PSU question

[milen007]

Hi Guys

i need help on psu.
attached below is the psu with:

according to duncan psud 2:
power transfomer is 137VAC
voltage across C1 is 189vdc
voltage across C2 is 62.8VDC

while in real circuit:
power transfomer is 137vac
voltage across C1 is 185VDC (good)
voltage across C2 is 137VDC (???)

my question:
why the voltage across C2 in psud2 and in real circuit is different? i can understand the few voltage different on C1, but there is 74.2v different across C2.

while psud2 needs R1=1.8K to drop the voltage in C2 to 137VDC. while in real circuit uses R1=10K

please help..

[milen007]

the psu modelled above is for tube buffer 2x6j1. details in link below:
Tube Buffer 6J1 / 5654 (6AK5)

while in the other case in link below:
http://www.diyaudio.com/forums/showt...42#post1464542

and the attachment is the model in duncan psud 2
they matches quite close
in psud2 result:
power transformer is 141VAC
voltage across C1 is 195 VDC
voltage across C2 is 175 VDC
voltage across C3 is 142 VDC

thanks

[richie00boy]

You need a load on the real circuit for voltage to drop as at the moment there is almost no current flow.

[zoranaudio]

Just be careful with that simulation, the real voltage is sometimes different in the real time, and you can damage the vacuum tube, anode voltage!
Make a real circuit with tube and than measure all voltages without speaker load and (static) dc current( biasing current) and with maximum speaker load>maximum input load or driving for full power.
Of course that with full load power this voltages will drop but it is important how much( good drop is around 10 % of maximum no-load voltage, if the voltage drop 20 or more % it is not good.
Anyhow you should never go with more anode voltages than the tube manufacturer rated that tube.
Once I destroyed both tubes in my Yaesu ft 101 (Ham radio transceiver) with increased voltages 50 or little bit more volts.
If voltages drop too much than is reccomended more powerfull transformer with more current and smaller dc resistance.
Be careful, and good luck!
Regards ,
Z37hwx, operator Zoki
http://www.qrz.com/z37hwx
73.

[milen007]

Originally posted by richie00boy
You need a load on the real circuit for voltage to drop as at the moment there is almost no current flow.

Hi richie00boy, the real circuit is loaded with 2x6j1. while the simulation is unloaded and the sim give out much lower voltage which puzzle me. if the real circuit that drop that much voltage then it maybe caused by the load. but in this case its the other way around.

Originally posted by zoranaudio
Just be careful with that simulation, the real voltage is sometimes different in the real time, and you can damage the vacuum tube, anode voltage!
Make a real circuit with tube and than measure all voltages without speaker load and (static) dc current( biasing current) and with maximum speaker load>maximum input load or driving for full power.
Of course that with full load power this voltages will drop but it is important how much( good drop is around 10 % of maximum no-load voltage, if the voltage drop 20 or more % it is not good.
Anyhow you should never go with more anode voltages than the tube manufacturer rated that tube.
Once I destroyed both tubes in my Yaesu ft 101 (Ham radio transceiver) with increased voltages 50 or little bit more volts.
If voltages drop too much than is reccomended more powerfull transformer with more current and smaller dc resistance.
Be careful, and good luck!
Regards ,
Z37hwx, operator Zoki
http://www.qrz.com/z37hwx
73.

Hi zoranaudio

i use sim as guideline only. on this stage of my limited understanding, i only use tested schematic. thx for the tips, i'll keep those in mind. Hence, use bigger PT

[milen007]

Anyone know how much the voltage drop by 10K resistor in 185VDC power supply? they drop 50VDC or 125VDC?

thanks

[zoranaudio]

Hi!
Well this question can be answered with 100% right value, this will be 100 % same in reality.
Very simple:OHM LAW
Voltage drop on the resistors depends from the current passing threw that resistors and the formula is:
Vr=Ir*R
where:
Vr is voltage drop on the resistor
Ir is the current passed threw the resistor in amperes (A)
R is resistance of the resistors (converted in OHMS)
So you need to know the value of the current in (A)- amperes and multiply Ir*R.
Example:
R=10K=10000 ohms
Ir=10mA=0.01 A
Vr=0.01*10000=100V.

Or for the 50 volts voltage drop on the 10K resistor the current will be half or 0.005A or 5mA.
Is it clear now mister?
You need to now the current load threw the resistor and calculate the voltage drop.

[milen007]

Hi zoranaudio.

Thanks, this is what i am looking for. thx

this tube buffer i am analysing looks like only draw 5mA. great