Unregulated power supply output impedance

[jackinnj]

Quite a while back Fred Dieckmann illustrated a fixture for testing power supply regulators and it can be employed to measure the impedance of your supply.



The MOSFET is biased into Class-A so must be heat sinked.

[AJT]

Originally Posted by Navyblue

Thanks again.

The power transformer is rated at 230V, when loaded it is around 200V. Secondary windings DC resistance is 220 ohms per winding.

Chokes' DC resistance is 90 ohms for both.

The chain of 390 ohms resistors, it draws about another 70mA.

If you wonder why the resistor chain is wasting so much current, I did this to cut down the supply voltage. If you wonder why I didn't cut voltage the usual way (putting them in series with the chokes), I observed a noticeable improvement in bass when comparing between the 2. I thought it interesting, and this made me look into the effect of Zout of PSU. But this also keep the load DC resistance low, so I plan to get rid of it.

Of course in theory I should just dump the power transformer, replace it with one with appropriate rating, and do without the resistors. but that cost money.

So total current draw is around 140mA, at 110V, it would be about 786 ohms overall.

so are you saying you have a traffo with 230-0-230 secondary?

since you have a choke input filter, raw B+ is around 200vdc....you need to lose 90volts..

to cut down on voltage, you can put additional RC in the B+ line......it will be more effective that way in bringing ripple voltage down....

you can also use an 0A2 to shunt regulate the B+.....

you can use higher valued resistors to bias your filament with dc voltage, say 270k to B+ and 82k to ground.....the junction connects to 2 470ohm resistors to each leg of the filament supply....this ensures that current drain is low...

do not dump your traffo just as yet....resistors are cheaper.....

[Navyblue]

Originally Posted by Tony

so are you saying you have a traffo with 230-0-230 secondary?

Something like that, specifically I used 2 traffos with 230V secondary)

Originally Posted by Tony

since you have a choke input filter, raw B+ is around 200vdc....you need to lose 90volts..

to cut down on voltage, you can put additional RC in the B+ line......it will be more effective that way in bringing ripple voltage down....

I did tried that before I got to this. A lot of series resistance audibly affected the bass. The bass has that rounded sound. By switching to this arrangement, bass sounded "normal" again. This led me to investigate about this.

Originally Posted by Tony

you can also use an 0A2 to shunt regulate the B+.....

I didn't know something like this existed. I tried looking for a schematic on how to use it, no luck so far. Does it work like a Zener diode?

Originally Posted by Tony

you can use higher valued resistors to bias your filament with dc voltage, say 270k to B+ and 82k to ground.....the junction connects to 2 470ohm resistors to each leg of the filament supply....this ensures that current drain is low...

Yes that is the "normal" way, I was just trying to kill 2 birds with 1 stone.

Originally Posted by Tony

do not dump your traffo just as yet....resistors are cheaper.....

True. But I kinda went another way. I intend to use another tube (EL84), its operating point is more suited to my traffo, still waiting for the parts to come.

[Navyblue]

Originally Posted by jackinnj

Quite a while back Fred Dieckmann illustrated a fixture for testing power supply regulators and it can be employed to measure the impedance of your supply.



The MOSFET is biased into Class-A so must be heat sinked.

Thanks. That looked like a handy tool. Unfortunately most things SS are a bit beyond me at the moment.

[DF96]

I see you have a choke input supply. The formula I gave you was for a cap input supply. For a choke input supply it is much simpler: provided that you are taking enough current the effective DC resistance of the supply is just R's plus choke resistance etc. No factor of 5 (i.e. duty cycle) in front of R's, and the 0.005/C term disappears because the cap is being continuously charged.

[AndrewT]

The effective 40Hz impedance of the last capacitor in the PSU is given as above by the formula
Z = 1 / 2 / Pi / Freq / C
for 1mF//120uF the C=0.00112F
Freq is stated as 40Hz.
Pi ~ 3.142
Z ~ 3.6ohms.
That is ~ 1/22 of the load resistance value. Well below 1/10 stated above.

All the other factors are just small corrections to that 3.6ohms calculated. eg, esr raises it slightly. Preceding Capacitance lowers it slightly. Transformer impedance is probably not relevant.

[Navyblue]

Originally Posted by DF96

I see you have a choke input supply. The formula I gave you was for a cap input supply. For a choke input supply it is much simpler: provided that you are taking enough current the effective DC resistance of the supply is just R's plus choke resistance etc. No factor of 5 (i.e. duty cycle) in front of R's, and the 0.005/C term disappears because the cap is being continuously charged.

Thanks. Does DC damping factor matter in this case?

Originally Posted by AndrewT

The effective 40Hz impedance of the last capacitor in the PSU is given as above by the formula
Z = 1 / 2 / Pi / Freq / C
for 1mF//120uF the C=0.00112F
Freq is stated as 40Hz.
Pi ~ 3.142
Z ~ 3.6ohms.
That is ~ 1/22 of the load resistance value. Well below 1/10 stated above.

All the other factors are just small corrections to that 3.6ohms calculated. eg, esr raises it slightly. Preceding Capacitance lowers it slightly. Transformer impedance is probably not relevant.

So it all boils down the the size of the last cap? And shouldn't amplifier's current draw somehow be factored in?

[DF96]

Originally Posted by Navyblue

Does DC damping factor matter in this case?

Not sure what you mean. Damping factor is an AC concept.

[Navyblue]

Originally Posted by DF96

Not sure what you mean. Damping factor is an AC concept.

Really?

I mean does R's + choke resistance need to be below a certain level with respect to the load resistance?

[DF96]

There is no rule. It all depends on how much voltage you are willing to drop, and how much heat you are willing to dissipate in the power supply. Generally, you would want the PSU resistance to be signficantly less than the load resistance.