Total Current Draw

[eeyore]

Hi, I am doing a new PSU (Tube rectified) for the Crack w/ Speedball, and was trying to determine the total current draw for both channels, and I was wondering if anyone knows what the Speedball is set at for both the driver and 6080 tube? For the PSU, I was thinking of choke input (common mode) followed by 2 RC stages, basically adding a choke before the first PSU cap, and keeping the down stream the same. Obviously needs a new transformer, and hence the question. With around 50mA of current draw, I am needing a 240-0-240 with an 80 tube. Any help would be great, thanks.

[Beefy]

Why not measure it exactly in your own amp? Determine the voltage drop across each of the resistors in the power supply, and their exact resistance, and you can calculate your current draw.

[eeyore]

I know, but I won't get the chance to get everything out until the weekend, and wanted to get started on the PSU design before then... I know, a little lazy...

[dbishopbliss]

I've always wondered how to determine the total current draw for existing amps.  Let's say I have a circuit with a power supply that looks like the one in the attachment (assume the rest of the circuit is attached, but not shown in the diagram).

A = 218V
B = 192V
C = 170V

Therefore:

voltage drop for R1 = 218-192 = 26V
voltage drop for R2 = 192-170 = 22V

Calculate current using Ohm's law, I = V/R.  So I1 = 26mA and I2 = 22mA. 

Does this mean my total current draw is 48mA?  That doesn't seem right since I determined these values using PSUD where the load is supposed to be a constant current of 20mA.  What am I doing wrong?

[Paul Joppa]

The voltage at A has a lot of ripple, and your meter is probably not measuring the average DC voltage - it may be the peak, or some approximation to the RMS. The measurement at B is better (has been smoothed by the second capacitor, and at C is best (very little ripple is left at that point).

DC current does not go through capacitors, so the DC output current is close to 22mA; 28mA is just a less accurate measurement of the same current.

[dbishopbliss]

For this example I used the "RMS" values provided by PSUD so there was no meter actually involved.  

Are the following steps correct to determine the current draw (somewhat accurately)?

• Measure voltage at B
• Measure voltage at C
• Measure Resistance of R2
• Calculate current using the following forumla: I = (Vb-Vc)/ R2

[JC]

Are you also assuming that C is where all of the current is being drawn from the power supply?  That there is 0 current being drawn at B and A, excepting that which is being drawn from C?

[dbishopbliss]

Are you also assuming that C is where all of the current is being drawn from the power supply?  That there is 0 current being drawn at B and A, excepting that which is being drawn from C?

Yes, that is what I'm assuming.  I should have specified the capacitors as 220uF, then it would be a copy of the original Foreplay I power supply (Except the original used a full-wave bridge rectifier instead of center-tapped, but close enough). 

[Grainger49]

David's formula ignores the charging current for the first two capacitors which is pretty small in comparison with the circuit itself.  As Paul Joppa points out there is also some charging to the third capacitor but little ripple.

[JC]

Well, I really don't know that much about PSUD; it has been on my "to do" list for a while!

But, if it allows for ripple current, which will be more prevalent in the earlier stages of the filter, that may account for the "extra".  IOW, the Voltage drop across R2 will probably give you a more accurate picture of DC current being drawn at C, than that across R1.  

Additionally, the diagram is indicating "2 Ohms" at each of the capacitors; presumably, the program is taking this into account as well, whatever it is referring to.

LOL!  I guess I may have just re-stated what Paul said!  Remind me to re-read his posts more thoroughly before I shoot my mouth off!

[dbishopbliss]

This has been a very useful thread for me (sorry for hijacking it).  I'm pretty good at playing with PSUD now, but I find myself guessing when it comes determining the load of existing circuits.  Therefore, my models have been basically useless.  I'm setting up a breadboard with a power supply "jig" that should let me get the voltage close using a series of resistors and wire jumpers.  Each resistor is a 13W wirewound so they should be able to handle any voltage/current that I'm going to be using.  Once I get the voltage close, I can calculate the current using the steps described above.  Then I can use that value in PSUD to better model the exact values needed.  I'll let you know how things go. 

[JC]

I will certainly look forward to hearing about your results.

I was just thinking how much simpler all of this might be if everything ran on batteries!

[Beefy]

Well back to the initial question then, I calculated my Speedballed Crack as drawing 72mA based on the voltage drop across the resistors. FWIW, both the first and second resistors gave very similar readings.

Before the Speedball, it was drawing 81mA.