How to calculate Tube Preamp output impedance?
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john65b
diyAudio Member

Join Date: Jan 2005
Location: Chicago
How to calculate Tube Preamp output impedance?

Hello all,

I have built the following 12B4 seen in the attached...only difference is i changed out the Cathode Resistor for three red LEDs. I remember getting between 7 and 9 volts across those LEDs for the bias. Sounds very nice.

Anyway, I would like to know if there's a simple way to calculate output impedance of a tube preamp - I read somewhere 1/gm, and maybe for a single tube pre setup as simple as the one attached, the 1/gm may be correct (gm of 12B4 is around .006 ohm??)

I also have a more complicated AN M7 Clone with two 5687 tube/channel and wonder how to calculate the output impedance for it too...(2/gm?)

I am concerned that with the low-ish input impedance of the Aleph5 or the Krell Clone I want to build, the two preamps may be too low for either of these to drive.

The UCD400 I have at 100K was no problem for either of these...
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 18th November 2007, 09:29 PM #2 7N7   diyAudio Member     Join Date: Jan 2003 Location: England Hello. The calculation 1/gm is for cathode followers. For a Common Cathode amplifier as shown in your diagram the method is to treat the anode load and the valve's anode resistance as resistors in parallel i.e. R1 x R2/R1+R2 Using the one voltage you have given and looking at the GE curves on Frank's I estimate the anode resistance at your operating point to be about 1.4k. This gives a result i.e. output resistance of just over 1k plus of course your 100-ohm resistor. Remember that anode resistance and gm are variables; the numbers given by the makers are when operating the valve close to bursting point so the numbers look good. Given gm for 12B4A is 6300 micromhos or 6.3mA/V. I would say that you should run more current through your 12B4, at present you are a bit too close to the "knee" in the curves. This will increase gm and probably reduce noise. Another tip is to float the heatr at about +40V; 12B4s are notorious for poor heater/cathode insulation. 7N7 Edit: Sorry overlooked the 5687 bit. A paralleled pair of identical valves will have half the output resistance of a single valve. __________________ Plug them in and light them up
 18th November 2007, 09:51 PM #3 john65b   diyAudio Member     Join Date: Jan 2005 Location: Chicago 1K +100R = 1.1kR on the my 12B4 Output Impedance. Cool. Should be OK with the Krell (22K Zin), but the Aleph may be a prob (11K Zin) I see we add that last resistor - I see lot of Tube Pre designs with a 1Meg resistor to ground at output - doesn't this raise the output impedance of the pre to a value most amps can't work with? Also, about the raising the current on the 12B4..,B+ is 220V and I think I am at 15ma. because I get around 90V at the plate. More current would be lower the Plate Resistor (currently 8.2kR) or change the LED from red to blue or add another Red (raise voltage)? Sorry, It has been a while since working on tube preamps... EDIT: Whoops, you added the 100R not the 100k to ground...I understand... And if I remember correctly the cathode end determines the current, so changing the plate resistor won't change current...correct? __________________ I Speak Spanish to God, Italian to Women, French to Men, and German to My Horse. Charles V
 18th November 2007, 10:24 PM #4 7N7   diyAudio Member     Join Date: Jan 2003 Location: England Changing the anode resistor will change the current as it will increase the voltage across the valve for a given value of cathode bias. However, best to keep the anode resistance high as this reduces distortion. Looking at the curves if I were you I would though reduce the value of the load resistor say to 6K or so. You will then be more on the characteristic So reduce the bias and increase the current; you probably do not need to swing many volts to drive your power amplifiers so there should be no problem. 7N7 __________________ Plug them in and light them up
Jeb-D.
diyAudio Member

Join Date: Nov 2005
Location: SoCal
Quote:
 I see we add that last resistor - I see lot of Tube Pre designs with a 1Meg resistor to ground at output - doesn't this raise the output impedance of the pre to a value most amps can't work with?
If I understand what you are asking. The 1M resistor is usually at the output to keep the capacitor output from floating. If it weren't there and you were using the preamp into a source selector switch it would cause a pop in the speaker whenever you select it as a source. 1M is low enough in value to keep the cap output at ground potential. However it is still high enough in value to where you won't have to increase the size of the coupling cap to prevent low frequency roll off.

 18th November 2007, 11:06 PM #6 john65b   diyAudio Member     Join Date: Jan 2005 Location: Chicago Yes, I got the 100R and the 100K mixed up. You now explained its purpose. __________________ I Speak Spanish to God, Italian to Women, French to Men, and German to My Horse. Charles V
 19th November 2007, 12:28 AM #7 Jeb-D.   diyAudio Member     Join Date: Nov 2005 Location: SoCal Your placement of the 100R and the 100K look fine as in your posted schematic. It will work. You only need that 100R if you are worried about protecting it from a shorted output, or if it is there to attenuate the signal into a desired load. Otherwise it can be removed. That 100k can be changed to 1M and the output coupling cap can be reduced to 1uF. Of course these are just suggestions, I don't intend to cramp your style.
 19th November 2007, 03:03 AM #8 mus   diyAudio Member   Join Date: Mar 2006 Location: bc Hi, How about active loaded CCS for the anode resistor, would it be have even lower output impedance?
 19th November 2007, 03:17 AM #9 Conrad Hoffman   diyAudio Member   Join Date: Mar 2007 Location: Canandaigua, NY USA Calc is good, and all the above is good, but you're dealing with some variables that aren't carved in stone. After the calc, why not load the preamp with a resistor to the point where you have some small but easily measured drop in signal, and use ohms law. Imagine the amp is a perfect voltage source feeding a resistor (the output impedance). With no load, the voltage will be the same on both sides of the imaginary resistor. When you load it, the voltage drops on the output. You now have a simple voltage divider where you can calculate the upper resistor. Do it at 20Hz and 20kHz, or whatever limits you like. When the load resistor pulls the signal down to half, that's also the output impedance, but IMO it throws everything in the circuit off, so I wouldn't go down that path. Just use a load that drops the signal 5% or so.
 19th November 2007, 04:39 AM #10 Bandersnatch   diyAudio Member   Join Date: Jul 2003 Location: Ann Arbor, MI hey-Hey!!!, Conrad has the proper idea to measure it. Keep in mind that the couplng cap is in series, and at low frequency it will add substantially to the output Z. Still it would be best to measure in a few different output frequencies just to know what's in there. 20, 200 and 2k cps for instance. cheers, Douglas __________________ the Tnuctipun will return

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