A beginners very first try of a very very simple headphone amp. Help.

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fatheaddrummer

Active member
Joined
Aug 10, 2013
Messages
30
Hi everybody,

As the title says, i am a very beginner. I startet learning about electronics about 1 year ago. I really want to get an understanding of those things, so i tried to build my first little headphone amp...

I chose the Power Supply from the G Pultec und the amp-parts values from the tubes datasheet and some i calculated for myself.
As the output Trafo is 12,8:1, in combination with 32Ohms Headphones it should be around 5k Anode resistor...

i am pretty sure, that there are lots of mistakes, so i a glad if anyone can tell me how to do it better.
It took me about 3 days, to do this Projekt. So please be patient with me :)

THANK YOU VERY MUCH! :)
 

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Did you build it and try it?

I don't see a problem.

C6 *could* be 10 times smaller. 300uFd ought to be plenty.

C5 could be a little bigger. As shown, it is -3dB at 20Hz, or part-dB down at 50Hz, slightly crimping the deepest bass.

This is, as drawn, a *5 Watt* headphone amp. That is WAY more power than headphones need, about 100X more. However it is very low-gain. It needs maybe 12V at the grid of the EL84 to make 5 Watts. With a more typical 1V line-level signal you will get more like 0.05 watts, which is fine for headphones.

You could probably run it on much lower B+ supply, like 100V,  and still have ample output for headphones. The EL84 would run "cool" and last forever.

You could use a "lesser" tube than EL84, but EL84 is about as cheap as any and a lot more available than most.

If you are in USA/Canada, the 60FX5 is interesting. It is higher gain, only 2V-3V to make a whole Watt with 110V supply voltage. Scaled down to a 60V supply it is still over 0.2 Watts and you can get that with about 1V input. 60V supply because the heater is 60V, so you need just one 60V DC supply for the whole thing.

I'm not sure where you find a suitable transformer with high primary inductance and rated for many mA of plate current.
 
From the missing junction points in your schematic, I assume you used WIRE to draw electric connections.
Always use NET to connect components in EAGLE, please. Otherwise expect some funny stuff when doing
the PCB  :eek:

Cheers,
Carsten
 
Thank you very much PRR and Carsten!!

To PRR:

No I haven't built it yet, i wanted you guys to have a look at it first :)

How do you get the Values for C5 and C6? I have the values from this formula: 1/2*pi*frequenzy*R....I took C6 a little higher because i didn't find a cheap one for the right value....is there a better way or an easier way to get those values?

"This is, as drawn, a *5 Watt* headphone amp. That is WAY more power than headphones need, about 100X more. However it is very low-gain. It needs maybe 12V at the grid of the EL84 to make 5 Watts. With a more typical 1V line-level signal you will get more like 0.05 watts, which is fine for headphones"

Could you for me explain this in Detail? How to get those 0.05 Watts? Where is the 1V line-level signal coming from?
How can I calculate Ra/ the transformer when I regulate B+ down to 100V?

To Carsten:
Do I need to set junction Points manually between every connection or is it enough to just use the NET function which automatically draws the junction points just between the connected wires?

Thank you very much!!

Best,
Christian
 
fatheaddrummer said:
To Carsten:
Do I need to set junction Points manually between every connection or is it enough to just use the NET function which automatically draws the junction points just between the connected wires?

As you already may have noticed, the junctions are set automatically when using NET to draw connections. So there should be no need to add them manually.

Cheers,
Carsten
 
> formula: 1/2*pi*frequenzy*R....I

And what did you use for F and R?

If I assume 20Hz and take your 6,800uFd, I get 1.17 Ohms.

What I *see* at C6 is R5, a 135 Ohm resistor. About 100X bigger than 1.17 Ohms, so we would think to use a cap 100X smaller, 59 or 68uFd.

Actually we must figure R5 in parallel with the cathode internal impedance. This is 1/Gm, where Gm is the transconductance at/near your actual operating point. However for nearly all self-biased tubes in nearly all situations, 1/Gm is very similar to the cathode bias resistor. So just double the cap, 136uFd. In practice, 100u to 220u, about 15V-25V, which *should* be very available and inexpensive (9 cents, 17,000 in-stock) (6,800uFd seems to cost $2).

> Where is the 1V line-level signal coming from?

I thought you knew? From a hi-fi preamp? From a monitor-out on a mixer? Perhaps an iPod, though their outputs tend to be a little under a Volt.

> How to get those 0.05 Watts?

If an amplifier needs around 10 Volts input to make full 5 Watts output (roughly true for SE EL84), and you only put 1 V of input signal in it, the output voltage will be just 1/10th of "full" and the output *power* will be 1/(10^2) just 1/100th of "full". 5W/100 is 0.05 Watts, 50 milliWatts, which is good loud headphone drive. And working a simple power amp at 1/10th of full voltage is *about* 1/10th of the distortion, so 5% THD comes down to much less than 1% THD.

The "optimum" load for a SE pentode does not change much with supply voltage. For a large reduction, say 250V to 100V, the current goes down faster than the voltage, so you might pick a load 1.6X higher, 8K. However your "32 Ohm" headphones may be 24 ohms or 100 Ohms, and the useful range of a tube amplifier is very wide, so great precision is not needed.
 
Dear PRR,

Thank you very much for your time and the detailed explenation!

I think, i just made a stupid calaculating mistake, when figuring out the value for c6. This is now clear for me.

> Where is the 1V line-level signal coming from?

I thought you knew? From a hi-fi preamp? From a monitor-out on a mixer? Perhaps an iPod, though their outputs tend to be a little under a Volt.


Yes I know where it is coming from, i just haven't completely got which signal you have been talking about, but know i also got what you meant!

Also I think that I misunderstood what you told me about the 5 Watts in your first post. I thought you meant that i drive with 5 Watts power even with an input just with 1 V.....and thought you suggest that i change something completely. Doesn't matter. As I am a very beginner and english is not my mothers language i am very slow in understanding but i am trying hard. I think that I got everything now! Thank you very very much!

I will recalculate the parts as you told me and then try to post a stereo version of this and then build it and try it out.


 
Maybe a stupid question:

Can I use one PSU like above in parallel to drive two EL84 - for stereo? Or do I need 2 PSU?

Thanks! :)
 
fatheaddrummer said:
Maybe a stupid question:

Can I use one PSU like above in parallel to drive two EL84 - for stereo? Or do I need 2 PSU?

Thanks! :)

As long as your transformer is rated to provide the extra current required then you can use just one power supply. You will need to change the value of R1 to account for the current being double in a stereo unit. 3K3 is probably too high for R1 anyway. The correct value will depend on the idle current you choose for the tube.

Cheers

ian
 
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