Measuring current (newb questions)

Hi Guys,

A couple newbie questions if you'd be so kind? ...I'm trying to measure the current drawn from a Quad Eight EQ module (I'm racking 4 of them, and trying to find out if this Fivefish PSU will be adequate for all 4). The EQ's are +18/-18. I put my DMM inline on the + between the PSU and the EQ. It reads 65mA. So is that the total draw or should I also be measuring - or ground too and then do some sort of equation?

Bonus questions...

1) Should I be passing signal through the unit while measuring? Like if I turn all the knobs up(boost), does that make the unit work harder?
2) The Fivefish PSU pcb has 4 sets of (+/-/ground) solder pads. Should I wire each EQ unit to each of those or should the units be daisy-chained? I thought I read somewhere that if units are powered in parallel, the voltage would be divided to each unit, so they should be wired in series. Any clarification?

Thanks in advance for any knowledge you'll share...

Typical approach is to wire a small value resistor in series with power supply (single digit ohms). Measure the voltage drop with VOM and calculate current from that. Measure the UUT as it will be used in situ,

JR

Thanks for the response, John. Not sure if I'm fully understanding, but here's how I interpreted your instructions...

+18dcv PSU ---> low value resistor --- (using DMM, I see 17.67dcv here) ---> EQ unit.

So if I'm doing this correctly, I see a voltage drop of .33 dcv using a 10 ohm resistor. *10 ohm is the lowest value resistor I have on-hand. 

To be honest, I'm not understanding why I'm adding the resistor. I also don't know what to do with the information or what the equation is to find the current. Ohms Law?  :-\ Warned you, I'm a newbie. Thanks!

jimkeaney said:

Thanks for the response, John. Not sure if I'm fully understanding, but here's how I interpreted your instructions...

+18dcv PSU ---> low value resistor --- (using DMM, I see 17.67dcv here) ---> EQ unit.

So if I'm doing this correctly, I see a voltage drop of .33 dcv using a 10 ohm resistor. *10 ohm is the lowest value resistor I have on-hand. 

Click to expand...

Ohms law is your friend... so if you know the voltage drop and the resistance you can compute the current.  I=E/R = .33V/10 Ohm =  33mA

To be honest, I'm not understanding why I'm adding the resistor. I also don't know what to do with the information or what the equation is to find the current. Ohms Law?  :-\ Warned you, I'm a newbie. Thanks!

Click to expand...

You should be careful you don't blow anything up..  ;D ;D

In theory you could just put the VOM on current scale in series to directly measure  a few tens of mA but i still prefer the dropping resistor approach. Some circuits may not like the intermittent power connection

A few things,, to be more accurate the resistor may not be exactly 10 ohms so measure it with your VOM while the circuit is not turned on.

Did you measure 18.0V on one end of the 10 ohm resistor and 17.67V on the other at close to the same time? More reliable to measure the DC drop directly across the 10 ohm resistor by probing across it.

be careful...

JR

Okay now I'm getting somewhere. I had one wrong pin on the EQ pinout so disregard my previous reading. I also toasted the 10R.  and blew a fuse in my DMM (exceeded 400mA)...  ;D Learning stuff!

I had a new fuse for the DMM, but unfortunately, I have no more low value resistors around to try. They're all like 2K+.  So if I simply break into the circuit between the PSU and the EQ (without a resistor), I'm measuring 75.6 mA with the DMM. That seems like a more appropriate current for something like this EQ (at least based on what other similar pieces of gear draw from my online research).

So you're saying this 75.6mA reading may not be super accurate. This Fivefish PSU is stated to provide 830mA. 4 eq's @ 75.6mA = 302.4mA, which makes me think I should be okay.

I really appreciate you taking the time to walk me through this stuff. Definitely going to add it to my notes. Still wondering about if I power these units in parallel from the PSU or daisy-chaining?

Be careful  but hopefully you are learning what not to do....

JR

You want to power the units in parallel - if you put them in series, each would only see 1/4 of the correct voltage, and the grounds would be at different voltages.

Right now you're only measuring the current draw on the positive rail.  If you want to know the total power being consumed, you need to measure the negative as well.  You might double check the specification for the power supply to see whether it's 800mA per rail or total.

So, is it like you put the DMM in series to the + rail and measure the current (say 80mA) and then you put in series with the - rail and measure again (say 40mA) so the total current draw would be 120mA?

warpie said:

So, is it like you put the DMM in series to the + rail and measure the current (say 80mA) and then you put in series with the - rail and measure again (say 40mA) so the total current draw would be 120mA?

Click to expand...

NO... current is current and power is voltage times current.

So 80mA of plus current  and 40mA of negative current is just that...

Total power on the other hand is the sum of positive current times positive voltage, plus negative current times negative voltage.

JR

dfuruta said:

You want to power the units in parallel - if you put them in series, each would only see 1/4 of the correct voltage, and the grounds would be at different voltages.

Right now you're only measuring the current draw on the positive rail.  If you want to know the total power being consumed, you need to measure the negative as well.  You might double check the specification for the power supply to see whether it's 800mA per rail or total.

Click to expand...

Ok, thanks for the wiring clarification. So I'm seeing 75.6mA on the positive rail and 75.6mA on the negative. So is 151.2 mA not the total current?

If I follow Johns equation... 
75.6mA times 18v(+) = 1360 ...PLUS... 75.6mA times 18v(-) = 1360  = 2.72amp total? I'm confused by Johns response to Warpie, saying "current is current and power is voltage times current". Mostly because Warpie didn't use the word 'power'.

jimkeaney said:

If I follow Johns equation... 
75.6mA times 18v(+) = 1360 ...PLUS... 75.6mA times 18v(-) = 1360  = 2.72amp total? I'm confused by Johns response to Warpie, saying "current is current and power is voltage times current". Mostly because Warpie didn't use the word 'power'.

Click to expand...

I believe what John said is:

P = (Icc*Vcc) + (Iee*Vee) => P = (0.0765A * 18V) + (0.0756* 18V) => P = 1.377W +1.377W => P=2.754Watts

But take it with a grain of salt 

jimkeaney said:

dfuruta said:

You want to power the units in parallel - if you put them in series, each would only see 1/4 of the correct voltage, and the grounds would be at different voltages.

Right now you're only measuring the current draw on the positive rail.  If you want to know the total power being consumed, you need to measure the negative as well.  You might double check the specification for the power supply to see whether it's 800mA per rail or total.

Click to expand...

Ok, thanks for the wiring clarification. So I'm seeing 75.6mA on the positive rail and 75.6mA on the negative. So is 151.2 mA not the total current?

Click to expand...

No it's only 75mA,  but from twice the voltage. I don't know if this will help but the same 75 mA that comes out of the + supply, ends up going into the - supply.  So it's the same current across twice the voltage not twice the current. It is however twice the power...

If I follow Johns equation... 
75.6mA times 18v(+) = 1360 ...PLUS... 75.6mA times 18v(-) = 1360  = 2.72amp total? I'm confused by Johns response to Warpie, saying "current is current and power is voltage times current". Mostly because Warpie didn't use the word 'power'.

Click to expand...

Power may be used causally by some ... When I said power I mean the power coming from the supply and consumed by the circuit.  Power in Watts is current times voltage.

75mA is 0.075A  so that times 18V is 1.36W  (or 1360 mW)

Sorry if this is confusing  If we use the water analogy, 75mA of water is flowing out of the + supply tap, and flowing through the circuit eventually returning into the - supply drain. (perhaps a little too simple?)

Good luck

JR

John, I appreciate you not giving up on me or losing your patience. The water analogy is perfect for a newbie like me. I've built some Pre-amp, EQ's and compressor DIY builds, but I don't actually understand the circuits I'm building. It's more like paint-by-numbers. I hope to change that over time. Thinking I might even try some courses. Wish I could find some course specific to audio circuits, since that's where I'm most interested.

I wasn't looking for the total power in wattage. Just the current (to compare to the PSU's total), but nonetheless I've picked up some good equations here that I will put in my notebook for future reference. Thanks again.

My understanding was that this supply has +, - and ground rails, in which case the current on the + rail is not necessarily returning on the - rail, for example in the case of a unit mixing dual supply opamps with single supply logic, or something like that.  That's why I suggested measuring both rails.

But that's not a concern here!

dfuruta said:

My understanding was that this supply has +, - and ground rails, in which case the current on the + rail is not necessarily returning on the - rail, for example in the case of a unit mixing dual supply opamps with single supply logic, or something like that.  That's why I suggested measuring both rails.

But that's not a concern here!

Click to expand...

With out looking at the schematic I can't be sure, but most op amps only have two power pins a + supply and - supply, so the majority of their current consumption from the + supply does indeed flow into the - supply.

Ground currents will mostly be related to audio signals so AC with no net DC current flow.

JR