Dropping string question

[Tdale]

I get the vacuum analogy..

So when you measure a certain amount of voltage after the last resistor in any given dropping string, you're measuring across the filter cap?

I have always learned that you have to measure voltage across some load, and what is the load between the last resistor and ground?

Tommy

[Tdale]

But if you measure 450V DC across a cap, which doesn't let DC through, you're basically measuring across a "resistor"?

[Structo]

Visualize a sine wave.

The center horizontal line through the wave is '0' or zero volts.

When you rectify AC current you only allow the voltage to flow when positive.
So now that voltage is positive to the '0'.

Lets just talk half wave rectifier for now.
So you have the positive parts of the wave now above zero.
The caps filter that AC content to ground leaving a straight line of Direct Current which is above zero or ground.

When you measure any DC voltage in an amp, the reference point is usually that '0' or ground.

So for pure DC which is a straight line above zero you are measuring the difference between that and '0' or the difference in potential.

You would really benefit from Merlin Belcowe's Power Supply design book.

[Firestorm]

There has to be a resistance between two points or they will be at the same potential. The resistance can be "infinite." Your meter has an internal resistance which, A) is how it works, and B) prevents it from shorting out the circuit.

[Tdale]

I'm afraid I have to give up here... I'm not able to express myself in a way, so that you understand what mean.

I get the rectifier thing etc.

I basically wondering why the voltage is the way it is at the different B+ taps.....

We all know that if you have a 9V battery, and put 3 equal resistors in series between + and -, there will be 9V across all resistors, 6V across two of them and 3V across one of them.

+ -----A----/\/\/\/----B----/\/\/\/----C----/\/\/\/-------- -

Between A and - there will be 9V
Between B and - there will be 6V
Between C and - there will be 3V

But most powersupplies look like this:

+----/\/\/\/----A----/\/\/\/----B----/\/\/\/----C-------- -

Between A and - there will be 6V
Between B and - there will be 3V
Between C and - there will be 0V, since there is no resistor there..

So, can anyone tell me where I'm wrong, because I must be....

Tommy

[FunkyE9th]

No it's not 0.

Yes there is no resistor there but there is a load (actually all nodes A, B and C are driving a load). The load are your tubes connected to those nodes and those tubes have some equivalent impedance and are referenced to ground.

Even if you pull out the tubes there is still a load but the impedance is very large (i.e. you can assume infinite).

So your drawing at nodes A, B and C should really have a resistor (i.e. the equivalent resistance of tube circuit) to ground.

Also the caps block DC so the behave like an open circuit under DC once they fully charge. Think of it as a switch that is open, so impedance infinite.

[Tdale]

Thanks!

I believe I asked about having to count the tube and the cap in the equation, but got no positive response on that.

Now it makes more sense to me!

Tommy

[FunkyE9th]

I forgot to clarify one more thing...

You cannot assume 6 or 3 volts on nodes A and B because the loads driven by node A, B and C (i.e. the current drawn by each load) affect the voltage drop on those resistors.

[Tdale]

I know, my drawing was a circuit with no loads connected, just to show the voltage drops clearly.