Thks again to everybody, without your advice I can't get it.🙂
For C2 I'm using a Panasonic FC 1000uF 63V, I can get anye better cap due the high voltage needed.
He integrated some analogue output stage for D1 with a symmetric V1 reg like Merlin's in one PCB. I am not familiar with more. Maybe you can ask CRT.
Rcruz, I am thinking to change C2 for two Elna Silmic II 470uF 63V to reach aprox. the 1000uF 63V.
For C1 BG F series 100uF 25V
For C4 Nichicon Muse Kz series 220uF 100V
All resistors in the reg. are carbon film Koa Speer.
Now I am calculating the exact value of R1 to reach the needed 200mA because with 12 ohms 2W & after 1 hour connected only reach aprox. 170mA so surely I need aprox. 17 Ohms 2W to get the 200mA I am using a big pot of 50 ohms 5W to get exact value.
For C1 BG F series 100uF 25V
For C4 Nichicon Muse Kz series 220uF 100V
All resistors in the reg. are carbon film Koa Speer.
Now I am calculating the exact value of R1 to reach the needed 200mA because with 12 ohms 2W & after 1 hour connected only reach aprox. 170mA so surely I need aprox. 17 Ohms 2W to get the 200mA I am using a big pot of 50 ohms 5W to get exact value.
Well done Mr. Mago!
Without your help I never resolved mine CCS problem, now & thks to you I resolved & I understand a little bit how operate a reg.🙂
keep the 12r that passes 170mA when warmed up.
Now consider your required 200mA. That is 30mA extra.
Just calculate what resistor value will pass 30mA with your voltage drop and add it in parallel to the 12r you already have.
hint:
expect between 63r and 74r.
hint2:
total parallel resistance should be 10r2
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Salas I haven't words to be grateful, without your kind help I never, never resolved the shunt problem🙂
Many thanks AndrewT, so for a total parallel resistance of 10R2 I have to parallel a 68R 2W to get 200mA, right?
Some calculations with ohm law:
I have a Voltage of: 62,6V
I need an Intensity of: 0,2A
R = V / I
R = 62,6V x 0,2A = 313 ohms
It's correct?
Real world (pot 50ohms 5W+DMM)said that I need 15 or 16 ohms to get aprox. 200mA???
What's wrong?
I have a Voltage of: 62,6V
I need an Intensity of: 0,2A
R = V / I
R = 62,6V x 0,2A = 313 ohms
It's correct?
Real world (pot 50ohms 5W+DMM)said that I need 15 or 16 ohms to get aprox. 200mA???
What's wrong?
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no. Volts / Amperes = Ohms.
Your calculations:
62,6V / 0,2A = 313 Ohm
use this pic for formulas
Your calculations:
62,6V / 0,2A = 313 Ohm
use this pic for formulas
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no. Volts / Amperes = Ohms.
Your calculations:
62,6V / 0,2A = 313 Ohm
use this pic for formulas
So 313 ohms is not the real value to get 200mA in mine shunt reg.😕
Your leds maybe strong, follow DVM. Its not critical to set that or this current right now. See if your phono works OK too first. When you sort out all, and get used to it, then you can experiment to find your best subjective setting VS heat dissipation.
Just stick a 10R 5W on each one for R1 and you will be OK. If your chassis will get too hot, go back to 12R, or 15R.
@merlin
measure voltage on R1. Then divide this voltage (Volts) with R1 value (Ohms) and you will get current through resistor (Amperes). You can not just stick your input voltage, then multiply with wanted current, this is not correct.
Several posts above is formula how to calculate output current. But it will be a lot easier for you to follow your DVM
measure voltage on R1. Then divide this voltage (Volts) with R1 value (Ohms) and you will get current through resistor (Amperes). You can not just stick your input voltage, then multiply with wanted current, this is not correct.
Several posts above is formula how to calculate output current. But it will be a lot easier for you to follow your DVM
Thks for the explanation, now I understand better.
Tomorrow I will use 12R 2W, so:
Voltage on R1 = 1,62V / 12R = 0,14A
I will follow Salas advice, 1st test the phono & after fine adjustment using 10R or other value to reach 0,2A
merlin el mago said:
Great 🙂
Now you can use formulas from my picture and calculate power dissipation on your 12R resistor:
Watts = Volts x Amperes
1.62V x 0,14A = 0,227W
Your resistor will need to dissipate 0,227W. Multiply this value by 4 or 5, to be into safe area and round to nearest higher standard value.
0,227W x 4 = 0,908W and with 1W resistor (or more) you will be on the safe side 😀
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