Final check - voltage at headphone jack

[Laudanum]

Finished the build. All resistance checks were fine and voltage checks were fine.
One question on the voltage at the headphone jack however.  When first turning on, voltage climbs to just below 9.8 volts then drops down to 0.  Well, actually, it jumps around the low millivolt range but if memory serves, this is normal.  So, am I ok with the 9.8 volts or do I need to troubleshoot?

Assuming all is well ... with it being normal to have dc voltage on the headphone jack at turn on, as a general rule should the amp be turned on for 30 seconds or a minute before plugging in phones?

Thanks

[Jim R.]

You should be fine.  When you plug in the headphones, the load will be even lower and thus the voltage as well.  This has come up before and the bottom line is that it's fine to plug the phones in before turning the amp on.  It would be a real pain if you had to disconnect the cans everytime you switched the amp off and on again.

Sounds like you're ready to go!

Let us know how it is,

Jim

[Laudanum]

Thanks Jim.   I figured if everything checked out OK as it did with the reistance and voltage checks, then eveything should be ok but didnt want to risk it. 
I'll be listening after dinner.

Thanks very much.

[Grainger49]

As the 100uF cap starts to see voltage on the input side it leaks a little voltage as it charges.  The 2.49k resistor will bleed it off.  But it bleeds with an RC time constant of 100uF X 2.49k ohms.  A headset will drain it faster with a lower impedance.

Make sense?

[Laudanum]

So ... a lower resistance (due to lower impedance of the phones) bleeds the caps leakage voltage faster than a higher resistance (that of the 2.49K bleeder resistors alone).

Does the decreased resistance decrease the leakage voltage (lower than 9Vdc) itself as well, or just bleed it off so fast that it cannot damage the drivers of the phones?

And since film caps dont leak like electrolytics, would that leakage voltage still be present, or lower,  with film output caps?

Figured if I was going to learn something I may as well make it a three-fer :-)

Thanks

[Paul Joppa]

The 9 volts results from charging the capacitor. It's the "jumps around the low millivolt range" stuff that is leakage. Just to be technical about is ...  :^)