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Damping factor of amplifier in LTSpice simulation

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Hi,
Nominal_load_impedance = 8 Ohms.
Amplifier_output_impedance = V / I = amplifier_output_voltage / load_current.

Nominal load 8 ohm, input shorted, see 366uV and -45uA:
Immagine.png
Immagine2.png

I've done it right now? Probably not :cry:
 

Since the open loop gain of an amplifier drops at high frequencies, the damping factor is highest at low frequencies that are the resonance of a woofer and/or its enclosure. You might not hear a difference between a damping factor of 20 or more.
Here is a good damping factor:
 

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Hi,

I've done it right now? Probably not
You missed to use an AC source to excite the output.
It seems you missed to calculate with AC voltages only.

Note: As written in the book and in my post: You need to know about your amplifier's nominal_load_impedance but there is no need that your measurement_resistor is of the same value. You might use 8Ohms, but you might use 800 Ohms...It won't harm the DF result.

Circuit:
Shorted_input -- amplifier (DUT) -- measurement_resistor -- excitation AC source (sine waveform).

The excitation AC source is important. It introduces (via the measurement_resistor) measurement_current into the amplifier output.
This current causes some (small) AC voltage at the amplifier's output.
With these both values you are able to calculate the amplifier's output impedance: V_amplifier / measurement_current.

And DF = nominal_amplifier_load_impedance / amplifier_output_impedance.
(Here nominal_amplifier_load_impedance comes into account)

Btw: the DF depends on excitation frequency. Thus you should do several tests with different frequencies (or use the specified frequency). Usually DF decreases with higher frequencies.

Klaus
 

Ok, putting the ac source at 1Vrms - 1kHz on the output after the nominal_load_resistor.
Immagine3.png
Result of voltage at V(out) and current on resistor
Immagine.png

Situation at V(out1):
Immagine2.png

Then what is the total DF? Voltage swing is 3uV, current is 130mArms.
 
Last edited:

Hi,

V_out is the amplifier output (voltage). --> V_out1 is not of interest.

Please verify your measurements:
Voltage swing is 3uV, current is 130mArms.
I can´t find 3uV nor can I see 130mA RMS. How do you come to those values? Do you have other informations than the scope pictures?

Is it asking too much to do the calculations on your own?
With these both values you are able to calculate the amplifier's output impedance: V_amplifier / measurement_current.

And DF = nominal_amplifier_load_impedance / amplifier_output_impedance.

Klaus
 

I can´t find 3uV nor can I see 130mA RMS. How do you come to those values? Do you have other informations than the scope pictures?

Is it asking too much to do the calculations on your own?

It's because I have not yet understood if I have done well the measurements that I continue to ask, I understood the formula, but I still do not understand if the measurement_current is the current that passes throught the measurement_resistor(or load), and if the V_amplifier is the V(out).

If it's so, then why i don't have 130mArms? In the image I have a swing of +130mA and -130mA. Then I have a swing at V(out) of 1uVrms (because the amplifier without input and without an output source voltage have 366uV of DC output).

It's that not true? I'm still confusing.
 

Hi,

R17 is directely connected to the amplifier output. Thus it should be obvious that R17_current = amplifier_output_current.
Not to offend you, but if this isn´t obvious to you then I recommend to stop designing amplifiers and start to learn electronics basics.

If it's so, then why i don't have 130mArms? In the image I have a swing of +130mA and -130mA.

Reading the scope picture shows a sinewave with -125mA peak and +125mA peak (the difference from 125mA to 130mA is not the problem)
Please learn basics: A sinewave with +/-125mA peak has an amplitude of 125mA .. has an RMS value of 125mAp/sqrt(2) is about 88.4mA RMS.

But this value doesn´t meet the 1VRMS you are talking about.

****
Then I have a swing at V(out) of 1uVrms (because the amplifier without input and without an output source voltage have 366uV of DC output).
Please explain where your 1uVRMS comes from. I can´t find it.

Don´t use DC voltages for DF calculations. You have to look at AC voltages only.

Klaus
 

This is the voltage of the source:
Immagine.png
Are you saying that is 1Vpk and not 1Vrms? That is 0.707Vrms?

Please explain where your 1uVRMS comes from. I can´t find it.
Then the 1uVrms is not rms but 1uVpk. If you see the graphic you'll find that the voltage swing betweet 365uV and 367uV. This is the voltage I obtain at V(out).
 

Hi,

Are you saying that is 1Vpk and not 1Vrms? That is 0.707Vrms?
* It is 1Vp (or 1Vpk like you write)
* it is 2Vpp
* it is 0.707V RMS
It´s not "me" that says this, it´s the common definition.
--> google for "RMS sinewave"

If you see the graphic you'll find that the voltage swing betweet 365uV and 367uV.
Maybe you have a different picture.
I refer to the upper picture of post#24.
There it is:
365.97uV ... 367.28uV which is 1.31uVpp = 0.655uVp = 0.463uVRMS
Don´t you agree? Even if you round it, it is not 365uV.

Klaus
 

Then wait, set voltage to 1.414Vp (to obtain 1Vrms), obtained:

- 175mAp = 123mArms
- 365.6uV|367.5uV = 1.9uVpp = 0.95uVp = 0.67uVrms

In calculation:

0.00000067Vrms / 0.123Arms = 0.00000544715

DF = 8 / 0.00000544715 = 1468657.922

absurd :sad: I have failed again.
 

Hi,

No, you did not fail.

Well done. Everything is correct now. (except the simulation results. 0.67uV at 123mA is not realistic - my assumption)
If possible: do the measurements on the real circuit.

****

The simulation shows an output impedance of about 5.5uOhms. Every piece of trace, wiring, every connection will cause way more series impedance.
But even if the simulation doesn´t care about wiring or connections.... it needs a huge open_loop_gain inside the amplifier to get such low output_impedance.

Klaus
 

Sorry but a DF of over 1.4million it's really impossible, right? Or should I look at this value in db?

The distortion is really low: 0.000030%@50W - 1kHz, at least in the simulation. I used double CFP differential input with a complete Wilson mirror and cc generators, a buffered VAS, two pole compensation and mosfet power output.

How much I can consider the simulation correct, knowing I'm using Bob Cordell's spice model and 5% tolerance resistor?
 

Your damping factor and distortion numbers are not correct. Build the circuit and measure them, the simulation is missing something.
 

The simulation isn't missing anything. This amplifier required 8 months of design, the initial distortion was 0.1% at 1kHz with the same power, now it's 0.000030%. It seems strange, but looking in different volumes such as Self's and Cordell's books, seems like I built a very nice amp: I do not have the precision tools to measure the real it, though.
 

Hi,

again: These values are not realistic. They are decades away from the real circuit.
Do tests on the real circuit.

I´m curious: What is the simulation DF with an excitement frequency of 100Hz and 10,000 Hz?

Klaus
 

I´m curious: What is the simulation DF with an excitement frequency of 100Hz and 10,000 Hz?

100Hz 1Vrms: 0.67uVrms | 137.82mArms - DF: 1645613.104
10kHz 1Vrms: 3.88uVrms | 128.67mArms - DF: 265298.962
 

The resistance of a short piece of wire or a drop of solder is much more than 8/1,645,613 ohms. The distortion that is inaudible is probably 0.003%, not 0.00003%.
 

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