DI/Reamp Theory Questions

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MrG said:
If I wanted to use this di/reamp box in either direction (200>50k or 50k>200) as a line level insertable tone box in any random AD/DA, presuming DAC is 100 ohms and ADC is 10k, how would you all recommend I create acceptable impedances on either side without losing gain, while keeping it passive?
Unfortunately you can't.
In the 50k:200 mode, you have a loss of 24 dB.
In the 200:50k mode, A) you cannot drive the nominal +4dBu into 200 ohms, so you have either to turn down the level upfront or add an attenuator B) the 50k winding loaded by the 10k imput impedance will result in a loss of about 10-15 dB.
 
abbey road d enfer said:
Unfortunately you can't.
In the 50k:200 mode, you have a loss of 24 dB.
In the 200:50k mode, A) you cannot drive the nominal +4dBu into 200 ohms, so you have either to turn down the level upfront or add an attenuator B) the 50k winding loaded by the 10k imput impedance will result in a loss of about 10-15 dB.

Thank you once again, abbey road d enfer. Thanks to your and other’s earlier explanations, I had figured this was the case, but wanted to ask you experts.

Another question, since it’s fitting for the title of my thread and I can’t find it at the moment in any threads here:

-What is the ideal input/instrument-side impedance of a passive DI? I’ve been using 50-60k and enjoying the results, but am toying with an 85k:250 option, and not sure if there are issues I’m not considering. I believe I read somewhere where Doug Williams mentioned 80k being good, if not better for these purposes.

The theory seems no problem to me, as I believe you all said an EGT output is closer to 1M, but I don’t know if I’ll get a lower output voltage or reflected impedance than is ideal.

Thanks much,
MG

 
MrG said:
What is the ideal input/instrument-side impedance of a passive DI?
There is no "ideal" value; it depends on the instrument and your expectations.
A typical bass would be pretty happy with only 50k interms of sound, however the taper of the volume pot will be altered in a way that may or may not be an issue.
A single-coil equipped guitar will want to see at least 250k for best results, when a humbucker-equipped will want to see about 500k.
An electroacoustic guitar with a piezo pickup and no on-board preamp will want at least one megohm (some say up to 10Meg).
OTOH, most instruments with an on-board preamp are content with only 10-20k.

I’ve been using 50-60k and enjoying the results, but am toying with an 85k:250 option, and not sure if there are issues I’m not considering. I believe I read somewhere where Doug Williams mentioned 80k being good, if not better for these purposes.
There is no significant difference between 50 and 80k. A 50k on a Fender guitar will work equally bad than an 80k with a Gibson.

The theory seems no problem to me, as I believe you all said an EGT output is closer to 1M, but I don’t know if I’ll get a lower output voltage or reflected impedance than is ideal.
There's no way you can get a xfmr with a 1Meg winding. You need an active stage.
 
Hey all,

I’ve just had another idea tonight and wanted to see what you thought.

In my idea of using the reamp concept as a line level insert - just the box - would it be possible to put resistor(s) in paralell to the hi-z (~50k) side to bring the load down to a healthy bridging value for the ADC input? If so, how do you think it would be best to do this?

Thanks much in advance.

Mark
 
MrG said:
In my idea of using the reamp concept as a line level insert - just the box - would it be possible to put resistor(s) in paralell to the hi-z (~50k) side to bring the load down to a healthy bridging value for the ADC input? If so, how do you think it would be best to do this?
There's nothing wrong with loading a modern line output with 50kohms, or even more; you don't need to do anything.
 
abbey road d enfer said:
There's nothing wrong with loading a modern line output with 50kohms, or even more; you don't need to do anything.

Thanks, abbey road. I may have misspoken - I’m talking about the hi-z 50k side being the output, the ADC 10k being the input.

Mark
 
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