# Understanding voltage drop

#### hoho

Hi, likely a very stupid question but since it is around high voltages, I need to understand it.

I am to build a variation of the preamp part of the Dumble ODS. Basically, from this as a source: https://thesubjectmatter.com/dumblearchive/Dumble #0094/094_schematic_12_132.pdf. Just the part with two 12ax7s (without PAB relay and without effects send-return).

So, I need just B+4 and B+5. Could someone explain me what causes the voltage drop between B+1/B+2/B+3/B+4/B+5? And the further voltage drop on the valve? I have found this variation with the voltage readouts: . What causes the drop from 400V to 300V between B+3 and B+4 (there is a resistor in between, but it doesn't seem like a voltage divider to me)? And further on, what causes the drop to 200V on the actual valve?

Basically, my idea was to use a supply like https://www.aliexpress.com/item/1005005909458258.html that provides 6.3VDC for filament and 300VDC for B+, and I plan to just cut out the part after 22k resistor after B+3 and connect this power supply there. But I'd like to understand what happens with the voltages before switching the power on...

Thanks.

#### pinholer

Paid Member
Voltage will drop if current flows through a resistor. Ohm's Law says V = I x R. So, multiply the current through the resistor by its resistance and you will get the voltage drop.

The valve's plate load resistor will likewise drop voltage.

#### hoho

Thanks! Does that mean that when I will measure the voltage on the empty socket I will be 300V and the voltage will drop only after inserting the actual valve?

#### pinholer

Paid Member
The voltage drop across the plate resistor should be zero if the tube is not in place. The actual voltage will depend on any other voltage drops, if any, in your circuitry. So, the voltage at each end of the plate load resistor should be the same. If not, there is current flowing somehow.

#### AllenB

Paid Member
and the voltage will drop
Voltage 'drop' is not defined as the Voltage 'reducing'.

The Voltage across the resistor is defined by it's current, so without the valve in place the current is different.

#### jan.didden

Paid Member
Thanks! Does that mean that when I will measure the voltage on the empty socket I will be 300V and the voltage will drop only after inserting the actual valve?
Without the valve there is no current through the plate R so there is no voltage across the plate R, as V=Ra*Ia and Ia =0.
Insert the valve and you will have Ia through Ra so across the plate R you will have V=Ra*Ia.
So assuming that the B+ side of Ra remains the same, the anode side of Ra will now be B+ - (Ra*Ia).
Basic Ohms' law.

So, multiply the anode resistor value in kOhms with the anode current in mA and that will be the anode voltage drop when you insert the valve.
(Assuming B+ will stay the same which it probably will not, so check that).

Jan

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#### hoho

Thank you very much for the clarifications.