Capacitor question

[random007]

Re: LM38xx typical application circuit and the Ci feedback to ground capacitor.

As I understand it, Ci is used with Ri as a highpass LC filter to remove any possible DC in the feedback loop.

Usually specified as a non-polar 22uF, there are various designs where Ci is polarised or only 10uF. Ri is nearly always 1K.

Also, the datasheet application notes say:

At this point, it is a good idea to ensure that the Gain-
Bandwidth Product for the part will provide the designed gain
out to the upper 3 dB point of 100 kHz. This is why the
minimum GBWP of the LM3886 is important.
GBWP ≥ AV x f3 dB = 13 x 100 kHz = 1.3 MHz
GBWP = 2.0 MHz (min) for the LM3886
Solving for the low frequency roll-off capacitor, Ci, we have:
Ci ≥ 1/(2π Ri fL) = 4.85 μF; use 4.7 μF.

...which is a bit beyond my level of understanding *ook ook - wanders off to look for a banana* : I'm assuming giving Ci a value between 10 and 20-or-so uF will suffice for a general purpose audio amp into an 8ohm nominal load.

So... I've the following options for Ci using components I have or can get easily:

1) Mitsumi 10uF MPT - good quality metallized polypropylene but bigger than the rest of the gainclone circuit put together (~ 5cm x 5xm x 3cm). Will long leads lead to undesirable effects here or doesn't it matter? An increasingly strange prospect is stacking two of these monsters in parallel for 20uF.

2) Vishay Sprague 22uF tantalum.

3) Electrolytic 10 or 22uF. Low ESR or low leakage (similar to tantalum) kinds available.

Comments as to which option sounds good (as it were) are very welcome.

[ampimp]

Hi
Electrolytics sound terrible as input cap. im not sure about the tant either.
10uf film cap on the input is plenty.long leads to the caps are not good.
Ian
ps it is mentiond in the data sheet about the type of cap to use.

[AndrewT]

Hi,
the Negative FeedBack capacitor acts as a DC block in the feedback path and this results in a DC gain for the amplifier of 1times (+0db).
The capacitor in conjunction with the resistor acts a filter to AC signals. It filters the feedback so reducing the LF AC gain gradually down to near 1times at extremely low frequencies.

If the feedback path develops a voltage across the NFB cap then that cap with AC voltage across it increases the LF distortion of the amplifier.
It should be sized so that it is not allowed to develop significant AC voltage across it.
This condition is brought about by ensuring the the input filter determines the bandwidth of the amplifier, not the NFB C+R.

For an AC coupled amplifier with DC blocking caps in both the input and in the NFB loop the NFB capacitor must be greater than sqrt(2) * Rin * Cin / RL
where RL is the lower leg feedback resistor.

If you want good low frequency performance from your amplifier then expect the NFB cap to be between 100uF and 1mF.

[random007]

Originally Posted by AndrewT

Hi,
the Negative FeedBack capacitor acts as a DC block in the feedback path and this results in a DC gain for the amplifier of 1times (+0db).
The capacitor in conjunction with the resistor acts a filter to AC signals. It filters the feedback so reducing the LF AC gain gradually down to near 1times at extremely low frequencies.

If the feedback path develops a voltage across the NFB cap then that cap with AC voltage across it increases the LF distortion of the amplifier.
It should be sized so that it is not allowed to develop significant AC voltage across it.
This condition is brought about by ensuring the the input filter determines the bandwidth of the amplifier, not the NFB C+R.

For an AC coupled amplifier with DC blocking caps in both the input and in the NFB loop the NFB capacitor must be greater than sqrt(2) * Rin * Cin / RL
where RL is the lower leg feedback resistor.

If you want good low frequency performance from your amplifier then expect the NFB cap to be between 100uF and 1mF.

Thanks very much. I understand now. Mostly.

Confused as to why many designs have an RC high pass filter on the input but omit Ci and why many don't filter input (presumably trusting input equipment not to generate DC or very low frequencies???) but do have Ci (eg lm3886 datasheet typical application design). I take it the latter design can cause LF distortion if very low frequencies are presented?

What about filtering input but not NFB - if low frequencies are removed at input, why also filter the NFB loop?

[random007]

(oops, in 2nd line of 1st post, I meant RC filter, not LC)

[macboy]

Originally Posted by random007

What about filtering input but not NFB - if low frequencies are removed at input, why also filter the NFB loop?

C_I is used so that you do not amplify the chip's input offset voltage. Input offset voltage varies chip-to-chip. In the LM3886 datasheet, the specified "typical" input offset voltage is 1 mV, with a maximum of 10 mV. If you set up the amp for a gain of 31.5 (30 dB, a little on the high side but this is for illustration) then without C_I, the amp will show typcially around 31 mV offset on the output, but up to a maximum of 315 mV. That is very high as you know. Adding C_I reduces the DC gain to 1 so that output offset voltage will be around ~ 1 to 10 mV.

The input capacitor ("C_IN") prevents DC in the input signal from being amplified and appearing across Ci. It cannot block the amp chip's own "input offset voltage". If you skip C_I, then any DC offset at the input will appear at the output, amplified by the gain of the amp (31.5 in my example), unless it is first blocked by C_IN. If you skip C_IN but use C_I, then any DC offset at the input will appear equally at the output (since C_I sets DC gain to 1). If you use both C_I and C_IN then the DC offset of your amp should be very low, in the single-digit mV range, maybe less.

[random007]

Originally Posted by macboy

C_I is used so that you do not amplify the chip's input offset voltage. Input offset voltage varies chip-to-chip. In the LM3886 datasheet, the specified "typical" input offset voltage is 1 mV, with a maximum of 10 mV. If you set up the amp for a gain of 31.5 (30 dB, a little on the high side but this is for illustration) then without C_I, the amp will show typcially around 31 mV offset on the output, but up to a maximum of 315 mV. That is very high as you know. Adding C_I reduces the DC gain to 1 so that output offset voltage will be around ~ 1 to 10 mV.

The input capacitor ("C_IN") prevents DC in the input signal from being amplified and appearing across Ci. It cannot block the amp chip's own "input offset voltage". If you skip C_I, then any DC offset at the input will appear at the output, amplified by the gain of the amp (31.5 in my example), unless it is first blocked by C_IN. If you skip C_IN but use C_I, then any DC offset at the input will appear equally at the output (since C_I sets DC gain to 1). If you use both C_I and C_IN then the DC offset of your amp should be very low, in the single-digit mV range, maybe less.

Thanks for the very clear explanation. Appreciated.

[random007]

Given my volume control will be 100K, and (if I'm reading the datasheet right, that determines input impedance), to get a filter happening at around 10Hz or less (which seems comfortably safe for audio: 6db per octave, starting -3db freq 10Hz, double frequency for every octave) any input capacitor larger than about 150nF is OK. Just using the basic 1/(2PiZR) calculation there but the result seems a bit optimistic considering most circuits use between 2.2uF and 4.7uF.

In any case, I've some 2.2uF 250V metalized poly caps lying around which should do the job.

[AndrewT]

Hi,
your 100k pot appears as a variable source impedance to the next stage.
It can be used as part of a filter but results will vary a lot.

The DC blocking cap is usually put in series with the line signal after the pot.
This isolates the variable resistance of the pot from the amp input.

If the maximum output impedance of the 100k pot is 25k (it varies all the way down to ~1r0) then the input impedance of the amplifier should be 10 to 20times that source impedance, i.e 250k to 500k.

Use a 10k or 20k pot in front of your power amplifier.

[random007]

Originally Posted by AndrewT

Hi,
your 100k pot appears as a variable source impedance to the next stage.
It can be used as part of a filter but results will vary a lot.

The DC blocking cap is usually put in series with the line signal after the pot.
This isolates the variable resistance of the pot from the amp input.

If the maximum output impedance of the 100k pot is 25k (it varies all the way down to ~1r0) then the input impedance of the amplifier should be 10 to 20times that source impedance, i.e 250k to 500k.

Use a 10k or 20k pot in front of your power amplifier.

Thanks so very much for the reply!

I've a lot yet to learn about amplifiers and low frequency AC circuit design.

Pity, I just bought this really gorgeous (and expensive) dual 23 pole stepped 100k log pot (not a pot, a switch, I should say) *sigh* I wasn't paying attention at the time of purchase - always a mistake!

Hey, anyone want to buy a beautifully built gold contact, make before break, 23 pole 100K logarithmic stepped rotary switch?

(((Thank goodness there's been a huge delay to construction after I damaged *scratch* a photo-sensitive PCB before etching and have had to wait for another one to be sent interstate. Design improved a lot since then. Just ordered a 10K log rotary switch. I'll post a construction history with photos, PCB plans (for what they're worth) and explanation once I finish and test it all works (not a lot of point if it doesn't!) - will have to say positive photosensitive PCBs are cheap, easy to use... and surprisingly hard to find... but THE bee's knees (or aardvark's armpits) for one-off PCB construction. An ink-jet printer and some transparencies and a UV bulb have worked well for me for a long time now. Currently using a 30 day trial edition of Altium for the PCB design - the PCB editor is driving me insane (it always thinks it knows best) though I appreciate the other conveniences and facilities of this multi-thousand dollar software package (aimed squarely at commercial digital circuit designers). )))