Measuring output impedance
post-211937
Thread Starter
Post #1 of 6

Joe Bloggs

Member of the Trade: HiBy / EFO technologies Co
His Porta Corda walked the Green Mile
Joined
Oct 21, 2001
Messages
11,034
Reaction score
2,784
Location
Hong Kong and Melbourne
Joined
Oct 21, 2001
Location
Hong Kong and Melbourne
Posts
11,034
Likes
2,784
A while ago I had 'Team 0 Ohm output impedance' in my sig. :p Today I tried measuring the output impedance of various headphone jacks around the house...

I measured the resistance between the left or right channel and the ground of a miniplug plugged into the headphone jack. Is this a correct way to measure?

Measured the Porta Corda running out of 3 somewhat stale 9V batts connected in series (giving 17V as of tonight):

when volume pot near zero, ~35 ohm for both channels
when volume pot near max, got <10 ohm for one channel and >20ohm for the other


Sony D-EJ725 pcdp headphone out:
Reading steady at ~335ohms for all volume settings, including when the player is off

Old Sherwood CDP:
Now this is pretty wild! Fluctuates from 15 to 18MOhm when playing what turned out to be something akin to white noise (data track of hybrid CD) but steady at 5kOhm when not playing.

Sherwood integrated amp:
~250ohm

The CDP and integrated amp results were really a surprise for me. I thought that 0 ohm output impedance was the norm rather than the exception for SS amplification circuits.

Edit: I realized that a proper measurement of output impedance required a lot more than that... but I was looking for a methodology to do a quick-and-dirty technical evaluation of headphone outs of various devices... will this do?

The readout for the Porta Corda above was with no input plugged it. Now I tried plugging in a pcdp and playing a song and looking at the multimeter 'resistance' reading. It fluctuated between 10 and 30ohms, but stayed around 30ohms. Now I don't know what these numbers really stand for since I get the idea the multimeter simply isn't measuring anything near what I want to measure, but would a headphone out with near zero impedance yield a similar multimeter result as the Porta Corda?

(The multimeter reading for the CDP fluctuated wildly when it was playing real music.)

THIS is the right way to do this right? I just want a quick and dirty shortcut


Quote:

From http://www.eatel.net/~amptech/elecdisc/outptimp.htm
For this test, you want to drive the amplifier to a sufficiently high level to produce large measurements (voltage, current...). The larger numbers will help increase the accuracy of the tests. You will drive the amplifier to a level which will be below clipping when loaded with the test load. You must be absolutely sure that the level of the input signal is not changed throughout the test procedure. Measure and make note of the output voltage of the amplifier with no load. Connect the 4 ohm load to the amplifier's output terminals and measure the output voltage again. Make note of it. Please note that the voltage measurements should be taken as close to the amplifier as possible. Since the resistor will change in value when it heats up, after reading the voltage quickly disconnect the resistor from the amplifier and measure its resistance, including the wire used to connect the resistor to the amp. Make note of the resistance.

Decipherin' the data

Method 1:
If the NO LOAD voltage was 20 volts and the LOADED voltage was 19.95 volts, the difference in loaded and no load voltages is .05 volts. If the resistor measured 4.15 ohms and we know that there was 19.95 volts across it we know that the current through the resistor was:

I=V/R
I=19.95/4.15, I=4.81 amps.
If we use the formula R=V/I
We get R=.05 (voltage drop in amplifier)/4.81
R=.01 ohms.
The output impedance of the amplifier is .01 ohms.

Method 2:
Output impedance=(Load resistance*(Vunloaded minus Vloaded))/Vunloaded
Output impedance=(4.15*(20-19.95))/20
Output impedance=.01

Note: If you are using a digital multimeter, you should check the frequency response of the meter by measuring the output voltage of the tone generator at various frequencies. Some multimeters are designed to measure only a small range of A.C. frequencies. If the voltages change significantly (more than 1 or 2 percent), you must take that into consideration in the various measurements you make. If you have a 'true RMS' meter, you are likely going to get accurate readings over the entire audio spectrum.


 
     Share This Post       
post-212148
Post #2 of 6

ServinginEcuador

Founder of the Head-Fi Pay-to-Post Program.
Joined
Sep 1, 2002
Messages
8,384
Reaction score
14
Joined
Sep 1, 2002
Posts
8,384
Likes
14
If you're output is solid state, you're readings are probably thru a capacitor or some other path other than the actual output one. When SS stuff shuts down it is effectively an open cirtuit and you are probably getting readings from some other part of the circuit other than the output.

This is going back a ways, but if I remember correctly you can't read accurate resistance measurements with an energised circuit. IOW, you can't get accurate reading of resistance when something is powered up unless you physically disconnect it from the ciruit under test. When a muitl-meter checks for resistance, at least all the flukes I used ot use in the Navy, sent out about 5 volts of electricity, and then read the current to determine the resistance.

Measuring current is even harder as you have to read current in series to the current flow as the muitl-meter provides a mere 5 ohm load when checking current, so it tends to fry things or pop the protection circuitry when you try to read it in parallel. IOW, you cause the meter to drop the whoel load voltage across so small an amount of resistance that the current is so high it can damage the meter.

The two formulas to know just about anything about basic electronics are these:

E=I*R E= voltage, R= resistance, I=current
P=I*E P=power

After that you can merely substitute values and come up with the answer to the resistance question by transposing the equation and factoring out what you want to get rid of. IOW, if you want the I of a circuit, you can take the first equation and transpose it to come up with I=E/R or I=P/E. Or, if you want to figure out power, but only have I and R you transpose the first equation to come up with the what I equals, which is E/R, then substitute the values for I into the second equation and you will come up with P=I* I* R or P=I2*R (the 2 here being square the current, not multiply by two.)

Now, back to your problem. Give me a little while to see if I can come up with a way to get an accurate measurement of output resistance. There is and must be an easy way, but it's not coming to me right now!!


Lord Bless,
doug p.
 
     Share This Post       
post-212202
Post #3 of 6

Joe Bloggs

Member of the Trade: HiBy / EFO technologies Co
His Porta Corda walked the Green Mile
Joined
Oct 21, 2001
Messages
11,034
Reaction score
2,784
Location
Hong Kong and Melbourne
Joined
Oct 21, 2001
Location
Hong Kong and Melbourne
Posts
11,034
Likes
2,784
I suppose I could wire resistors across the L/R and ground on another miniplug and start taking voltage measurements instead of 'resistance', right?

As for what I did last night, I was looking for a way to tell apart those headphone amps that are speaker amps with a resistor wired in series...
 
     Share This Post       
post-212246
Post #4 of 6

ServinginEcuador

Founder of the Head-Fi Pay-to-Post Program.
Joined
Sep 1, 2002
Messages
8,384
Reaction score
14
Joined
Sep 1, 2002
Posts
8,384
Likes
14
Quote:

Originally posted by Joe Bloggs
I suppose I could wire resistors across the L/R and ground on another miniplug and start taking voltage measurements instead of 'resistance', right?


You won't want to add any resistance into the signal path to measure voltage, it will throw off the measurements. Just connect across the mini plug using the leads of the multi meter. There's enough resistance in there to accurately and safely measure voltages up to at least 115v AC.


Quote:

As for what I did last night, I was looking for a way to tell apart those headphone amps that are speaker amps with a resistor wired in series...


Gotcha! Sounds like you have a receiver or amp w/ a headphone jack that you are trying to figure out how it's wired. One way to do that is to connect one end of the meter to either the left or right channels of the headphone jack, and use the other end to probe the red/positive speakers connections in back of the amp and check for continuity. Your meter should emit a noise when you get continuity, so just listen for it once you start probing. (You can verify this by switching the meter on, setting it to resistance, and touch the meter leads together. It should make a noise when you touch them.) And always remember to have the amp OFF when doing this as the output posts of a normal amp are deadly if you happen to get your finger in the wrong place!!


Lord Bless
 
     Share This Post       
post-212425
Post #5 of 6

Joe Bloggs

Member of the Trade: HiBy / EFO technologies Co
His Porta Corda walked the Green Mile
Joined
Oct 21, 2001
Messages
11,034
Reaction score
2,784
Location
Hong Kong and Melbourne
Joined
Oct 21, 2001
Location
Hong Kong and Melbourne
Posts
11,034
Likes
2,784
It doesn't beep... but I don't understand how the multimeter can test for this 'continuity' given that the circuit is powered up...


Incidentally, it says there's continuity between the left and right channels of the Porta Corda headphone out but not between either and the ground
 
     Share This Post       
post-212456
Post #6 of 6

ServinginEcuador

Founder of the Head-Fi Pay-to-Post Program.
Joined
Sep 1, 2002
Messages
8,384
Reaction score
14
Joined
Sep 1, 2002
Posts
8,384
Likes
14
Quote:

Originally posted by Joe Bloggs
It doesn't beep... but I don't understand how the multimeter can test for this 'continuity' given that the circuit is powered up...


Incidentally, it says there's continuity between the left and right channels of the Porta Corda headphone out but not between either and the ground


Joe,

I should have been more specific when I said to read the resistance from the speaker output to the headphone output. You must do this with the circuit off, as is teh case whenever you are measuring resistance. You should get continuity between them as IF the speaker output is connected to the headphone jack via a resistor. I'm not very familiar with this setup, but the signal path of the headphone jack should include a small amount of resistance in the path to reduce the signal down to the headphone level. Either that or it picks the signal off before it is amplified to full power and sent to the speakers. That way the signal would be much smaller. But, like I said, I am not sure exactly how they do this particular setup. Just trying to help.


OBTW - not getting continuity from left to ground or right to ground when the circuit is de-energized is a good thing. As I said the path to ground should include some form of solid state device. Quite possibly what you are reading from the left to right is some path in the circuit, and fine.


Lord Bless
 
     Share This Post       

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top