Author Topic: Fairchild signal amp gain calculations  (Read 1332 times)

Flundran

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Fairchild signal amp gain calculations
« on: October 21, 2011, 07:36:52 AM »
I've been calculating the gain in the Fairchild 660 signal amp, wich according to the manual is 14dB. When I do my calculations I get about 18dB. I thought I might as well ask here, and maybe someone will shime in and tell me what I'm doing wrong.  ;)

I use this formula:       Av = (mu * Rp) / (Rp + ra)
   
Av = gain   
mu = amplification factor, which is 17 for the 6386
Rp = plate resistor
ra = anode resistance, which is 4250 ohms  for the 6386

To find out the plate resistor (Rp) that every tube is seeing, I am thinking like this: If the output transformer secondary is loaded down by 600 ohms, the reflected resistance on the primary side is 48.6 kohms (because of the 9:1 ratio). Since the tube stage is a long-tailed pair, each side is seeing half of this, that is 24.3 kohms. Is this correct? But then since the amp has 4 tubes in parallel, this resistance is multiplied by 4. The plate resistance that every tube "is seeing" is 97.2 kohms.

If I put this value for Rp in the formula I get a gain at about 16. Due to the long tailed pair, this value is then divided between the two anodes, so the gain is actually 8. That equals 18dB...

Thanks!
/F



 

 


MeToo2

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Re: Fairchild signal amp gain calculations
« Reply #1 on: October 21, 2011, 09:54:48 AM »
I think the operating point in the Fairchild is so far off what is the typical operating point quoted on the datasheet (quoted at plate current 9.6mA class A) that the values for mu and plate resistance (small signal model) are pretty meaningless. IMHO The two tubes are more of a short tailed pair class AB push pull amp than a long tailed pair. You'd almost certainly have to have a look at the composite load lines of the large signal model to get a better picture.

See e.g. http://greygum.net/sbench/sbench102/composit.html or
http://www.freewebs.com/valvewizard1/pp.html
« Last Edit: October 21, 2011, 11:53:48 AM by MeToo2 »

abbey road d enfer

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Re: Fairchild signal amp gain calculations
« Reply #2 on: October 21, 2011, 01:45:48 PM »
I've been calculating the gain in the Fairchild 660 signal amp, wich according to the manual is 14dB. When I do my calculations I get about 18dB. I thought I might as well ask here, and maybe someone will shime in and tell me what I'm doing wrong.  ;)

I use this formula:       Av = (mu * Rp) / (Rp + ra)
   
Av = gain   
mu = amplification factor, which is 17 for the 6386
Rp = plate resistor
ra = anode resistance, which is 4250 ohms  for the 6386
The values of 17 for amplification factor and 4250 for Rp are given for a bias of -4.8V (9.6mA).
Remember the 6386 is a vari-mu tube. The input stage is pre-biased from the -17V supply, so the idle bias current is probably lower, thus the amplification factor is lower too.
Quote
To find out the plate resistor (Rp) that every tube is seeing, I am thinking like this: If the output transformer secondary is loaded down by 600 ohms, the reflected resistance on the primary side is 48.6 kohms (because of the 9:1 ratio).
that is correct. Since the tube stage is a long-tailed pair, each side is seeing half of this, that is 24.3 kohms. Is this correct? But then since the amp has 4 tubes in parallel, this resistance is multiplied by 4. The plate resistance that every tube "is seeing" is 97.2 kohms. [/quote] That is also correct.
Quote
If I put this value for Rp in the formula I get a gain at about 16. Due to the long tailed pair, this value is then divided between the two anodes, so the gain is actually 8.
I don't understand that. Within your parameters, each 6386's has a gain of about 16, period. Each leg of the OT's primary sees that gain.
The only reason the gain is not 24dB is because the operating point is much lower than what you use.
Who's right or wrong is irrelevant. What matters is what's right or wrong.
Star ground is for electricians.

Flundran

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Re: Fairchild signal amp gain calculations
« Reply #3 on: October 21, 2011, 02:19:03 PM »
Hi MeToo2 and thank you for your input.  :) Those links are great, I have started to read about composite load lines for the first time. What is the difference between a long tail pair vs. a short tail pair? You and abbey road d enfer are  right about those mu and plate resistance values that I took from the datasheet being useless in this particular case...

You'd almost certainly have to have a look at the composite load lines of the large signal model to get a better picture.

Are those to be found in a datasheet? Or do you mean the composite load line I will be drawing myself?

Quote
If I put this value for Rp in the formula I get a gain at about 16. Due to the long tailed pair, this value is then divided between the two anodes, so the gain is actually 8.
I don't understand that. Within your parameters, each 6386's has a gain of about 16, period. Each leg of the OT's primary sees that gain.
The only reason the gain is not 24dB is because the operating point is much lower than what you use.

Ah, ok. I read a text somewhere about cutting the gain 1/2 in push-pull, but I think I misread that part. Thanks for your input also!  :)

I drew a couple of loadlines just now. They indicate that this amp stage (at the chosen bias point) is operating in class A. Is it correct?  I chose a biaspoint of -18V based on a handwritten note in 660 schematic found at the DIY group Gmail account.





« Last Edit: October 22, 2011, 05:35:22 AM by Flundran »

mjk

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Re: Fairchild signal amp gain calculations
« Reply #4 on: October 22, 2011, 10:52:01 AM »
OK, let's take your -18V bias load line as an example and estimate small signal gain. Let's take a big enough swing to reduce the error reading lines off the chart, say 20V peak grid-to-grid. It is indeed class A operation, but highly asymmetric. For one grid going +10V, the plate swing is about 56V. For the opposite side the grid swings -10V and it's plate swings about 44V. Adding the two plate swings gives about 100V plate signal for 20V grid signal. Pretty close to 14 dB of gain.
« Last Edit: October 22, 2011, 11:07:02 AM by mjk »

Flundran

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Re: Fairchild signal amp gain calculations
« Reply #5 on: October 22, 2011, 11:35:58 AM »
Wow, that was simple.  :) It does seem to add up perfectly. Thanks for the explanation mjk!

MeToo2

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Re: Fairchild signal amp gain calculations
« Reply #6 on: October 27, 2011, 07:03:57 AM »
OK, let's take your -18V bias load line as an example and estimate small signal gain. Let's take a big enough swing to reduce the error reading lines off the chart, say 20V peak grid-to-grid. It is indeed class A operation, but highly asymmetric. For one grid going +10V, the plate swing is about 56V. For the opposite side the grid swings -10V and it's plate swings about 44V. Adding the two plate swings gives about 100V plate signal for 20V grid signal. Pretty close to 14 dB of gain.
Exactly. You've got it. Good explanation.

Hi MeToo2 and thank you for your input.  :) Those links are great, I have started to read about composite load lines for the first time. What is the difference between a long tail pair vs. a short tail pair? You and abbey road d enfer are  right about those mu and plate resistance values that I took from the datasheet being useless in this particular case...

Are those to be found in a datasheet? Or do you mean the composite load line I will be drawing myself?

The values on the datasheet assume a particular single ended class A circuit operating at a particular operating point .

You need to use your own load line on the composite characteristics at the operating point in your circuit, or exactly as already MJK stated (as he's already taken into account the circuit topology in his explanation).

The difference between a theoretical long tail pair and short tail is that a long tail pair assumes that the total current through the tail is always pretty much constant irrespective of voltage (perfect constant current sink, so that the common cathode voltage is largely irrelevant.) So if one tube sources more plate current, it automatically robs current from the other tube because the total plate current for both tubes is kept constant by the long tail. A short tailed pair does not behave as a perfect current sink. If one tube sources more current, it also affects Vcathode on both tubes and thus Vg (and thus the total plate current for the 2 tubes also varies)



Small signal models try to predict AC characteristics at one DC operating point and assume that the gain and other transfer characteristics can be characterised by a single linear parameter: e.g. mu = dVo/ dVi.

Large signal models (looking at the actual tube curves and then taking a delta over two or more points) do not make this assumption about linearity. As MJK said, this circuit operates asymmetrically.

A vari-mu tube is a remote cut-off triode, which changes all sorts of parameters across its operating range, not just mu, but transconductance and plate resistance too. The curves are all over the place, which means linear small signal approximations start to break down pretty quickly. Any value of mu or other parameters is only valid at that one DC operating point. Change the DC operating point (by decreasing the value of Vg and/ or Ip or when going into compression) and you will get very different results. That's why it acts a compressor. Because it's non linear.

Also assuming Class A can also be misleading, depending on the load and input signal. A tube can only "turn off," and it cannot source negative current. Which means during class A operation the tubes see a different load impedance than in class B operation (once one tube turns off) That happens at different times for different output voltages and different loads, so the circuit can operate as class A up until a certain amplitude of input and then class B for the peaks. Sine wave in does not mean sine wave out. Bye bye linear model. Hello distortion.

regards,

opacheco

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Re: Fairchild signal amp gain calculations
« Reply #7 on: July 31, 2012, 08:22:45 PM »
OK, let's take your -18V bias load line as an example and estimate small signal gain. Let's take a big enough swing to reduce the error reading lines off the chart, say 20V peak grid-to-grid. It is indeed class A operation, but highly asymmetric. For one grid going +10V, the plate swing is about 56V. For the opposite side the grid swings -10V and it's plate swings about 44V. Adding the two plate swings gives about 100V plate signal for 20V grid signal. Pretty close to 14 dB of gain.

What was the LOAD RESISTANCE used for these calculations??....2 x 9^2 x 600Ohms = 97.2 K Ohms for a pair of 6386?????.....How about the anothers 3 Pairs of them??, How these 3 anothers Pairs affect this Result??

Opacheco.
« Last Edit: July 31, 2012, 08:29:12 PM by opacheco »
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mjk

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Re: Fairchild signal amp gain calculations
« Reply #8 on: July 31, 2012, 11:31:02 PM »
OK, let's take your -18V bias load line as an example and estimate small signal gain. Let's take a big enough swing to reduce the error reading lines off the chart, say 20V peak grid-to-grid. It is indeed class A operation, but highly asymmetric. For one grid going +10V, the plate swing is about 56V. For the opposite side the grid swings -10V and it's plate swings about 44V. Adding the two plate swings gives about 100V plate signal for 20V grid signal. Pretty close to 14 dB of gain.

What was the LOAD RESISTANCE used for these calculations??....2 x 9^2 x 600Ohms = 97.2 K Ohms for a pair of 6386?????.....How about the anothers 3 Pairs of them??, How these 3 anothers Pairs affect this Result??

Opacheco.

Here again is the load resistance calculation from the first post of the thread:

"To find out the plate resistor (Rp) that every tube is seeing, I am thinking like this: If the output transformer secondary is loaded down by 600 ohms, the reflected resistance on the primary side is 48.6 kohms (because of the 9:1 ratio). Since the tube stage is a long-tailed pair, each side is seeing half of this, that is 24.3 kohms. Is this correct? But then since the amp has 4 tubes in parallel, this resistance is multiplied by 4. The plate resistance that every tube "is seeing" is 97.2 kohms."

I see no problem with this analysis for *worst case* load line steepness; when we're driving a 600 ohm load. This is the load line represented on the plate curve.

In push-pull each side sees 1/2 the load resistance, or 24.3 kohms. Because there are 4 tubes inparallel, each tube sees 4x the resistance per side, or 97.2 kohms

Can you point out a specific problem or issue?

opacheco

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Re: Fairchild signal amp gain calculations
« Reply #9 on: August 08, 2012, 03:35:12 AM »
OK, let's take your -18V bias load line as an example and estimate small signal gain. Let's take a big enough swing to reduce the error reading lines off the chart, say 20V peak grid-to-grid. It is indeed class A operation, but highly asymmetric. For one grid going +10V, the plate swing is about 56V. For the opposite side the grid swings -10V and it's plate swings about 44V. Adding the two plate swings gives about 100V plate signal for 20V grid signal. Pretty close to 14 dB of gain.

What was the LOAD RESISTANCE used for these calculations??....2 x 9^2 x 600Ohms = 97.2 K Ohms for a pair of 6386?????.....How about the anothers 3 Pairs of them??, How this 3 anothers Pairs affect this Result??

Opacheco.

Here again is the load resistance calculation from the first post of the thread:

"To find out the plate resistor (Rp) that every tube is seeing, I am thinking like this: If the output transformer secondary is loaded down by 600 ohms, the reflected resistance on the primary side is 48.6 kohms (because of the 9:1 ratio). Since the tube stage is a long-tailed pair, each side is seeing half of this, that is 24.3 kohms. Is this correct? But then since the amp has 4 tubes in parallel, this resistance is multiplied by 4. The plate resistance that every tube "is seeing" is 97.2 kohms."

I see no problem with this analysis for *worst case* load line steepness; when we're driving a 600 ohm load. This is the load line represented on the plate curve.

In push-pull each side sees 1/2 the load resistance, or 24.3 kohms. Because there are 4 tubes inparallel, each tube sees 4x the resistance per side, or 97.2 kohms

Can you point out a specific problem or issue?


mjk, very Clear explanation!!......but in conclusion, what we get for these push pull 4 Tubes 6386 preamp ??.....what mean this 97.2Kohms per each tube???....for the tube performance, is this a high or low impedance???

Opacheco.

   
« Last Edit: August 08, 2012, 11:51:28 AM by opacheco »
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MeToo2

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Re: Fairchild signal amp gain calculations
« Reply #10 on: August 08, 2012, 12:53:22 PM »

mjk, very Clear explanation!!......but in conclusion, what we get for these push pull 4 Tubes 6386 preamp ??.....what mean this 97.2Kohms per each tube???....for the tube performance, is this a high or low impedance???

Opacheco.
 
When the Fairchild is in deep compression, the 6386 (a remote cutoff triode) is running very far to the left of its normal operating curve (large -ve grid), so you need a very high effective load impedance [and hence running 4 tubes in parallel] to avoid loading the output too much and thus affecting the output voltage when the load impedance changes (for example at different frequencies). Other designs of tube compressor (like the Altec, gates etc.) have a buffer amp on the output precisely to avoid loading the tube that is compressing.
« Last Edit: August 08, 2012, 01:02:28 PM by MeToo2 »

opacheco

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Re: Fairchild signal amp gain calculations
« Reply #11 on: August 08, 2012, 01:17:27 PM »

mjk, very Clear explanation!!......but in conclusion, what we get for these push pull 4 Tubes 6386 preamp ??.....what mean this 97.2Kohms per each tube???....for the tube performance, is this a high or low impedance???

Opacheco.
 
When the Fairchild is in deep compression, the 6386 (a remote cutoff triode) is running very far to the left of its normal operating curve (large -ve grid), so you need a very high effective load impedance [and hence running 4 tubes in parallel] to avoid loading the output too much and thus affecting the output voltage when the load impedance changes (for example at different frequencies). Other designs of tube compressor (like the Altec, gates etc.) have a buffer amp on the output precisely to avoid loading the tube that is compressing.

MeToo2, ok......but what is the optimal impedance for this tube then for this particular use in orden to avoid loading the output too much??.....97.2Kohms per side tube???...where I can find this data or how I need to do in order to find this data??

Thanks for your very clear explanations!
Opacheco.
I love the Vaccum Tubes Sound!!!

mjk

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Re: Fairchild signal amp gain calculations
« Reply #12 on: August 10, 2012, 10:56:58 AM »
OK, let's take your -18V bias load line as an example and estimate small signal gain. Let's take a big enough swing to reduce the error reading lines off the chart, say 20V peak grid-to-grid. It is indeed class A operation, but highly asymmetric. For one grid going +10V, the plate swing is about 56V. For the opposite side the grid swings -10V and it's plate swings about 44V. Adding the two plate swings gives about 100V plate signal for 20V grid signal. Pretty close to 14 dB of gain.

What was the LOAD RESISTANCE used for these calculations??....2 x 9^2 x 600Ohms = 97.2 K Ohms for a pair of 6386?????.....How about the anothers 3 Pairs of them??, How this 3 anothers Pairs affect this Result??

Opacheco.

Here again is the load resistance calculation from the first post of the thread:

"To find out the plate resistor (Rp) that every tube is seeing, I am thinking like this: If the output transformer secondary is loaded down by 600 ohms, the reflected resistance on the primary side is 48.6 kohms (because of the 9:1 ratio). Since the tube stage is a long-tailed pair, each side is seeing half of this, that is 24.3 kohms. Is this correct? But then since the amp has 4 tubes in parallel, this resistance is multiplied by 4. The plate resistance that every tube "is seeing" is 97.2 kohms."

I see no problem with this analysis for *worst case* load line steepness; when we're driving a 600 ohm load. This is the load line represented on the plate curve.

In push-pull each side sees 1/2 the load resistance, or 24.3 kohms. Because there are 4 tubes inparallel, each tube sees 4x the resistance per side, or 97.2 kohms

Can you point out a specific problem or issue?


mjk, very Clear explanation!!......but in conclusion, what we get for these push pull 4 Tubes 6386 preamp ??.....what mean this 97.2Kohms per each tube???....for the tube performance, is this a high or low impedance???

Opacheco.

   

OK, let's look at the plate resistance vs. load resistance to see what the damping factor wrt the load is. As MeToo2 points out, when the load resistance is too small relative to the plate resistance there will be gain error. There are no hard and fast limits but we can evaluate the performance.

At the -18V bias point the plate resistance of a single 6386 is about 15K ohms. At -40V the plate resistance is about 25K ohms. Taking the worst case of these, the damping factor is about 4 (ratio of load resistance to plate resistance). This means that between light load and 600 ohms load the output voltage will drop by about 20%, resulting in about -2 dB max gain variation from light load to 600 ohm load when in deep gain reduction. The gain variation over load with no GR will be about -1 dB. So the difference in the GR curve from no load to 600 ohms load will be about 1 dB i.e. there will be about 1 dB more gain reduction with 600 ohms load vs. no load, due to the varying non-zero plate resistance of the 6386.

There will also be a little additional symmetric distortion at higher GR when driving a 600 ohm load but I leave that as homework ;-)
« Last Edit: August 10, 2012, 11:58:56 AM by mjk »