Dropping string question

[Tdale]

This is a little embarresing...but here we go.

I'm having some trouble understanding some of the power supplies. I know the basics of a dropping string...if you have 9V between + and -, and put three equal resistors in series between them, you will measure 3V across each one, or 6V if you measure across two of them.

What I don't get is this. The last B+ (usually B4 or B5) is taken after the last resistor in the string. The voltage is usually also the lowest of the string. But how can that be? Since it's taken after the last resistor, sholdn't the voltage be 0....since that resistor is grounded through the filter cap?

Here is an example: http://ampgarage.com/forum/download/file.php?id=14586

Where are the two voltages measured between? Between ground and B-tap or between + and B-tap?

Anyone care to help me here?

[TheHandsomeOrk]

Caps block DC and allow AC to pass

[Tdale]

I know, what i Asker about is how the voltages Are measured, or sverd. I have read solen, and i Think the reason is that i forgot to add the plate resistors in the equation...

[greenbottle]

Have a look at the example you cited(Dumbleator).
There are two stages:- the top one has 200 and somthing volts
the bottom one has 300 and somthing volts.

Each of these stages are biased on so that the input signal causes a variation of voltage on the grid. So each stage has a quiescent current.

Lets say for the sake of the argument that the top stage is drawing "X" current and the bottom one is drawing "Y" current. The current through the last resistor is "x" current but the current through the first resistor is the total of "x"and "y". These voltage drops are a result of the Quiescent currents.

Basically Ohms Law.

When a signal is applied to a stage you get an AC component.
The capacitors are to a degree transparent to this AC component .( see the post above)

Hope clarifies things re your inquiry. If not, keep asking questions.

The answers help fill the holes in other peoples knowledge who are not game enough to ask to ask the questions.

Alan

[Tdale]

Hmm. I think my biggest problem is how to explain what I don't understand

The two voltages, 256 and 368; Can they be measured by using the multimeter, measuring between ground and each point?

Basically what I'm trying to understand, is how you can get 256 volts AFTER the last resistor, and not before.......

Tommy

[Tdale]

Let me try..

Do I have to look at the whole circuit, from dropping resistor, through plate resistor, through the tube, and then to ground, to understand how the voltage is divided? If so, that makes more sense...

Tommy

[passfan]

What the handsomeork is trying to say is the last resistor drops to 256 or so volts dc instead of 0 because a cap only passes AC. Since it blocks DC the dc voltage never sees ground. The AC ripple (which represents itself as noise riding on your DC voltage, and is a bad thing) does see ground and is drained off. That, as well as current reserve, is what the filter caps mainly do.

[surfsup]

To get the voltage after, or through, a resistor... V = I*R

I = current
R = resistance

For the first R, it is 15k, 2W resistor so R = 15000

You need to know the current draw, as someone said, fro this resistor, it sees the current from the first triode, and the second, so both currents would be summed to give a total.

The voltage drop would then be V = IR = I*15000

For the second, you have a drop of 368-256 = Vdrop = 112

V = IR (here we will sum for I)
V = 112 = I*100,000
I = 112/100000 = 1.12mA

If we assume both triodes to pull about 1.25 mA, you have a total for both of 2.5mA and putting that into equation 1 you have:

V = IR = 0.0025*15000 = 37.5 Volts dropped.

So since the supply node there should read roughly 368V (nothing is ever going to be perfect), you have a supply after rectfication of about 405V

This would tell you, using that style of rectification, what transformer V you would need based off your mains Voltage.

[surfsup]

BTW, what is this dumbleator? It looks like an outboard FX loop. If it is I think I gotta build that!

[Tdale]

So when you're measuring those two voltages, you'r really measuring the voltage across the plate resistor...to simplify it..?

Tommy

[Structo]

And yes, to measure the various DC nodes of B+ you set your meter to read DC volts, clip your negative meter lead to chassis ground and touch the red probe to the node you want to measure.

The Dumbleator is a bit different in that the cathode follower (the first stage) doesn't use a plate load resistor.
So you have to make sure the voltage at the power supply is correct.

Another way to think of a power supply B+ string is that it is a kind of ladder and the voltages usually get succeedingly smaller as the string goes along.

[markr14850]

Please remember that the voltages we're talking about here are lethal.

Capacitors can hold a charge for quite a while after the unit is off and unplugged.

If you are new to electronics, please use caution. Maybe start with lower voltage circuits.

[Tdale]

I've made several amps, and know about the filter caps etc. But since I just follow schematcs, and don't design them, I never really put too much thought into what actually happenes in the power supply section.

My whole "point" is that to measure votage, you need to measure across some resistance. And when the voltage is 256V after the last resistor in the string, I assumed that there were 256V between that point and ground, with no resistor in between. But there is of course a plate resistor in between, which I forgot to add to the equation.

But in a cathode follower there is no plate resistor..but there is a cathode resistor, with a much higher value than the usual 1.5K, which the voltage is measured across...

Or am I not getting this still?

[Firestorm]

Another way to think of a power supply B+ string is that it is a kind of ladder and the voltages usually get succeedingly smaller as the string goes along.

I always liked the central vacuum analogy: the rectifier is the central vac and the B+ wiring is like the pipes going through the house. The rectifier is trying to suck every electron off every cathode in the amp. When you suck away negatively charged electrons, the remaining charge is relatively positive. The resistors are like flow restrictors that limit how much "suction" is available at a given point.

[Firestorm]

My whole "point" is that to measure votage, you need to measure across some resistance.

Not really. Voltage is just the difference in potential between one point and another. If you connect your rectifier to the main filter cap and make no other connection, you will still read, for example, 450VDC at the cap anode relative to ground. The cap is 450 volts more positive than ground. (Or 450 volts less negative).

Also remember that the tubes themselves are a high resistance connection to ground: the anodes pull electrons from the cathode, which can be at ground potential (in which case the cathode will read 0), or more commonly a bit above ground potential if there is a cathode resistor. A small value resistor will result in a small positive voltage; a large resistor (like in a cathode follower circuit) will result in a large voltage.