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 Posted: 24 Dec 2010, 02:21

Joined: 27 Nov 2010, 02:18
Posts: 34
Location: Arroyo Grande, CA, US
I studied EE in college back in the '90s, and have a grasp on the general concepts, but am a bit out of touch with details.

I'm trying to size some DC blocking caps and am trying to remember the math behind RC frequency response. I know, in general, the higher the impedance (either by resistance or capacitance), the lower the frequency response. That is, if you're cutting off too much low end, you can either make the cap bigger, or increase the impedance on the load (both of which have side effects, so you have to be careful with which one you choose.)

I went in search of the specific numbers to back this up and found this article which links to this page of a ton of math that eventually boils down to:

The cut-off (read: half point) frequency of an RC circuit is: f_c = 1/(2*Pi*R*C)

Which agrees with my general understanding: Increase either the R or the C and the frequency response lowers. So, let's say I was building a simple tube headphone amp which is notoriously bright and lacking a bit in low end, and I wanted to calculate what the frequency response of the DC blocking caps on the front and back end were.

The front end is a 2.2uF cap, to a 100kOhm resistor to ground, and the grid of the tube (which has "a very high impedance," probably higher than 100kOhm, so we'll ignore it here.) Our equation above would give us:

f_c = 1/(6.28*0.0000022*100000) = .72Hz

Uhh.. Really? Even if the grid were 10k, the response would still be 7.2Hz. Am I doing my math correctly here? Specifically, that the DC blocking cap is _NOT_ the source of the missing low end?

Similarly, the output drives a 470uF cap to your headphone load (likely lower than the 1k bleed cap, which we'll ignore) which is 50 ohms in my case. This gives us:

f_c = 1/(6.28*0.000470*50) = 6.8Hz

10x the input, but still less than 1/2 the response of the headphones, and 1/3 the response of my ears.

Does anyone see anything wrong with my math here? That .72Hz seems suspect, but the 6.8Hz seems reasonable. Thanks for your help.

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 Posted: 24 Dec 2010, 16:05
 Editor

Joined: 28 May 2008, 21:53
Posts: 4580
SmittyHalibut wrote:
The cut-off (read: half point) frequency of an RC circuit is: f_c = 1/(2*Pi*R*C)

That gives the corner frequency of the RC filter which is the point where the filter has attenuated the signal by 3 dB.
Here is a good article on input caps for head amps: http://tangentsoft.net/audio/input-cap.html
Cheers

_________________

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 Posted: 25 Dec 2010, 00:01

Joined: 27 Nov 2010, 02:18
Posts: 34
Location: Arroyo Grande, CA, US
Gio wrote:
Here is a good article on input caps for head amps: http://tangentsoft.net/audio/input-cap.html

That is a GREAT article. I knew about the high-pass effects of the DC blocking caps, but didn't think about the phase distortion or the cap distortion itself. Thank you for pointing me to that.

Quick summary: there are reasons other than the high pass qualities of a DC blocking cap to choose a large cap, on the order of a decade larger (10x), than just frequency response would dictate.

-Mark

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 Posted: 25 Dec 2010, 01:07

Joined: 27 Nov 2010, 02:18
Posts: 34
Location: Arroyo Grande, CA, US
Here's a silly idea. That article talks about eliminating DC blocking caps, assuming you can first eliminate your DC offset voltage.

I'm talking about using PCM2900B from TI/Burr Brown as one of the inputs to my mixer. The chip generates a common reference voltage on pin 14, Vcom. On a single ended supply, it is typically tied to ground through a 10uF electrolytic cap.

What if I build a transformer isolated, linear power supply, and instead of grounding the negative supply line, I ground the Vcom pin. The signal output will then be centered around Vcom, which is ground. (I want two DACs, so I'd build two such supplies.) My DACs are now DC coupled to the mixer section, which is entirely passive until it hits an OPA2227 opamp at the output stage. Ok, so we build a third supply, with a grounded center tap, so we end up with +/- 2.5vDC (or 6vDC, or whatever... large enough to provide good headroom).. That allows us to DC couple the opamp.

(There is the third input of the iPhone. I can pretty much guaranty (and will verify) that it's already DC blocked; I just plan to ground it's sleeve to our new generated ground, and not provide any additional DC blocking caps.)

The only remaining question is the tube amp. I can make Yet Another(tm) isolated supply for B+, but how do I tie it to ground? The circuit I'm using has a DC blocking cap, then a 100k resistor to ground (or B- in this case), then directly to the grid. There is no explicit bias voltage set to which I can tie my ground. Would DC coupling the grid to the output of my mixer magically center the tube's supply where it needed to go, using the 100k to B- and the bleed current in the grid, to Do The Right Thing(tm)?

And if so, what will the DC offset at the output look like? The circuit does have the bias pot at the top, which we normally adjust to center the output. I wonder if we could just center it around our ground and do away with the output caps too.

It sounds like these two things: centering the grid around ground, and centering the output around ground, are potentially two mutually exclusive things.

Am I just going a bit over-board with the DC coupling of the system here, or is there some sense to this?

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