I don't need a multimeter to tell me how volume controls work. I have a suggestion: if you wish to demonstrate how well the channels are balanced why not measure channel balance? I'm sure you could give some detailed instructions, for the benefit of those who don't know Ohm's law.
Gee Whiz Chris,
This is a farce.
http://en.wikipedia.org/wiki/Farce
Either someone else designed this circuit, or you did it by randomly trying things until it worked. But it's clear you don't have the any idea what's going one with it, beyond poking in multimeter leads and taking readings.
You have already given us voltage readings across the pot in post 116.
You can't have a voltage difference across a resistor without current. This is basic stuff.
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Re Post 136, Thats fine, you do not have to help/ contribute, I understand.
Cheers / Chris
Answer: Virtual Ground inverting op amp circuits are explained as since the positive input of the op amp is grounded, the op-amp will do everything it can to keep the negative input at ground as well. and the op amp inputs draw no current.
The negative inputs of the op amps are placed at the upper and lower end of travel of the pot. Measuring actual circuit, its schematic described and provided at post 108 Resistance probe terminal Adj to Op amp ground above 10 meg ohms ( my hand held DMM ranges out ) Resistance probe terminal Adj to gnd also above 10 meg ohms,
It performs extremely well as an attenuator with NSL32SR2.
I have AD5228 and DS 1869 digital pot ic, which have very real restrictions, on current through upper lower and wiper of 1ma . I have some appointments today, but happy to prove the point of their use to apply in this circuit.🙂
Cheers / Chris
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You're placing words in posts that have no meaning in the context of your circuit.
Your op-amp outputs do not have a direct feedback path through a resistor to the minus inputs, i.e. the classic I/V converter.
In fact, all your op-amp outputs go to the input on the first LM317, U1, in this case wired as a voltage regulator.
Do you understand that a voltage regulator gives a fixed output voltage, regardless of changes to the input, as long as the input voltage stays above the dropout voltage value? So the op-amps outputs are basically doing nothing, and have no connection to the minus inputs. Which means all your op-amps do nothing. Go ahead - connect +12V or whatever your supply voltage is to the U1 LM317 input instead. Then remove both op-amps from the sockets. It should work the same.
You may also want to wonder, where is the current that goes to the LEDs coming from, if not from the volume pot? There is no other possible source.
...suggest a better way...
You have already been given the equations to use, and examples in this very thread, here
http://www.diyaudio.com/forums/anal...t-resistor-current-control-5.html#post2955763
The total resistance of series + shunt tells you nothing about channel balance. What it does tell you is the load is on the upstream component.
It is the RATIO of shunt over series plus shunt for each channel that tells you about the channel balance. Roughly, it is
Vout = Vin x R1 / (R1 + R2)
Vin = voltage input
Vout = voltage output
R1 = shunt resistor value
R2 = series resistor value
This does not include the output impedance of the source, or input impedance of the following load, which will alter the actual numerical results, but not that much in terms of channel balance.
more here
http://en.wikipedia.org/wiki/Voltage_divider
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Nice explanations, BFNY. I doubt Chris will understand them though, which is why I can't be bothered anymore. As DF96 put it:
One minor detail, though - apparently he's actually managing to draw some current from the opamp's inverting inputs. Yes I know it's a JFET gate, but if you pull it's voltage far enough below the opamp's V-.......
One minor detail, though - apparently he's actually managing to draw some current from the opamp's inverting inputs. Yes I know it's a JFET gate, but if you pull it's voltage far enough below the opamp's V-.......
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I suspect you do not understand. However, were you looking for help in describing how to measure channel balance? If so, BFNY has kindly given you the answer. All you have to do is turn it into detailed instructions for those who are Ohmically challenged, as you did for measuring the total resistance.Chris Daly said:
I was waiting to see if he knew that, now you have told him!BFNY said:
I too suspected that his circuit contains no virtual grounds, but the feedback path (if it exists) is so convoluted that I couldn't decide without spending more time than this nonsense deserves. By the way, I can't remember whether his circuit uses a dual-polarity supply. Just checked: I can't see a negative supply. Could Chris tell us how he achieves a DC virtual ground at the inverting input of an opamp with a single polarity supply?
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hi BFNY
Yes there is another voltage source, that you are not considering. It is the inherent voltage given up by a single rail connected op amp. .860v at inverting inputs in 2 and 6 in the case of TL072 when a voltage is connected to pin 8 with 3,4,5 grounded, and .5 less than rail at op amp outputs.
In a simple demonstration,to prove an attenuator is possible, construction of a breadboard circuit will take about 30 mins.
You will need: a power supply of 12 volts
a LM317
2x 120 ohm resistors
a TL072
2x BC547
2x Red LED
A 50k pot or trimmer ( a screwdriver needed to adjust )
some test leads.
2x 1n4148 diodes( we just use the metal legs )
cutters
the attached schematic LED.demo
With a DC power supply of 12v+ . We connect Vin+ to a LM317, next to limit current to 20ma we solder a 2x 120 ohm in parallel between V out and Adj . ... a simple PSU capable of 20ma is made, suitable for driving LED's and to show how the circuit would work if it had NSL32SR2 in place instead. No caps are shown for PSU as this is a simple demonstration, not a circuit to connect to other equipment. if you want to connect caps then do so with 100uf from vin to Gnd and 10 uf from Adj to Gnd.
We need to solder Pin 3,4 and 5 of our op amp which are all ground, and connect negative ground of PSU. and we then solder a connection for Pin 8 suitable for connecting a jumper lead from Adj of the LM317 to connect power.
Construction needed first:
If we solder the outputs pins 1 and 7 together by using the legs of our 1n4148, and a connect a potentiometer wiper to that point, the other two potentiometer leads one goes to pin 2 and the other to Pin 6 which are the negative inputs. Now connect voltage and measure voltage from Pin 2 and Pin 6 which is either end of the potentiometer upper and lower, we get equal voltages. of 11.50v and no measurable current. A current of 2ma - which is quiescent current of a TL072. however can be measured between ground and voltage supply ground.
Next we assess what happens when loads occur on the negative input.
so connect two LED's with cathode to ground. We need to make the load current occur downstream rather than through the high impedance section - which can occur if forced ie with a component directly connected to ground, and where some posts have attributed load through the pot.
see the schematic:
We need to connect a transistor to each op amp negative input. Here we use a current steer diode, using a BC547 with CB joined, ( as is also shown on the schematic at post 108 with 3 such transistors ). connect the anodes of each LED to each Emitter, a measurement of current made directly across the transistor CB to E, now shows it now contains most of the current . so virtual earth gnd is on CB and current is on the emitter, we have turned the tables so to speak.
so the circuit provides us with an explanation that current in actual use occurs at the load not at or through the high impedance op amp., which performs as a virtual earth.
Next with the 2x LED in place, and transistors that steer current in place, if we move the potentiometer we see one decreasing and the other increasing.
So we have achieved a circuit that enables high impedance isolation yet control of the load, and we have achieved up and down control with a single pot. Neat !! 🙂
the attached schematic, shows LED as a thyristor oregano library sadly has no LED, but part is labeled as RED LED.
with this schematic and assembled breadboard circuit you can simply measure current and voltage and resistance to confirm the larger schematics validity. ... what fun.
Cheers / Chris
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By connecting the V- supply pin to the non-inverting input instead of ground. How else were you going to do it?😉
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Thanks, Chris, you have done us all a favour. That circuit, and your 'explanation' of it, confirms that as we suspected you do not have one tiniest clue what you are doing.
Please go away and read Horowitz and Hill, or some other good book on electronics. Stop selling junk on ebay to gullible fools. Start designing simple circuits which actually work. We are willing to help, but first you have to realise just how profound is your ignorance of electronics. Sorry to be blunt, but there is no kinder way to say it.
Please go away and read Horowitz and Hill, or some other good book on electronics. Stop selling junk on ebay to gullible fools. Start designing simple circuits which actually work. We are willing to help, but first you have to realise just how profound is your ignorance of electronics. Sorry to be blunt, but there is no kinder way to say it.
thanks for the detailed explanation.
Do you need the diode connected transistors in the LED circuits?
Where does the current for the LEDs come from, i.e. which components pass the LED currents?
Do you need the diode connected transistors in the LED circuits?
Where does the current for the LEDs come from, i.e. which components pass the LED currents?
Yes diode connected transistors are needed. If a connection to ground with say an LED, is made without the transistor, current then flows through summing point 🙁. The virtual earth only performs its function when diode connected transistor is in place.
If we measure current across CB, E we get ( depending on pot position ) lets say 16ma
then voltage across the same point is .22v, which is the diode junction voltage drop.
source of current for the LED's is a deep question, can you help me on that one🙂, my thoughts are its the inverting properties of the op amp, with the voltage component, provided by the single rail voltage connection.
Cheers / Chris
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Ah, maybe that explains it. I was puzzled by the inappropriate use of technical terms, like early on referring to a diode-connected BJT as a "band gap".
Yes, the inability to comprehend Ohm's Law was the tipoff that this was software rather than a human. It's sophisticated, though; it can draw schematics with randomly-connected parts. Props to the developer, it's quite good.
@Chris
That explanation doesn't seem to make any sense. Perhaps there's a language barrier. It might help if you write in your home language, then use Google translate. It does a surprisingly good job sometimes.
That explanation doesn't seem to make any sense. Perhaps there's a language barrier. It might help if you write in your home language, then use Google translate. It does a surprisingly good job sometimes.
unsure how to answer, the what is not specific.
The voltage source is the single rail connection, and nothing else.
Cheers / Chris
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