What are the series pair of 270k resistors and 300uF caps just after the rectifier and just before the standby switch for? Is this a bleeder circuit? How would you model the voltage drop across this in PSUD, as a current tap using ohms law?
I have a power tranny that I want to drop the b+ on a bit and was wondering if this "bleeder circuit" would do that anyways and save me the work.
Thanks, ER
#124, bleeder circuit in PS?
Moderators: pompeiisneaks, Colossal
Re: #124, bleeder circuit in PS?
The two large caps are wired in series to form "one large cap" that filters the raw pulsating B+ voltage coming off the rectifier. Wiring these caps in series does two things.
1. It doubles the voltage rating of the "cap" so the voltage rating of the individual caps add. If the two caps were rating for 300V each, when put in series that series caps could safely see 600V.
2. It halves the capatinence, so 150mF if both caps were 300mF.
The resistors are balancers so both caps see the same amount of voltage. it would be pointless and defeat the purpose of series caps if one caps saw 300V and the other say 140V for example. The voltage across each cap needs to be equal, that's where the two balancing resistors (of the same value) are used.
The resistors do also act as bleeders when the amp is off. They are a voltage divider essentially, you ll find half the B+ at the node where the two resistors connect. They don't act as any significant load when the amp is on though. They do provide a place for voltage to "bleed" out when the amp is off.
You want a linear voltage divider so both R1 and R2 are equal.
Voltage drop across R1 =
(R1/R1+R2)*Vcc (Vcc is just the B+)
1. It doubles the voltage rating of the "cap" so the voltage rating of the individual caps add. If the two caps were rating for 300V each, when put in series that series caps could safely see 600V.
2. It halves the capatinence, so 150mF if both caps were 300mF.
The resistors are balancers so both caps see the same amount of voltage. it would be pointless and defeat the purpose of series caps if one caps saw 300V and the other say 140V for example. The voltage across each cap needs to be equal, that's where the two balancing resistors (of the same value) are used.
The resistors do also act as bleeders when the amp is off. They are a voltage divider essentially, you ll find half the B+ at the node where the two resistors connect. They don't act as any significant load when the amp is on though. They do provide a place for voltage to "bleed" out when the amp is off.
You want a linear voltage divider so both R1 and R2 are equal.
Voltage drop across R1 =
(R1/R1+R2)*Vcc (Vcc is just the B+)