Couple PSU questions

[daz]

Been trying to ask questions elsewhere because i don't want to dominate the talk here with my never ending questions. But you guys are simply better than any other forums, so i had to start another thread here due to lousy or lack of answers elsewhere.

So anyways, i have 2 questions. one, with a sag resistor right after the rectifier i was told i can determine the amount of current being drawn from my PT by dividing the resistor's value (180R) by the voltage drop across it. I don't see how that can be determined w/o the voltage in the equasion, but this is what several people told me. It drops 22v across the 180R, so 22/180 gets me 122ma this correct? Seems too low with a pair of EL34's, but i'm just clueless about this. second, why is it i have a 22v drop across the resistor yet when i measure my plates with and w/o the resistor (i have a defeat switch on the R) i het about a 50v drop instead of the 22 i measure across the 180R?

[defaced]

I think I understand your question so I'll take a stab at answering it.

I don't see how that can be determined w/o the voltage in the equasion

Voltage is in the equation, it's the voltage across the resistor. The B+ voltage doesn't make any difference. Consider two nodes connected by your 180R resistor with various voltages with respect to ground:
22v - 0v = 22v
372v - 350v = 22v
100000022 - 100000000 = 22v

In all of these scenarios, your 180R resistor will draw 122ma because the voltage of interest isn't between the nodes and ground, but the difference between each node. I posted a thread about choke voltage rating the dealt specifically with this difference.

What does the B+ do when you switch in the resistor?

[loudthud]

To illustrate the power of Duncan's PSU designer, I submit two simulations. What I can't show due to the size of my laptop's display is the data in the window at the lower left of the program's window. For each parameter that you can display, the program computes the Maximum, Minimum Mean and RMS values across the simulation (time) window. The time window should be an integral number of cycles. I used a resistive load to simulate the effect of the current going down as the voltage goes down. In the first sim I used an RC filter of 180 ohms and 40uF and in the second sim I just changed the resistor to 1 ohm. Note that in the simulation with the 1 ohm resistor, the voltage at the output of the transformer starts to distort at the peak and the peak voltage is reduced.

With 1 ohm sag resistor

V(C1).....V(C1)......V(R1).....V(R1).....V(T1)
Mean......Pk-Pk......Max.......Mean......Peak

376........15.3........0.567.....0.075.....383

With 180 ohm sag resistor

V(C1).....V(C1)......V(R1).....V(R1).....V(T1)
Mean......Pk-Pk......Max.......Mean......Peak

335........11.2.......51..........12..........387

When you measure across the Sag resistor, you meter probably takes the Mean (average) value. The peak value (51V) is what reduces the voltage at the output, but the average value (12V) is what your meter reads. Note that switching in the Sag resistor reduces the B+ by 41V, but across the resistor you only measure 12V.

The answer to your question is that because the current through the Sag resistor is pulsed, normal rules about DC voltage drops don't apply.

Edit: I may not have given a clear answer to your question. The amount of voltage that the B+ drops due to the sag resistor is a function of the peak voltage drop across the resistor. Some is due to the slightly different load that the transformer sees, it distorts a little with no sag resistor. The current can be measured by measuring the voltage across the sag resistor but your meter must take the average reading and not the peak.

[daz]

Ok, i guess this is all just gonna be over my head. But the one thing that caught my attention was....

In all of these scenarios, your 180R resistor will draw 122ma....

that worries me because i assume you are saying that with this resistor my PT is now being asked to put out 122ma above and beyond what it's putting out with no sag resistor? I don't know what the total the amp draws without the R, but whatever it is i imagine that when you add 122ma to it i would be past the PT's rating of 250ma. On the other hand, i never felt it get hot. I just want to be sure that adding that resistor doesn't put a strain on the PT. If it does, what value would you guys suggest i use instead? Would going to a 100R be much safer? sadly, before i this i wanted to change it out for a 270R i have because the sag the 180 provides isn't quite as much as i wanted. Guess thats out of the question now eh?

[Chuck H]

The no tech explaination is like water running down a stream. Bolders in the way would represent resistors. The bolders do absorb some energy but the flow isn't greater because of it. The water on the other side of the bolder has to stabilize before it can continue downstream. But it's the downhill force of gravity that determines the overall amount of flow. Not the bolders. The gravitational force pulling the water downstream represents your tubes, their bias conditions and operation. The resistor is just in the way.

Thats why in your application it's called a sag resistor. If you used a big enough resistor your amp would stop working all together. Your amp is drawing current through that resistor. The resistor, being a passive device, can't draw any current. The load on your PT is the same with or without the resistor. Well...Not exactly. The resistor does change the amps behavior and that changes the current drawn during variables in operating conditions. Thats why you put it in there. But the current drawn from the PT is BASICALLY the same with or without the resistor.

Chuck

[daz]

Now THATS answer I needed! Thanks.

[defaced]

that worries me because i assume you are saying that with this resistor my PT is now being asked to put out 122ma above and beyond what it's putting out with no sag resistor?

I could have worded that better. A better way to say it would be that in each case there is 122ma of current flowing through the resistor. Chuck is right, in this case, the resistor isn't changing the total because it's forcing the voltage to sag. If the voltage held constant, like in a regulated power supply, then current would drop off with added resistance.

Basically all the resistor is doing is changing the V-I slope of your power supply. The greater R, the greater the slope of the V-I curve, the more voltage dropped for a given current draw.

[daz]

It's a good thing too because the 180R didn't really create much sag, and it was noticable but hard to tell at times A/B'ing it with the switch. So i was disappointed thinking i couldn't raise it so it was much saggier. I just put a 270R in and now it's much easier to tell. High gain tones especially are nice and springy.

[black_labb]

just out of curiousity, do you have a cap in before the resistor? i usually put a cap in before the resistor to increase the filtering on the dc. by using a big cap at the input, and then smaller ones after the resistor you will get plenty of sag, and still plenty of dc filtering. this may not be neccesary if you have a fair bit of filtering through the system already. it also means you have a stiffer supply when the resistor is shorted.

[daz]

No, but with a 270R in there as of yesterday there is plenty of sag now.