guitar DI circuit

[circuithead]

Hi, I'm trying to create a guitar input circuit which will feed a mic pre of around 1000ohms input impedance..

A couple of things I'm not 100% sure of...   the output impedance of the emitter follower..  Am I right that it'll be around the same as the emitter resistor..  in which case it's a little high..  yet I don't want to draw too much current..   Most guitar DI circuits online have a resistor around this value or higher in some cases.. so I'm not sure on this.  Is there a way of lowering output impedance without drawing too much current in a simple circuit like this.. 

Also, is there any glaring mistakes or bad practices that I'm not aware of that I have done?  I've not breadboarded yet.. 

Any advice greatly appreciated!

[K5HJ]

The emitter resistor is selected to set the current in the follower.  The output impedance is determined by the driving impedance divided by the beta.
In your case, the output impedance is very low since the follower is driven by another low impedance follower.

Randy

[circuithead]

Thanks.  So perhaps I could save on current a little and choose a higher value resistor.   However I'm unsure on how to calculate this...
Perhaps if we say a 100k output impedance guitar is feeding the input..  3.3M I believe input impedance?  Then the output impedance of that first stage would be 3.3M || 100k  / hfe ?  Or am I way off here?

[Richard Crowley]

Yes, your output impedance is rather high.
Your output is also unbalanced, which somewhat defeats the purpose of using a conventional (balanced) mic input.

Is the destination (the mic input) equipped with phantom power? 
There are many good circuits that use phantom power to eliminate the need for batteries.

The circuit used for most electret condenser microphones is phantom powered and provides a balanced, low-impedance output.
If I were designing a phantom-powered guitar DI box, I would probably clone the classic "Schoeps circuit".
This is the "Dorsey" version of the Schoeps circuit. 



You could simply eliminate the whole "35V" circuit components and connect directly to C6 which has a super high impedance.

[circuithead]

Thanks..  the Schoeps circuit is definitely something I'll experiment with, thanks for the suggestion.   I'm going to build a few different circuits I think as they're very simple to breadboard and experiment with.  It's more to gain knowledge rather than fix a need, so if possible I'd love to finish & improve my initial circuit first. 

I'm going to be placing it in the same box as a micpre so unbalanced output is ok I think.. 

Looking again at the schematic.. is R6 needed at all?  And for that matter.. is Q3 actually needed? 

I guess my ultimate question is how do I work out the output impedance of the emitter follower.   In spice, it's showing up at around 4 ohms at 1k..  rising to around 80ohms at 20Hz.
Without Q3, the output Z of Q2 reads 85ohms rising to 114ohms at 20Hz.   

[Richard Crowley]

The term "DI" implies that you are using it at the other end of a mic cable, etc.
But if you are just building it into the gear along with other circuits, then just use a standard high-impedance instrument input circuit.
There are probably 100s of them out there to study.  Your circuit seems quite a bit overly complex for a simple high-impedance instrument input.
And if you are in the same box with the rest of the circuits, then emitter followers and output impedance, etc. aren't really much of a concern.

R6 is a key component in both AC and DC operating points of the differential output pair.
Q3 is the stage that takes the very low-level, high impedance signal and actually "generates" the balanced/differential signal for Q1 and Q2 to buffer.
But since you clarified what you are designing, then none of the "DI" suggestions really apply as you aren't making a "DI".

[circuithead]

Should have clarified.. R6 and Q3 refer to my original schematic, which you seem to agree is far too complex..  am I on the right lines to get rid of those do you think?

Perhaps DI implies balanced out yes.  High impedance input feeding a mic pre I guess is more correct..

The reason I want to clarify impedance is so I'm feeding the micpre input with the correct output impedance from the circuit.  Ideally I'd like to switch between mic input or guitar/keyboard or whatever.. 

When you say my output impedance is high, how did you calculate that?

[Richard Crowley]

I really think you are on the wrong track completely with your original circuit. Quibbling over a component here or there doesn't seem worth the effort.

It is NOT clear why you are even "feeding a mic pre"?  That also seems like a poor choice in this case.
Since we don't see the whole picture, we don't really know what you are trying to do here.
It also is not clear why you would need to "SWITCH between mic input or guitar/keyboard"?  Why not just have both?

Upon review of the dialog, I'm not even sure of which "output impedance" we are talking about anymore.

[circuithead]

Sorry for the confusion, I'll try to be clearer on the goal here..  So eventually I will build a lovely micpre with lovely transformers and such things to use for microphones when recording into my PC audio interface..  Now, it would be nice to have a guitar input rather than having to use a DI box so I thought I'd work out a circuit which would go into the front end of the mic pre, and as such I'd be able to use the nice mic pre(tbd).     

The ultimate goal is to learn electronics but as I'm interested in recording that's the project I've decided on so as I'm not just reading books but instead reading books and applying some of what I learn.

So with this circuit, I need a high impedance input so as not to load down the guitar.  And a low impedance out so the mic pre sees it like a normal mic signal.
Am I really way off with what I'm thinking?  Any pointers to the right direction appreciated ?!   

When I say output impedance..  I mean where it says "out" in my original circuit.   Or if I was to get rid of Q3..  so the emitter of Q2.    Where were you thinking when you said it was rather high?

[Richard Crowley]

But if you have access to the INTERNALS of the gadget, you don't have to feed the guitar signal into the MIC INPUT!
Mic inputs are optimized for balanced, and low-impedance.
Guitar inputs are optimized for exactly the OPPOSITE, unbalanced and high-impedance.
A "DI" box takes the unbalanced, high-impedance signal from an instrument and creates a balanced, low-impedance output that is compatible with a mic input.

HOWEVER, there is NO good reason to run your instrument input INTO the mic input circuit!
And there are MANY reasons why you would NOT want to do that!

If you take some more time to study how commercial gear is designed, you will see that there is some intermediate signal level/impedance that is used internally, AFTER the mic preamp circuit, and AFTER the instrument input circuit.  That INTERMEDIATE point is your target for the mic preamp circuit, and also for the instrument input circuit.  Since we don't see the whole picture here, it is very difficult to explain this because we are seeing only PART of your system.

If you don't look at the WHOLE picture, you will end up making unfortunate compromises based on incomplete knowledge.

[circuithead]

It's pretty common in recording circles at least where I'm from..   A couple of companies spring immediately to mind..  Great River, Pacifica, Vintech..  all have DI inputs before the mic input transformer..  Saves using a DI, yet still you get the full flavour of the mic pre. 

Hence I thought it would be a nice simple thing to tackle first as most designs I've seen are 1 or 2 transistors..   

So it's essentially a DI box in the same case as the mic pre.. and as such I wasn't going to do it balanced. 
Yes, mic inputs are optimized for low impedance and therefore this circuit I'm trying to come up with should have a low impedance output.. 

Yes, I may hate the sound.. find that it's too noisy.. find that balanced is needed even in the same box.... or that I prefer the sound of the guitar bypassing the whole mic pre..   but hopefully I'll have learnt a lot along the way!

So design goals..
high impedance input..  low impedance output..  Is the only problem with my circuit that the output is rather high?  How high is rather high?

[Richard Crowley]

The problem is that you are operating into a rather low impedance load (the mic preamp input).
So virtually ALL the output impedance of your instrument stage will tend to attenuate the signal that gets into the mic pre.
And you need all the clean gain you can get (unless you are really going for noisy "grunge").

[john_p_wi]

I think you are over complicating this.  I have built numerous pedals and chip amps for guitars.  One of the most important features is a good buffer circuit first to keep the guitar from loading the circuit down.  I would suggest something like this:

Jfet buffer (2n3819) then to an opamp TL082 "low to mid gain" stage then buffer stage with a low value pot for volume and low impedance out.

or with discretes:

Jfet buffer (2n3819) to J201 gain to Jfet buffer (2n3819).

The buffers will be a little less than unity, the middle gain stage will help with the signal level.  Should take less than 1/2 hour to breadboard all and test.

[K5HJ]

Thanks.  So perhaps I could save on current a little and choose a higher value resistor.   However I'm unsure on how to calculate this...
Perhaps if we say a 100k output impedance guitar is feeding the input..  3.3M I believe input impedance?  Then the output impedance of that first stage would be 3.3M || 100k  / hfe ?  Or am I way off here?

I ran a Spice analysis of your schematic and the output impedance measures about 4 Ohms.  Should be able to drive a low Z microphone input.

Randy