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poly_poly-man
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 Posted: Thu Feb 21, 2008 7:56 pm Post subject: Virtual ground circuit? |
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Well, I'm building a cmoy (headphone amp), and I've already posted a bit to the headwize forums about it, but I figure, I'm too impatient to wait for a slow forum like them, and I know a bunch of you guys are engineers and such.
So I'm currently trying to decide how to build my virtual ground circuit (get 0, 9, 18v out of 18 input, or 0, 4.5, 9v out of 9v input, etc.). It needs to be able to source my Grado headphones (32ohms) and a few components with a little headroom.
Basically, I'm looking to modify this design. I have some extra op amps lying around. The only issue is, according to the datasheets, the op-amps I have can only source 35-40mA (the headphones, I've calculated, take about that). So I thought that maybe I could use the op-amps I have, being dual parts, in parallel. I drew this circuit up, pretty badly (I admit), but will it give me (almost) double the current?
Any other ideas for virtual grounds?
Thanks for your input!
poly-p man
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syscrash
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| Posted: Thu Feb 21, 2008 8:43 pm Post subject: |
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I'm not a pro EE, but where are the inputs and outputs of those circuits?
Perhaps V+ and V- are meant to be your inputs.
What about the outputs? Maybe the op-amp outputs are the ones you want,
but which of the two outputs will you use?
Confusing... and maybe not what you want.
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poly_poly-man
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| Posted: Thu Feb 21, 2008 8:48 pm Post subject: |
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| syscrash wrote: | I'm not a pro EE, but where are the inputs and outputs of those circuits?
Perhaps V+ and V- are meant to be your inputs.
What about the outputs? Maybe the op-amp outputs are the ones you want,
but which of the two outputs will you use?
Confusing... and maybe not what you want. |
The battery is the input - either 9 or 18v (two 9v batteries in series).
The V+, V- and VGND go off to the rest of the cirduit, which does the actual amplification.
See http://tangentsoft.net/audio/cmoy-tutorial/ for more information on what I'm building.
The reason I need + and - (relative to VGND) is that the op-amps later down the line need to pull from both rails (has to do with the fact that audio is AC, methinks).
poly-p man
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syscrash
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| Posted: Thu Feb 21, 2008 9:01 pm Post subject: |
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The schematic at http://tangentsoft.net/audio/cmoy-tutorial/misc/cmoy-tangent-sch.pdf is much more clear.
I think the top part only provides power to the op-amp. Your diagram really seems to confuse this.
From what I can tell, you take the +9V and -9V outputs of the power section and connect them to the
appropriate pins of your op-amp. The actual audio input and output is shown in the "amplifier section".
The virtual ground circuit only powers the op-amp (since it expects one +V and one -V).
I have a new question =D
| Quote: | | get 0, 9, 18v out of 18 input, or 0, 4.5, 9v out of 9v input, etc. |
What does that mean exactly?
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poly_poly-man
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| Posted: Thu Feb 21, 2008 9:29 pm Post subject: |
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That schematic that you link to shows the power section with a VERY SIMPLIFIED virtual ground. See http://tangentsoft.net/elec/vgrounds.html for more than you ever wanted to know.
My circuit is basically a replacement for JUST THE POWER SECTION of that scematic. The reason I'm using op-amps is that they serve the function of a buffer very well. The resistors R1 and R2 in my design are voltage splitters, which seems like a dumb idea (any variation in the batteries unbalances the supply), but the buffer (the op-amp) keeps it constant.
What I mean by the "get ..." is that, assuming the battery in my circuit (or the power section of the schematic) is 10v (okay, so it's a wall-wart  ), V+ will be 10v, V- will be 0v, and Vgnd will be 5v. This, of course, calling VGND 0v, leaves me with V+ being 5v, and V- being -5v. The op-amps in the amplifier section need this split power-supply because they're weird like that.
(Gawd, explaining to you is making me understnd this circuit more and more ). Basically, my question is, if http://tangentsoft.net/elec/bitmaps/vgrounds/vfb-opa.png can source 30mA, can my circuit source about 60mA?
poly-p man
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syscrash
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| Posted: Thu Feb 21, 2008 9:32 pm Post subject: |
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Ah now I see, you want to use two op-amps to make a virtual ground
so you can... power another other op-amp?
What are you going to use to power the two op-amps in your virtual ground circuit?
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Naib
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| Posted: Fri Feb 22, 2008 2:44 am Post subject: |
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the usual method is to use an OMAMP across the batteryy with a precesion votage divider to make a virtual-earth to then feed the amplification stages
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padoor
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| Posted: Fri Feb 22, 2008 3:16 am Post subject: |
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the negetive feedback R3 is not required. as it provides some gain for any error in the voltage splitter the voltage unbalance can become more. if you work within 20 mA current levels of the other load ckt it may work satisfacory.
if you dont need dc coupling 2 equal capacitoors should provide you ac ground.
a regular push-pull pnp-npn transistor output stage with one driver negetive feed back should define the center voltage more accurately and power full.[we call it a box car unit]
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poly_poly-man
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| Posted: Fri Feb 22, 2008 5:56 am Post subject: |
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| syscrash wrote: | Ah now I see, you want to use two op-amps to make a virtual ground
so you can... power another other op-amp? |
See, now you're getting it 
| Quote: |
What are you going to use to power the two op-amps in your virtual ground circuit? |
the resistors (R1 and R2)...
It's quite alright, I got a reply back from the other forums, and he says that my design is good but keep R3 (the resistor right after the two op-amps) at a low value, like 100ohms or so. I still wouldn't mind more feedback here...
@padoor:
Are you thinking of something more like this?
That's basically a redesign of the TLE2426 circuit, the differences being that that can source more power, but is less accurate.
poly-p man
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padoor
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| Posted: Fri Feb 22, 2008 7:25 am Post subject: |
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in 'this' ckt shown instead of R2 a driver transistor is used base connected to out put center point through suitable resistor to get 1/2 vcc at the junction of output transistors.the other thigs can remain same.the negetive feedback to the driver transistor will keep 1/2 vcc constant indepedent of current through the Q1 and Q2., and they being emitter followers the out imperance will be low and can manage larger current requirements.it is like a typical audio output stage.
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zatalian
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poly_poly-man
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| Posted: Fri Feb 22, 2008 1:10 pm Post subject: |
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...and cost more money. That does essentially what mine does, except with an extra IC, some passives and some transistors.
Thanks for the circuit design, tho. I will consider it in future designs.
poly-p man
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