That schematic will need not a 1.0µH inductor, but a 0.7µH inductor, and it must be air core! Either that or a 0.68µH air core is okay, but 0.7µH is better...either that or no inductor, which works okay.
AndrewT said:...The statement at the top of the schematic seems suspicious:- setting R6=R2//R3 But R3 is DC blocked by C1. I think R6=R2+-variation to trim output offset.
The feedback tapping AFTER the half Thiel network seems odd.
Should it be BEFORE the Thiel? i.e. at node 3?
How far away are C5 & C6 from the supply pins1&4?
Thanks, AndrewT. I'm planning on skipping some of these caps (C2 and prolly C7) as well as the aircore/resistor output setup since my driver is like 10 inches away from the amp board so capacitive loading should not be an issue.
I'm also skipping the trim pot (R1) and going for fixed values here.
I believe that C2 is there to keep out some kinda HF oscillation but overall this amp design seems almost "overprotected."
I'm thinking about bringing the resistor values closer to the original "high impedance" design for this amp since I'm not still 100% sure of the interaction between the left/right summing resistors that I'm using on the overall gain formula. It's a long story that I know you're familiar with 😀
AndrewT said:...Is the F-3db frequency=1/[2Pi *4.7uF*1k0]=34Hz?
and phase error upto 340Hz?...
I'm actually going with a 1uF cap here at C1.
Hi Carlost,
changing C1 on it's own to 1uF will convert the input filter into a single pole bass cut filter starting at 159Hz (-3db) and having a significant effect at 300Hz (-1db) and lessening effect all the way up to about 1500Hz.
The higher impedance figures from the original diagram will restore some of the lost bass and mid. But 1uF is NOT a good substitute.
changing C1 on it's own to 1uF will convert the input filter into a single pole bass cut filter starting at 159Hz (-3db) and having a significant effect at 300Hz (-1db) and lessening effect all the way up to about 1500Hz.
The higher impedance figures from the original diagram will restore some of the lost bass and mid. But 1uF is NOT a good substitute.
For some reason, in the inverting amp design, a cap on the input is important to reject DC coming in. You know from previous posts that I will be hooking up an iPod and combining L-R channels through summing resistors. The LM3875 will then be driving a single little 4 ohm Audax coaxial in a BB4 "boombox" alignment.
What would you suggest is a good value for this cap? Is this cap even optional in your mind for my application?
On the R values issue, my thought was that not knowing the exact interaction of the input summing resistors with R3, that I would keep R2 high enough so that errors in predicting the interaction would not be terribly magnified in the gain formula (i.e., the littler the numbers, the bigger the OOPS).
My thoughts last night were:
Summing resistors L-R - 2K each
R3 - 2K
R2 - 100K
R6 - 4K
This...I think...will yield:
Av = 100K / (2K + 2K) = 25
My thought is that the interaction between the summing resistors will be like two series resistors (2K for the summing resistor + 2K for R3) giving a combined effective value for R3 of 4K.
Then the value for R6 would be equal to the parallel values of R3 and R2 (4K x 100K / 4K + 100K) = 3.85K rounded to 4K.
Am I on the right track? Xmas is coming 😀
What would you suggest is a good value for this cap? Is this cap even optional in your mind for my application?
On the R values issue, my thought was that not knowing the exact interaction of the input summing resistors with R3, that I would keep R2 high enough so that errors in predicting the interaction would not be terribly magnified in the gain formula (i.e., the littler the numbers, the bigger the OOPS).
My thoughts last night were:
Summing resistors L-R - 2K each
R3 - 2K
R2 - 100K
R6 - 4K
This...I think...will yield:
Av = 100K / (2K + 2K) = 25
My thought is that the interaction between the summing resistors will be like two series resistors (2K for the summing resistor + 2K for R3) giving a combined effective value for R3 of 4K.
Then the value for R6 would be equal to the parallel values of R3 and R2 (4K x 100K / 4K + 100K) = 3.85K rounded to 4K.
Am I on the right track? Xmas is coming 😀
On the C1 question, witness the inverting GainClown design from Platenspeler.com. This also has an input cap...but the value is 2.2uF.
BTW this guy's website was about as straightforward and written "for dummies" like me...I hope it's right 😀
http://www.platenspeler.com/background/opamps/uk_opampgain_1.html
An externally hosted image should be here but it was not working when we last tested it.
BTW this guy's website was about as straightforward and written "for dummies" like me...I hope it's right 😀
http://www.platenspeler.com/background/opamps/uk_opampgain_1.html
Hi,
a chip amp is basically an opamp with a high current output stage and components suited to the higher rail voltages found in typical chipamps.
All opamps amplify as best they can all frequencies from DC to RF.
if you want to amplify the DC then omit both DC blocking capacitors.
If you want to amplify the AC only then put in the DC blocking capacitors.
This applies equally to inverting and non-inverting topologies.
The inverting combines the input DC blocking cap and the NFB capacitor into one and save one capacitor.
But in return it has a lowish input impedance.
For a summing amplifier (inverting with two input resistors) the gain from EACH input is Rfb /Rin, ALWAYS.
The sum from the output is VoutA + VoutB. These voltages can be different.
You have not shown the summing resistors! Where do you plan to put them with the attenuator?
Do you plan to add an RF filter? Careful it does not work as easily on an inverting opamp.
a chip amp is basically an opamp with a high current output stage and components suited to the higher rail voltages found in typical chipamps.
All opamps amplify as best they can all frequencies from DC to RF.
if you want to amplify the DC then omit both DC blocking capacitors.
If you want to amplify the AC only then put in the DC blocking capacitors.
This applies equally to inverting and non-inverting topologies.
The inverting combines the input DC blocking cap and the NFB capacitor into one and save one capacitor.
But in return it has a lowish input impedance.
For a summing amplifier (inverting with two input resistors) the gain from EACH input is Rfb /Rin, ALWAYS.
The sum from the output is VoutA + VoutB. These voltages can be different.
You have not shown the summing resistors! Where do you plan to put them with the attenuator?
Do you plan to add an RF filter? Careful it does not work as easily on an inverting opamp.
Thanks, Nordic but that was too cryptic for my little head...what am I to plug in to that spreadsheet? What value of Hz am I striving for?
AndrewT said:...For a summing amplifier (inverting with two input resistors) the gain from EACH input is Rfb /Rin, ALWAYS...
Thanks and in my case Rin is 4K (2K from the summing resistor + 2K from R3)?
Hi,
for effective summing the L&R 2ks should meet at the virtual earth (ground).
Each could then have it's own DC blocking cap (IF the source has NONE).
Maybe it's not important but a 2k input impedance and 4u7F removes quite a bit of bass.
If there is ANY DC blocking either at source OR at the amp input then the DC current flow from the chip amp input is BLOCKED and R6 then equals R2, not R2//R3.
To minimise output offset due to input offset current you should make R6 adjustable or at least easily changed. But this method of offset adjustment only works well if the chipamp does not have a compensated input bias current topology. I have asked and to date no-one has come back to confirm which way a typical (or any specific type of) chipamp is assembled.
for effective summing the L&R 2ks should meet at the virtual earth (ground).
Each could then have it's own DC blocking cap (IF the source has NONE).
Maybe it's not important but a 2k input impedance and 4u7F removes quite a bit of bass.
If there is ANY DC blocking either at source OR at the amp input then the DC current flow from the chip amp input is BLOCKED and R6 then equals R2, not R2//R3.
To minimise output offset due to input offset current you should make R6 adjustable or at least easily changed. But this method of offset adjustment only works well if the chipamp does not have a compensated input bias current topology. I have asked and to date no-one has come back to confirm which way a typical (or any specific type of) chipamp is assembled.
AndrewT said:Hi,
for effective summing the L&R 2ks should meet at the virtual earth (ground)...
AndrewT:
Isn't this in effect happening? The summing resistors are tied to the inverting input (virtual ground).
Or does the cap in the way create an issue?
How would I have any clue about the amount of DC coming from an iPod?
Thanks again.
AndrewT said:...If there is ANY DC blocking either at source OR at the amp input then the DC current flow from the chip amp input is BLOCKED and R6 then equals R2, not R2//R3...
This flies in the face of everything I have read about bias resistor values. I thought this was a really simple concept. Right from Maarten's webpage:
"Input Bias
According to the literature, the resistor R3 in the inverted design should be chosen such that the Input Bias Current I_b is optimal. It is advised to choose R3 according to the following formula:
R_3 = (R_1*R_2)/(R_1+R_2), which is the value of R1 and R2 in parallel."
http://www.platenspeler.com/background/opamps/uk_opampgain_1.html
AndrewT said:...The feedback tapping AFTER the half Thiel network seems odd.
Should it be BEFORE the Thiel? i.e. at node 3?...
Looking at the traces on the actual PCB, it seems that the feedback loop is before the output filter.
Hey carlos, no it just tells you the frequency of the highpass filter... created by the resistor and capacitor on the input you plug in the capacitor value in nf. 1uf = 1000nf and the resistor in kohms 1kohm= 1000ohm
Nordic:
Thanks...am I looking to target a high value above the range I care about like 20,000 Hz or the lower cutoff value below the frequency I care about like 20 Hz? In speaker building, we call F3 the low frequency cutoff. I'm not sure how it's used here.
Playing around with the spreadsheet you sent me, a 2K resistor in line with a 4uF cap shows an "fu" of 20 Hz which would be fine. Jack's use of 4.7uF is not entirely bad then...is it?
Which resistor do I care about...the summing input resistor or R3? In my case both are 2K but which one matters?
Thanks...am I looking to target a high value above the range I care about like 20,000 Hz or the lower cutoff value below the frequency I care about like 20 Hz? In speaker building, we call F3 the low frequency cutoff. I'm not sure how it's used here.
Playing around with the spreadsheet you sent me, a 2K resistor in line with a 4uF cap shows an "fu" of 20 Hz which would be fine. Jack's use of 4.7uF is not entirely bad then...is it?
Which resistor do I care about...the summing input resistor or R3? In my case both are 2K but which one matters?
Hi,
pin 7 is the virtual earth (ground).
F(-3db) is a useful way to assess the filter effect. Whether it is the low pass for the treble end or the high pass for the bass end and irrespective of speaker or amplifier or any source unit.
For a normal amplifier and speaker combination, setting the F(-3db) at 20Hz will sound quite bass light. Some will go one octave lower, many go to two to three octaves lower and I recommend a full decade lower.
For an ipod into 4inch driver, 20Hz(-3db) could well be very acceptable. Only you or the user will know after listening and experimenting.
The resistor value AFTER the final DC blocking cap affect the high pass filter. But if you have two series connected caps (one in the source and another in the amplifier) the combined effect of the staggered turn-over frequencies and the effective total capacitance into the final resistor gives a 2pole roll-off with a more complex phase and response roll off that I cannot calculate accurately. A simulator with an experienced user would get a good estimate for the combined effect.
Measure the DC output from the source. Maybe even open it and see if a cap is the last component before the output connector.
A magnifier will probably be necessary.
pin 7 is the virtual earth (ground).
F(-3db) is a useful way to assess the filter effect. Whether it is the low pass for the treble end or the high pass for the bass end and irrespective of speaker or amplifier or any source unit.
For a normal amplifier and speaker combination, setting the F(-3db) at 20Hz will sound quite bass light. Some will go one octave lower, many go to two to three octaves lower and I recommend a full decade lower.
For an ipod into 4inch driver, 20Hz(-3db) could well be very acceptable. Only you or the user will know after listening and experimenting.
The resistor value AFTER the final DC blocking cap affect the high pass filter. But if you have two series connected caps (one in the source and another in the amplifier) the combined effect of the staggered turn-over frequencies and the effective total capacitance into the final resistor gives a 2pole roll-off with a more complex phase and response roll off that I cannot calculate accurately. A simulator with an experienced user would get a good estimate for the combined effect.
Measure the DC output from the source. Maybe even open it and see if a cap is the last component before the output connector.
A magnifier will probably be necessary.
Thanks, AndrewT...we're getting there 😀
I'm sorry but Pin 8 (the negative or inverting pin) is the virtual ground right? Pin 7 goes RIGHT TO ground...nothing virtual about it 😀 Pin 8 is always driven to 0 reference (the "virtual" ground) because in a differential amp, Pin 8 will always strive to match Pin 7 (the "real" ground).
So how does having resistors on both sides of the input cap affect the gain formula?
Thanks for all the great help, Nordic and AndrewT.
I'm sorry but Pin 8 (the negative or inverting pin) is the virtual ground right? Pin 7 goes RIGHT TO ground...nothing virtual about it 😀 Pin 8 is always driven to 0 reference (the "virtual" ground) because in a differential amp, Pin 8 will always strive to match Pin 7 (the "real" ground).
So how does having resistors on both sides of the input cap affect the gain formula?
Thanks for all the great help, Nordic and AndrewT.
- Status
- Not open for further replies.
- Home
- Amplifiers
- Chip Amps
- Capacitor Choices...What's Important?