuit I will be using.Sorry I am so dumb, but i still do not understand how to calculate the value of R1.
Thanks
Alan.
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File Type: gif 37vshunt.gif (6.8 KB, 278 views)
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Old 7th November 2012, 12:52 PM   #5540
AndrewT is offline AndrewT  Scotland
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Vr1+VgsQ1 = Vledstring.
Arithmetic/algebra allows you to re-arrange the equation (or equality) to:
Vr1 = Vledstring - VgsQ1.

Vr1 also equals the current through the resistor times the resistance of R1, i.e.
Vr1 = R1 * Ir1
and re-arranged to give
R1 = Vr1 / Ir1

Put that back into our earlier re-arrangement and we get

R1 = [Vledstring - VgsQ1] / Ir1

You need Vledstring and VgsQ1 and Ir1

Let's assume, until you can measure your circuit, values for these knowns
Vledstring ~ 5.7Vdc
Ir1 = CCS current = 100mAdc = 0.1Adc
Vgs ~ 4Vdc

Put these into our equation and we get
R1 = [5.7 - 4.0] / 0.1 = 1.7 / 0.1 = 17r nearest e24 value is 18r.

Adopt 18r, or 20r, or 22r, or 24r, or 27r, or 30r, or 33r, or 36r.

That now allows you to directly measure that guessed at voltage of 1.7V across the R1 resistor.
Once you have that voltage you can calculate the actual CCS current.
If it's not big enough, then ADD second resistor that gives the extra current you want to give a total CCS current ~ =100mA.

Again lets suppose you measure 1.71Vdc across a 24r resistor in position R1.

CCS current = 1.72Vdc / 24r = 0.0716666 repeating ~ = 72mAdc.
You need an extra ~28mAdc (=0.026Adc).

1.72Vdc / 0.028 ~ 61r4.
Adopt 62r (nearest e24 value).
add this in parallel to the R1 already installed.
Remeasure Vr1. calculate the new CCS current. Is it close enough to your target?

Before closing let's check dissipation in R1.

Pq = V^2 / R = 1.72*1.72 / 24 = 0.123W
Use a 500mW or 600mW resistor.

If you had adopted a 18r and your Vr1 were found to be 1.95Vdc then Pq = 1.95^2 / 18 = 211mW. a 500mW resistor just about works but will run hot, too hot to hold with a finger pressed on.

that's the nice thing about using resistors in parallel. Each resistor runs considerably cooler than a single. eg two 36r in parallel for an effective 18r will dissipate 1.95^2 / 36 = 106mW. A 500mW resistor is OK.
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Last edited by AndrewT; 7th November 2012 at 01:01 PM.
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