1

I have a regular dynamic microphone (e.g. Shure SM57) and a regular power mixer with a PA system. When I connect my microphone to the mixer with a regular XLR-XLR cable I get a satisfactory volume at 25% value of that channel volume (assuming I have some master volume on).

Now, I want to use the jack input instead of the XLR input on the mixer. The reason is that at some later stage I would like to use an effect pedal (which only takes jack input). I am using Audix t50k to convert signal from XLR to the jack input. Thus I have microphone -> XLR-XLR cable -> Audix t50k -> jack input on the mixer. And in this case I have to put the volume to 75% value of that channel volume to get the satisfactory volume as before.

As far as I understand, this is an expected behavior. What do I have to do to still use 25% of that channel volume while using impedance matching transformer?

5

The problem you are looking at is that the 1/4" jack input is set for Line Level, and the Mic is producing the lower Mic level. The Impedance transformer is merely adjusting the impedance levels to match, not boosting the mic signal.

To have the same volume control when putting the mic through the line level input, you will have to pre-amp the mic signal up to line level. A mic pre-amp that outputs line level will also handle the impedance matching for the line level input, so you wouldn't use an impedance transformer with it also.

Single channel mic pre-amps can be purchased starting around $25 USD, and range up to high end in the thousands of dollars. Rolls, ART, and Behringer are some manufacturers that I remember make some inexpensive pre-amps.

1

You don't say what mixer you are using, but the mic inputs usually have a trim/gain adjustment up near the connector that is in addition to the channel volume. Often the 1/4 in on the mic channel has a different adjustment range for the same channel when compared to the XLR input. There may be a switch or it may be auto-sensing.

For an example, I looked at the Mackie mix8 manual online (chosen randomly) and when connected via the XLR, the gain/trim is 0dB when fully turned down, but when using the 1/4 on that channel, the same gain knob fully turned down attenuates (lowers) 20dB so -20dB. For XLR, the knob range is 0 to +50dB, but the same knob is -20 to +30dB when using 1/4 cable. For 0dB using the 1/4 inch input, you'd need to be at the "12 O'Clock" position

So check your manual or look carefully at the ranges listed on the trim knob.

EDIT

Looking at your mixer, I do not see trim or gain control separate from the channel level. The manual has a technical specifications section that says that the mic inputs (assuming they mean XLR) are 50dB gain, and the line inputs (presumably the 1/4 inch) are 20dB. If there is no control, it is probably a fixed setting, so your line-in is -30db as compared to the mic. An effects pedal with output level controls may be able to add back enough gain to compensate for the drop. Alternatively, a mic pre-amp box etc might work.

  • Thanks for the answer. I have added link to the mixer. I think your explanation makes sense. However, I am looking for a way to still have 25% of the volume on the input channel while using the impedance transformer. Is that possible? How would I boost the signal? – Maksim Sorokin Oct 6 '17 at 16:57
  • see edits above – Yorik Oct 6 '17 at 17:32
1

The Audix t50k is a specialist item designed to feed a mic into a high impedence guitar input. Is the effects pedal you intend to use designed for guitars? (We're having to drag full information out of you rather painfully, aren't we!). If so, the Audix may well output a signal it can cope with. You then have the problem of getting a guitar-type signal into a mixer-amp that only offers Mic inputs (on XLR) or Line inputs (on 1/4" jack).

A guitar-type signal is quite unsuited to a Line input. But modern equipment is very forgiving. It seems that you can get a result by simply turning the knob up to 75%. Before we suggest even more equipment, can you explain why you're unwilling to do this?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.