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Total recall power wiring help

Started by zachlovescoffee, December 19, 2021, 04:34:34 AM

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Zerro

#15
Stabil or stabilizator - yes it is another word for your regulator, thouhg it doesn't regulate any values, only makes one value firm, independently at input values. I am sorry for this confusing you. The "butt" of stabil is here pin 1, sitting at ground. Against this pin you will read all values. So, at input pin 2 you will have some voltage from power supply - you said that you use cca 34V. So here you will have cca 30V (after input resistor). At pin 3 will be output, against ground. So, here 15V. All according picture I attached. BTW, if you use somewhere 34V, filtering caps must have by cca 25% voltage values higher. Here for 34V at input you may need use caps for 40V or higher.
"Nudíte se? Kupte si našeho cvičeného ježka! Pobaví vás svými veselýmí kousky!"

zachlovescoffee

I just took the readings as followings:

Pin 3 across pin 1: -15.09vdc
Pin 3 across pin 2: -29.08vdc

I think that looks right. Do you think I should still increase the size of R57 in this case? Shall I move on to taking IC readings bows?

madbean

Quote from: zachlovescoffee on December 21, 2021, 01:41:52 AM
I just took the readings as followings:

Pin 3 across pin 1: -15.09vdc
Pin 3 across pin 2: -29.08vdc

I think that looks right. Do you think I should still increase the size of R57 in this case? Shall I move on to taking IC readings bows?

What wattage is your 240R resistor?

zachlovescoffee


Zerro

#19
Unhapilly I don't know what is comperhensive current consumtion of all vehement. So, I don't know what power is lost at this R57. Maybe Madbean will know. If your power has 34V, and after this resistor you have 29V, at this resistor is lost 5V with some current, I just don't know. This will give power lost in heat (current x voltage) Let's say that 100mA is proper value. With 5V lost at R57 it will give 500mW lost in heat. R57 must have mimimally 2x higher calibration - so here it is 1W. But if you will change polarity (by mistake), that protecting diode D1 will make shortage of this at ground, to protect circuit from bad polarity. Not to be D1 damaged, this resistor must lower current thru this diode at safe value. If you will use 34V, and diode can withstand 200mA (catalogue value), you will need 160 Ohm minimal value. But better raise this value for some safety at cca 270 Ohm. (Simplyfied, power lost at this resistor 160 Ohm when shortage, is 34V x 200mA= 7W! When resistor is 270 Ohm, lost power is cca 4W. For a moment it can protect your system. Dependig at max. current that power source can really give, and it's anti-shortage protection, of course!

I hope I didn't bore you with my article :@) I recommend to use 270 Ohm, 4W.
"Nudíte se? Kupte si našeho cvičeného ježka! Pobaví vás svými veselýmí kousky!"

zachlovescoffee

I got a 270 ohm @ 2 watt resistor. But, can someone confirm that the voltages I posted most recently appear to be correct?

I'd like to keep moving forward with the build if I am within the margin of safe operation.

jimilee

#21
According to the build doc, the 10r is supposed to be 1/2 watt and the 240r is supposed to be 1watt. 2 is plenty good. I'll dig out my power supply, can you clarify what to voltages you are referring? There are a lot listed.

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zachlovescoffee

Thanks! My resistors should be in spec. Readings from the regulator:

Pin 3 across pin 1: -15.09vdc
Pin 3 across pin 2: -29.08vdc

Zerro

"Nudíte se? Kupte si našeho cvičeného ježka! Pobaví vás svými veselýmí kousky!"

danfrank

It's been a long time but the total recall consumes areound 50-75ma depending on how the delay pot is set. It is DEFINITELY under 100ma consumption.

Zerro

Quote from: danfrank on December 26, 2021, 01:35:40 AM
It's been a long time but the total recall consumes areound 50-75ma depending on how the delay pot is set. It is DEFINITELY under 100ma consumption.

Thank you, so resistors I recommended are sufficient.
"Nudíte se? Kupte si našeho cvičeného ježka! Pobaví vás svými veselýmí kousky!"