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Old 26th April 2010, 09:51 PM   #41
gfiandy is offline gfiandy  United Kingdom
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I recomend simulating this circuit. Then you should be able to look at the trade offs at least in the opamp. As you reduce the input impedance you are increasing the amplifer gain and the input noise of the opamp will at some point become the dominant factor rather than johnson thermal from the input resistor. Also as the gain goes up problems with instability may become an issue due to parasitic capacitance. At the limit (i.e if you continued to reduce the impedance) the output would be driving into the virtual earth as a current amp. I am not sure what the effect of this would be, but clearly you will start to approach this as you reduce the impedance.

I am unclear as to the benifit of this balanced input. It provides an asymetrical load to the cartridge as one end is refered to a virtual earth, approx impedance 0 Ohm and the other is refered to ground through the combination of 51R and 6.19K.

I would think a true balanced approach would have a symetrical load L*2 ( whatever your chosen load is) on both sides and use two virtual earths then combine the signals in another stage. However this is just conjecture for your interest, I haven't really thought it through to a conclusion.

Regards,
Andrew
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Old 27th April 2010, 12:10 AM   #42
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This is my balanced in- balanced out solution.
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Old 27th April 2010, 08:41 PM   #43
FdW is offline FdW  Netherlands
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Quote:
Originally Posted by gfiandy View Post
I recomend simulating this circuit. Then you should be able to look at the trade offs at least in the opamp. As you reduce the input impedance you are increasing the amplifer gain and the input noise of the opamp will at some point become the dominant factor rather than johnson thermal from the input resistor. Also as the gain goes up problems with instability may become an issue due to parasitic capacitance. At the limit (i.e if you continued to reduce the impedance) the output would be driving into the virtual earth as a current amp. I am not sure what the effect of this would be, but clearly you will start to approach this as you reduce the impedance.
That is true, and it is exactly what I am doing, but as the exact data for the cartridge is not know (other than that the supplier notes an internal resistance of 5ohm and a input impediance for the preamp of 100ohms) I have no way of getting more figures on that. That was why I posed the ‘general’ question about the issue.

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Originally Posted by gfiandy View Post
It provides an asymetrical load to the cartridge as one end is refered to a virtual earth, approx impedance 0 Ohm and the other is refered to ground through the combination of 51R and 6.19K.
The configuration, as given, gives a symmetrical (what ever that means for a cartridge with 2 wires per channel) load of 102.2 ohms (2 times 51.1). Look up the wiki reference a few entries back.

Regards,

Frans de Wit.

Last edited by FdW; 27th April 2010 at 08:43 PM.
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Old 28th April 2010, 01:34 AM   #44
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I agree with FdW and Gerhard regarding load.

Unfortunately the opamp input sees the cartridge signal in series with 102.2 ohms. The signal should be in parallel with 102.2 ohms for lowest input noise. To realize this with DC-coupled balanced input requires more active devices than one op amp, for example:

http://upload.wikimedia.org/wikipedi..._Amplifier.svg
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Old 28th April 2010, 07:29 AM   #45
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I use exactly that at the moment and it performs great.
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Old 28th April 2010, 07:51 AM   #46
FdW is offline FdW  Netherlands
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Quote:
Originally Posted by nuvistor View Post
I agree with FdW and Gerhard regarding load.

Unfortunately the opamp input sees the cartridge signal in series with 102.2 ohms. The signal should be in parallel with 102.2 ohms for lowest input noise. To realize this with DC-coupled balanced input requires more active devices than one op amp, for example:

http://upload.wikimedia.org/wikipedi..._Amplifier.svg
The wiki gives no formulas for Zin (not for differential and not for common mode). So I looked it up, the Zin you are referring to is called 'Common mode input impedance' and this one is not important for the cartridge (it is for the common mode signal). It is the differential impedance that is important for the cartridge and it is (in the given case) 102.2ohm.

The information and the formula's can be found on:
Differential Amplifier Common Differential Modes

Regards,
Frans.
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Old 28th April 2010, 08:05 AM   #47
FdW is offline FdW  Netherlands
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Quote:
Originally Posted by Joachim Gerhard View Post
I use exactly that at the moment and it performs great.
In this case you did build a high Zin differential amp, and then lowered the deferential input impedance of the circuit by adding resistors parallel to the input of the amp. This works fine, but was not selected by me, due to the fact that I like a as low as possible components count (and this configuration adds at the least two opamps, and one in the signal path).

The configuration that I selected has a low differential input resistant and thus does not need the two ‘extra’ opamps. This is not the best solution in regard to common mode rejection, but it is more of a compromise. I used the following reasoning; the ‘standard’ circuit has no common mode rejection at all, the selected one has (using the .1% resistors) around 80db (or more), and that is a good thing, especially as no extra components are needed (compared to the ‘standard’ solution).

Regards,
Frans.

BTW; There is no such thing as a free lunch.

Last edited by FdW; 28th April 2010 at 08:13 AM.
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Old 28th April 2010, 06:12 PM   #48
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Hi FdW -

In the example I posted, and using the definitions in your link, Z1,Z2 and Z3 are infinite with ideal op amps, in reality they will be limited by the op amp input characteristics but will be much higher than 100 ohms to well above 20kHz.

In the example I posted, the cartridge and its 100 ohm load would be connected between pins V1 and V2, which places the cartridge in parallel with the 100 ohms. Gerhard's design works like this.

Have you consider transformer step-up for MC?
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Old 29th April 2010, 02:38 PM   #49
FdW is offline FdW  Netherlands
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Quote:
Originally Posted by nuvistor View Post
Hi FdW -

In the example I posted, and using the definitions in your link, Z1,Z2 and Z3 are infinite with ideal op amps, in reality they will be limited by the op amp input characteristics but will be much higher than 100 ohms to well above 20kHz.

In the example I posted, the cartridge and its 100 ohm load would be connected between pins V1 and V2, which places the cartridge in parallel with the 100 ohms. Gerhard's design works like this.

Have you consider transformer step-up for MC?
Simulation results for the LME49710 show, CMR@10Khz 70db and CMR@30Khz 60db(better than). These figures are good enough for me (given the simple circuit).

A sample audio file can be found at:
Diy ADC's

Yes I did consider adding a step-up and rejected it, I know there are many pro's (and even con's ) but these counted the most; transformer CMR (not the best there is [or you need a very, very, very… spectacular expensive transformer]) and transformer noise. Plus it adds an extra component to the equation, and as before, for me, simpler is better!

Regards,
Frans de Wit.
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Old 1st May 2010, 12:50 AM   #50
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Trust me, most manufactures have no idea how to best load their cartridges. I build everything from a virtual 0 Ohm input to 47kOhm loading, balanced and un-balanced and all circuits worked just fine when optimised.. Over the weekend i build a low noise version of your input stage and i will report how it performs against my INA.
I am using the OPA1611 that waited to be put into good use.
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