Output Stage Current Requirements for Electrostatic Headphone Amps

[bobkatz]

Here's a crazy thought. If the reflected impedance on the primary side actually is 12 k ohms, and the peak power is even as low as 1 watt, then the peak voltage on the primary would be 109 volts. That's crazy. So hopefully I'm way off base on my reflected impedance calculation. 

[simmconn]

I see a few problems in your calculations. First, the load impedance is the characteristic of the phones, which is listed in your headphone’s spec sheet. It is not determined by the output voltage or current capabilities of an amp, which on the other hand are the amp’s design choices.

Second, the reflected impedance to the amp is affected by many factors (and follows square of the turns ratio in an ideal case). In this case, the parasitics can have a significant effect on the load impedance the amp sees. Take a measurement and you may be surprised.

Third, impedance is a vector. Using a scalar value is convenient but let’s not forget its vector nature and mostly reactive for the estat phones. The 1W of driving power would then be apparent power and not real power.

Last but not least, may I suggest that we limit questions and comments regarding one project to a single or a couple of threads only? 

[bobkatz]

On 8/1/2023 at 6:18 PM, simmconn said:

I see a few problems in your calculations. First, the load impedance is the characteristic of the phones, which is listed in your headphone’s spec sheet. It is not determined by the output voltage or current capabilities of an amp, which on the other hand are the amp’s design choices.

Second, the reflected impedance to the amp is affected by many factors (and follows square of the turns ratio in an ideal case). In this case, the parasitics can have a significant effect on the load impedance the amp sees. Take a measurement and you may be surprised.

Third, impedance is a vector. Using a scalar value is convenient but let’s not forget its vector nature and mostly reactive for the estat phones. The 1W of driving power would then be apparent power and not real power.

Last but not least, may I suggest that we limit questions and comments regarding one project to a single or a couple of threads only? 

Thanks.

Dear @simmconnResponding to your Last response first: I honestly thought this was the best single place to have this discussion, to continue a discussion started by @JimL.

If I'm going to be way off base not using vector quantities, then I'm definitely not qualified to proceed because vector algegra is not in my skill set. Anyway, thanks, I forgot that reflected impedance is RL x (Np/Ns) squared, oops, sorry. 

Also, the difference between "apparent power" and "real power" is over my head, so if I mess up below, just tell me I'm full of it and I'll shut up. I'd like to stick to simple scaler algebra as a first approximation if it's not too annoying. 

Anyway, I'm trying to use JimL's calculations of output stage current to judge the actual load, as opposed to the simple answer of the impedance of a pure capacitor. 

So, if the secondary has a nominal 3 ma., with a nominal power of 1 watt (per Gilmore) at 106 dB SPL, then 109 dB would be 2 watts, etc. I'll pick 8 watts at 115 dB max level as a worst case. 

Next, if it was 3 ma at 1 watt, then isn't it it 8.5 ma at 8 watts?  Because of i squared R.

OK, now can we calculate the rough, approximate reflected impedance load on the source amplifier knowing these quantities: secondary current, power and turns ratio of 33:1? I just want to have a rough idea of what kind of load the power amplifier is expected to drive. 

When I get to building this box I will measure the actual quantities and actual impedance loads. I guess we'd call it Practical versus theoretical physics. Just like in the Oppenheimer movie 🙂

Edited August 3, 2023 by bobkatz
i squared R (oops)