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Hi Mr. DEB,
1) In the Tilt Filter, U1 and U2 pins 8, 9 and 10 can be replaced by a piece of wire.
2) The power supply has transistors, diodes and resistors that do nothing. The power supply can be just two 220uF capacitors, one for +9V and the other for -9V.
3) The pots on the LM3915 ICs have a value that is too high (then the LED current and brightness are very low). Added limiting resistors are also needed.
Replace the pots with a fixed 1k resistor for a full-scale sensitivity of 1.25V and 12mA through each LED.
4) The outputs cannot drive many models of headphones because they have low current opamps, not high current driver amps. Maybe U1 and U2 pins 12, 13 and 14 must be headphone driver ICs, not opamps.Try it and see (hear?) if it is loud enough with the headphones you will use.
The TL074 and many other good opamps oscillate when they directly drive the capacitance of a cable. Adding a 100 ohm resistor in series with each output decouples the cable's effect but then the output to low impedance headphones will be very low.
will make changes. perhaps add an amp?
the power supply is a rail splitter or virtual ground.
here is a link to design that someone else built **broken link removed**
thanks for your help.
will make changes etc.
the board is getting crowded at 3.75" x 3.25"
includes the two 10 led displays.
the pots on the lm3916 are from the data sheet?
the op amp that contain pins 8,9,10 are acting as a buffer so the displays won't interfer with the output sound quality. the Lsig and R sig go to pins 5 of the lm3916's(inputs)
You don't need a "rail-splitter". You have two 9V batteries. The rail-splitter is used when there is only a single battery.
The buffer opamp was needed to isolate the switching noise before when the display used multiplexing. This simple circuit does not use multiplexing so the buffer opamps are not needed.
took out the buffers, added an amp stage using op amps.
gained 20db.
with the buffer circuit I only had a 10db gain.
Debating on the power supply mods.
would clear some pcb real estate (getting crowded)
the resistor for the led displays? not going to have bar mode.
the pot circuit is directly from data sheet?
Hi Mr. DEB,
Is the low gain of the input high enough to peak the display?
Won't the gain of 11 at the output cause clipping distortion?
Why do you have a pot at pin 8 of the LM3916?
A line-level signal averages about 150mV, with peaks to 0.85V.
Your input has a gain of up to 2.1, the filters have an average gain of nothing and the LM3916 needs an input of 1.25V to light its highest LED.
With the input volume controls at max the displays will have the highest LEDs light on the music peaks.
But with a gain of 11 at the outputs their peaks will try to be 19.6V which results in severe clipping distortion since with new 9V batteries the unloaded output has a max peak of only 7.8V.
Add some gain at the input and reduce the gain at the output.
On the LM391x, pin 7 is the output of the adjustable voltage reference. Its voltage is connected to pin 6 and sets the input voltage level that will light the highest LED.
You need pin 8 to be grounded so that the sensistivity is as high as possible.
The resistance from pin 7 to ground sets the current in the LEDs. You have a 15k pot (which is in parallel with the 12k divider inside the LM3916) so the current in the LEDs will be very low (about 1.9mA) when the pot is turned down. Turning the pot up causes the sensitivity to drop until the voltage reference saturates. Then the sensitivity is reduced about 15dB and the LED current is increased to about 9mA.
just changed some resistor values--you were dead on --LOTS OF CLIPPING with the 10K resistor on output.
the top circuit looks better but only 6db.Bottom circuit shows clipping.
still trying to figure out what you mentioned about the lm3916 / led / 1.25v??
listed on the pic are the voltages at input op amp and output
I need to add the peak detector to the simulator circuit. Connect to output or just after the filter?
The top circuit has a gain of 2 for the input opamp. Then the 150mV input signal has a peak of 212mV and the opamp doubles it to 424mV.
When the filters are set flat without gain then their peak output is also 424mV.
The output opamp has a gain of 4 then a loss of 0.75 so it would amplify the peak to 1.27V. The max sensitivity of the LM3916 is 1.25V so they match well.
But is a peak voltage of 1.27V too loud or too faint in your headphones?.
Maybe the input opamp should have a gain of 6 and then the volume control can be set so that when the peak detectors are fed from the filters the LEDs peak. Then the output opamp can have the gain of 4 and another volume control can be used for the sound.
added peak detectors, the input has 50k/100k resistors. tried several different combos but get clipping
Now I assume the peak detector is putting out 4.2E-23V??what is E
Volts?
To much for the lm3916?
I forget lots of times that the meg is capital M not a small m.
The K ohm is a small k . if input a capital I get an error message about the decimal point or ??
Will change input to peak detector.I think after several trials it goes before the capacitor?
how does everything else look?
Still deciphering the lm3916 input==the sim says the peak detector is outputing 4.2E-23v??
The input of the peak detector circuit should be fed directly from the output of an opamp without a coupling capacitor.
Now your simulation shows the capacitor in the peak detector normally charging quickly and discharging slowly but it looks like the voltage reading is wrong because at 1kHz the capacitor should reach the peak voltage of the signal in about 10ms. Your trace has a duration of 5ms.
I find that changing the diode to the right one has some effect
Changed from a 1n4148 to 1n914
the multi meter readings are 22.89fv on the dc scale
48.41uv on the ac scale
?????
LOST in Idaho
The circuit will work exactly the same with a 1N4148 diode or a 1N914 diode.
Fairchild has them together on the same datasheet.
The transistor and diode in the peak detector will allow the 10k resistor to charge the capacitor to the positive peak of the signal voltage. then the capacitor is discharged by the 1M resistor slow enough for you to see its peak.
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