measuring current using 1ohm resistor. How?

[riotubes]

Hi, I want to measure the current before an HV regulator. The regulator's V+ input is fed from the plus lead of a smoothing cap located after a diode bridge. The HV reg gnd input is fed from the neg lead of this smoothing cap.

Where do I place the 1 ohm resistor to measure current before the reg? In line with the V+ rail (resistor lead to plus lead of cap with the other R lead to reg's V+ input? Or is it across the plus and neg leads of the smoothing cap? Oh, and how do I size the wattage of the resistor? Under no load, the HV reg measures 178V on input, 70V on output and under load, 138V on input and 50V on output with about 28-33mA running through it. Thanks!

[bear]

This is a job for Ohm's law!

E = I x R

The 1 ohm goes in series with the "load". If you want to see the total draw of the regulator plus any load (if a load is applied) then it goes in series with the B+ lead before the regulator.

If you only want to see the draw of the thing being run by the regulator, then you put the 1 ohm between the regulator and the load (thing being run by it).

We then measure the voltage across the 1 ohm resistor and apply ohms law to find the current: E/R = I

How large a resistor? P = I x E

You can substitute there like for I... P = E/R x E or P = E^2/R for example.

Since you stated the max current draw and the voltage you can just pop them into the power equation and then pick a resistor wattage that is above that... higher power is always safer.


ok?

_-_-bear

[AndrewT]

yes, it's ohm's law.
30 to 50mA through 1r0 does not need a power resistor.
50mA is just 2.5mW and can be passed by any 0.125W or greater resistor.

But if you want accurate results you must measure the test resistor accurately.
5% tolerance is not much good if you need 0.5% in your readings.

50mA passing 1r0 will show 50.0mV +-0.1mV if that is the accuracy of the resistor and multimeter combined.
A 1% multimeter on the 200.0mV scale can read no better than +-2mV.
A 5% resistor can read no better than +-2.5mV on 50mA.
The total tolerance of these combined values will be 4.5mV.
Your reading could be anywhere between 45.5mV and 54.5mV for an actual 50mA current.

[riotubes]

thanks guys!

One elementary clarification. by "in series" you mean that the resistor should be "in line" with the V+ rail? I don't need a 2nd 1ohm resistor in the gnd rail, just the plus rail? Again, the plus and neg leads of the smoothing cap (located after a diode bridge) is feeding the HV reg board. So I just solder one lead of the resistor to the plus lead of this cap and the other lead of the resistor to V+ input on the HV board?

[AndrewT]

Originally posted by riotubes
One elementary clarification. by "in series" you mean that the resistor should be "in line" with the V+ rail? I don't need a 2nd 1ohm resistor in the gnd rail, just the plus rail? Again, the plus and neg leads of the smoothing cap (located after a diode bridge) is feeding the HV reg board. So I just solder one lead of the resistor to the plus lead of this cap and the other lead of the resistor to V+ input on the HV board?

yes, "in series" means the output current passes through the series resistor.

But I have to ask: are you confident you have the skills/knowledge to work with HV and tube/valve gear?

[riotubes]

Thanks Andrew for the clarification. I'm definitely learning as you can tell. The way I keep myself out of trouble is that I *always* ask before doing anything, and I make sure to clarify. Several times if necessary. I always use someone elses circuit that is completely spec'd. I'm comfortable measuring voltage, but haven't measured current except by using ohm's law. Thanks again.

[riotubes]

In my parts bin, I found a 1 ohm 5% resistor and a 10 ohm 1% R. Would the latter be better to measure the current? The DVM is a fluke 87 model.

[AndrewT]

10r 1% will let you measure to 500mV +-5mV
If the fluke is 0.2% then on 2000mV scale is +-4mV.
Total error range +-9mV, for a 500mV reading is far better than +-4.5mV for a 50.0mV reading, more than ten times better.

Do the numbers, rather than guess.

[riotubes]

Thanks very much Andrew. I'll post my results sometime this week. Much appreciated!

[bear]

Btw, not sure what ur doing with the regulator(s), but dropping 100volts in the regulator seems a bit excessive... you will have a relatively hot regulator there... of course if it is a low current deal anyhow, maybe the heat won't be too much.

_-_-bear