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Old 17th August 2008, 02:07 PM   #1
awedio is offline awedio  United States
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Default Impedance, the 10x rule, losses & bandwidth

As a theoretical question:

What are the disadvantages, if any, of using matched impedances between stages when factors such as parasitic capacitance or inductance are negligible?

For example - if a tube gain stage has a 600ohm output Z and is driving a 600ohm input Z of a solid-state stage, where the SS stage does not have a Miller capacitance and there is no long cable to add capacitance, then I can only see two potential problem areas:

1) we "lose" half of the gain of the tube stage as a result of the power transfer characteristic of the 1:1 Z

2) if the stages are capacitively coupled then the capacitor must be quite large in order for us not to lose bandwidth in the low bass driving the 600ohm input

What happens to the high frequency bandwidth? If the stages are transformer coupled, then bandwidth degradation is solely dependent on the xformer itself, right?

What would be the advantages, if any, of lowering the output Z of the preceding stage to 60ohm, following the 10x rule?

Is the 10x rule predicated solely on getting 90% power transfer and easing capacitance issues, driving cables, etc.?
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Old 17th August 2008, 02:23 PM   #2
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Don't forget distortion. That 600R output Z stage loaded with a 600R load will give impressively large distortion figures.

In general, voltage amplifier stages perform best when the load is as high as possible.
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Old 17th August 2008, 02:36 PM   #3
awedio is offline awedio  United States
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SY,

Thanks for the quick reply. This is the kind of clue I'm looking for - but do you really mean:

-*TUBE based*- "voltage amplifier stages perform best when the load is as high as possible" ???

This is the kind of thing throwing me - matched impedances are regularly used with digital, RF applications, etc. and distortion problems are never mentioned. So are you specifically saying that *tube* circuits have trouble with low impedances? Or would the same hold for a SS circuit with 600ohm output Z- distortion problems?

If we ramp up the theoretical model 10 times, using 6000:6000ohm as an example, what are the implications?
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Old 17th August 2008, 02:51 PM   #4
SY is offline SY  United States
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Well, you posted this in the tube forum so I kinda assumed.

RF apps are different. Generally tuned stuff and more concerned with getting maximum power transfer than linearity. After all, we make class B and C transmitters, right?

In digital, impedance matching is important from the standpoint of reflections; this is not an issue at audio frequencies. And linearity is not a big deal when you're just switching on and off.
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Old 17th August 2008, 03:31 PM   #5
awedio is offline awedio  United States
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Well, you're right, i did post this in the tube forum! So, back to tubes - IOW choosing a lighter load, 6000R, what if the output Z of the preceding tube stage is also 6000ohm? Is the 1:1 Z-ratio simply not copasetic when tubes are involved?

Or, isn't there tube-based studio gear out there, even if it is transformer-interfaced, running 600ohm inputs yet not relying on a step-down or impedance-matching xformer but a 1:1 xformer?
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Old 17th August 2008, 03:42 PM   #6
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In general, no. It's a complex thing, many books have been written on the subject, but I can give you 95% of it while standing on one leg:

Assume a triode. The equivalent circuit for an undegenerated triode with a plate load resistor looks like a voltage source in series with the plate resistance, rp, in series with the load resistance, RL. The output is taken across the load resistance, so you end up with something that looks like the voltage source with a divider across it. Clearly, the output is proportional to RL/(RL + rp), with mu being the coefficient.

Now, RL is fixed. On the other hand, rp is a function of the tube, it's the slope of plate voltage versus plate current. It varies quite strongly with plate current. So as the plate current swings, so does rp and thus the divider equation becomes non-linear. The lower the load, the more the required current swing and the greater the (variable) rp contribution to the transfer function. Double whammy.

As RL gets much bigger than rp, not only does the gain get higher, but the divider ratio tends toward unity, i.e., the tube runs more linearly.

Whew, let me put my foot back down.
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Old 17th August 2008, 03:53 PM   #7
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Maximum power transfer occurs with matched impedance - you want this with power output stages and extremely low signal levels (to minimize noise). Voltage amplifier stages will be optimized for different parameters - maximum voltage swing, lowest distortion. This is true regardless of the type of active device. We aren't concerned with the efficiency of power transfer in a voltage amplifier stage - it's near zero.

A cathode follower (or emitter follower or source follower...) with an output impedance of 600 Ohms can't deliver a large voltage swing at low distortion into 600 Ohms. It can drive a high impedance load with significant capacitance or a cable.

The 10X "rule" is a good way to assure that the load isn't affecting the source characteristics (which is specified as a voltage output after all).
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Old 17th August 2008, 04:49 PM   #8
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Very enlightening, gentlemen. Thank you!

I needed the tools to visualize what's really going on in between the two stages, and you have given me just that. The textbook-style explanations I had read so far were unable to.
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Old 17th August 2008, 08:39 PM   #9
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Here are some things to consider about the 600 ohm system...
It is balanced, therefore it's common-mode noise rejection is significiant in audio.... The common return and the shield are seperate unlike typical single ended, this is important..
The lower Z allows longer cable runs...with signifantly less loss than the typical single ended.....

Chris
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Old 17th August 2008, 09:36 PM   #10
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Hi,
consider voltage output ability and current output ability separately.

If your output stage can deliver 20Vac (~=60Vpp=30Vpk) then can it deliver that voltage into a 100k load?
This is where output current comes into the equation.
Iout=Vout/Iload eg, 20Vac =30Vpk.
Ioutpk=30/100k=0.3mApk. Is the output stage capable of delivering 30Vpk AND 0.3mApk.
Now change the load to 10k.
The output stage must now be capable of delivering 30Vpk AND 2.83mApk.

Now, put in 600ohm as your load.
Is the output stage capable of delivering 30Vpk AND 25mApk. I suspect not.
It is often instructive to look at both voltage and current (separately) in many circuits, even SS and input circuits.
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