Power filtering calculations

[PBE6]

I'm building a new pedal for a friend who will be using wall-wart power. I found some tiny 1000uF capacitors, so I'm going to use them in my filtering scheme. The design will be essentially the same as the one included in Jack Orman's simple mixer:


I will however be modifying it to use 1000uF instead of 47uF caps, and also splitting the 100 ohm series resistor into two 47 ohm resistors each with their own 1000uF cap. Are the following calculations correct?


filtering on Vcc+ (47 ohm-1000uF x2):

freq = 1/(2* pi * 47 ohm * 1000 * 1E-6 F) = 3.39 Hz

hum freq = 60 Hz

single filter reduction = 20 * log(3.39/60) = -25.0 dB

double filter reduction = 2 x single = -50.0 dB


Additional filtering on Vref (10k-1000uF x1):

freq = 1/(2 * pi * 10 * 1E3 ohm * 1000 * 1E-6 F) = 0.0159 Hz

single filter reduction = 20 * log(0.0159/60) = -71.5 dB

total reduction = -50.0 dB + -71.5 dB = -122 dB


This seems like plenty of hum reduction, even if the signal itself will be boosted by 20-30 dB. Does that sound right? Are there any drawbacks to using such high value capacitors?

Also wondering about voltage drop with the series resistors if I have an LED. Is this how you calculate the voltage drop due to the two 47 ohm resistors?


Vcc+ = 9 V

Vdrop(LED) = 1.5 V

V = 9 - 1.5 = 7.5 V

R(LED) = 4700 ohm
R(filter) = 47 + 47 = 94 ohm

V = I * R --> I = V / R = 7.5 / (4700 + 94) = 0.00156 A = 1.56 mA

Vdrop(filter) = I * R(filter) = 0.00156 * 94 = 0.15 V


Is the above calculation correct?

Thanks!

[PBE6]

Actually, I just had a second thought about the voltage drop part. Maybe this is more correct?

Vcc+ = V(series) + V(LED) + V(4k7)

Vcc+ = 9 V

V(LED) = 1.5 V

Therefore:

V(series) + V(4k7) = 7.5 V

Since all resistors are in series, we can calculate the voltage drops using a voltage divider equation:

V(series) = 7.5 * 4700 / (94 + 4700) = 0.98 V


This seems more correct than the calculation in my previous post. Any thoughts?

[slacker]

You should get the same answer with either method, looks like you've just done 4700/4794 = 0.98 and not multiplied it by 7.5, also what you've calculated there is the drop across the 4k7 which is about 7.35, leaving about 0.15 across the 94 Ohm.
Don't forget you need to include the current draw for the actual circuit as well as just the LED, unless you're assuming that it's insignificant compared to the LED.

[PBE6]

Ah yes! Thanks!

My mistake, the 0.98 number was the multiplier, but I forgot to multiply it by 7.5 V.  And I switch the 94 and 4700 too. As noted, the proper calculation should be:

V(series) = 7.5 * 94 / (94 + 4700) = 0.15 V.

Much more reasonable!

I typically do assume that the LED takes most of the current, but you're right that may not always be the case.

How about the filtering?

[blackieNYC]

My, that is a big cap though.  Do you think it will actually make a diff over 47uf?  I'm not bein a smart@ss, just curious. Draws a little juice when you power it up, I would imagine.  A pedal board full of those could challenge a wall wart, theoretically.  Coincidently, I saw a Studer console schematic today with 100uf series coupling caps.  Never noticed that in an audio path before. It's certainly flat down to 20.

[PBE6]

The 100 ohm-47uF combination gives a cutoff of 33.9 Hz, which according to method provides 5.0 dB of attenuation. The 10k-47uF combo provides another 45 dB, so altogether that's 50 dB of rejection on Vref. If a boost is applied in my circuit, this rejection will be reduced by the same amount. The design could be sufficient or not, depending on the boost and S/N ratio.

(Provided my calculations are correct, anyway.)

How do you calculate the power drawn by the resistor/cap combo?

[blackieNYC]

Usually just a concern in power amps and higher current devices, but the DC filter caps are a dead short to ground until charged. The bigger, the badder.  This is the inrush current that causes us to use slo-blo fuses, for one.  If someone can clarify this, or knows how to calculate the power drawn, that would be interesting info.

[PRR]

> hum freq = 60 Hz

If your wall-wart isn't total junk, it will be full-wave rectification and 100Hz/120Hz ripple. So you get another 6dB per stage.

Do you need that much?? What kind of crap are you feeding it? I would put raw (rectified) AC to a TL07x with less capacitance than that.

Also note that the TL07x designers were aware that pin 7 might not see silk-smooth power. It has 80dB rail-rejection at DC (AC rejection not specified; probably 60dB at 120Hz).

But if you have cute kilo-uFd caps laying about your bench, go ahead.

> Additional filtering on Vref (10k-1000uF x1)

A full analysis would use the true node impedance, 5K; and also the voltage reduction (2:1). It comes to the same dB result though.

> total reduction = -50.0 dB + -71.5 dB = -122 dB

At Vref, yes, *in theory*.

Actually in ptactice it is hard to get better than 30dB-40dB per stage because of parasitic losses we call "negligible" in non-OVERkill design.

> drawbacks to using such high value capacitors?

Rough-calc, 10K charging 1,000uFd will take *ten seconds* to charge to 63% of final value. It may take a couple seconds after turn-on to get any sound at all. It may be many seconds before a maximum-level signal is delivered un-clipped. Probably not a problem. I never saw a musician plug-up and play that fast.

> DC filter caps are a dead short to ground until charged.

So draw a short. With the drawing at top of thread we have 100 Ohms in front. This will dissipate (9V^2)/100r or 0.8 Watts in the first instant. Time-constant is 100r*1000uFd or 0.1 seconds; at that point the cap will be charged-up to 63% and the resistor voltage will be down to 37%. Then (3.33V^2)/100 is 0.11 Watts, falling again in the next 0.1 Seconds. A half watt part will take double-watts for far more than a tenth-second. The monster cap is not yet large enough to endanger things.

> the signal itself will be boosted by 20-30 dB

With that op-amp plan? Or some other?

Invertings amps can have high gain, or fairly high input impedance, but not both (not without absurd values and high hiss). 30dB gain suggests non-inverting connection.

[PBE6]

Thanks PRR! All good info, as usual.

[PBE6]

Finally got around to boxing things up, but I'm having him problems.

My pedal consists of a stock Orange Squeezer feeding into a parametric EQ. I used a 100 ohm series resistor with a 1000 uF cap to ground to filter 9V DC coming from an AC/DC adapter, but there is still a noticeable background hum at 120 Hz. Changing the Orange Squeezer volume doesn't have any effect on the hum level.

The cutoff frequency for my filter is 1.6 Hz. For 120 Hz hum, I should be getting almost 38 dB reduction, or roughly a 13-fold reduction in volume. How am I still hearing noticeable hum? Did I do something wrong, or is this adapter just a piece of junk? (Unfortunately I don't have another one to test it against.)

Halp!!

[PBE6]

I think I halppped myself, adding another 1000 uF cap from Vref on the EQ to ground fixed them hum. Yay!

As a side note, I also forgot to add a pull down resistor to the output, which was making the pedal pop. Adding a 1.5M resistor from output to ground has helped immensely.

Couldn't have done it without the advice in this thread..thanks all!