transistor recommendation for input buffer

[fuzzy645]

Hi -

Having an issue with my easy drive build that input impedance is pretty darn low.  Therefore, I want to add a buffer on the input.

I was looking at either of these 2 from Jack Orman (see images below)

1.  They show a 10 uf on the output.  Would that be INSTEAD of or IN ADDITION TO the .1 uf DC blocking/coupling cap on the input of the easy drive. I assume I would not want to caps directly in series with one another so I assume this 10 uf would replace the .1 uf on the Easy Drive input, or do I treat the buffer circuit as a 100% independent object from the Easy Drive, and just string them together in series??

2. Which transistors would you recommend for either of these?

3. Any recommendations on which one sounds better (if any)?

3. Any other thoughts?

BTW, I was using this method I found on the net to approximately measure input impedance and I was shocked to discover the easy drive only measured a frightfully low 7K input impedance.  Here is the link: http://www.zen22142.zen.co.uk/Theory/inzoz.htm

Perhaps I should leave the easy drive alone.  After all, if I never measured the impedance (as per the link above), I wouldn't be the wiser :-).   The thing sounds great as is.  

Thank you.

Orman buffer using bipolar transistor:


Orman buffer using JFET:

[Earthscum]

If you insert this before your booster, then you can get rid of the 10u cap. It is there, and large, so you don't get bass roll-off at heavier loads if you are simply using it to drive longer lengths of cable.

Just about any generic transistor should work equally as well in this circuit. I prefer darlington (MPSA13) buffers, or Jfet (MPF102) buffers.

If you use an MPF102, you can copy this circuit, essentially. Only thing you would want to change is taking out the 220k from V+ to base (which will now be gate). You may increase the other 220k (to ground) to something around 470k. The 3.3k resistor should be a good value (I use 2.2k, but that's me). (follow the schem, lol... I don't do the half-supply thing with Jfet buffers, unless I fear the signal has a chance of hitting ground or V+, doubtful with straight guitar signal. I wasn't paying attention that the bottom schem was a jfet).

One thing, though, is you may consider putting in a "high cut" tone control between the buffer and the Easy Drive. Something like Mark Hammer's SWTC. I think that would make for a pretty awesome little circuit. You can then cut some of the highs you DON'T want, rather than relying on impedance to not cut tones you do want. It's basically taking control of the pre-shaping, if you will, that way you can tune in that "mojo" sound.

[fuzzy645]

Thanks!

I just wired the first circuit with an extra 2N5089 I had laying around and a pair of 220K resistors + 3.3K and measured the input impedance and got about 25K....still not very high.  I suppose the choice of transistor will make a big difference.  I will try the MPF102 that you recommended.

[blueduck577]

measured the input impedance and got about 25K....still not very high.

Almost so low that something doesn't seem right.  Though the input impedance of this circuit is dependent on hfe, the biasing resistors should dominate with even a moderate hfe.  You can't expect an input impedance of more than 110k with the circuit as it stands. 25k is very suspect, especially with the minimum hfe of the 2N5089 spec'd at 400. Check your method and/or wiring.

[azrael]

Search for Gus' bootstripped EF circuits on here. EF means emitter follower.

[John Lyons]

Search for Gus' bootstripped EF circuits on here. EF means emitter follower.

+1 

[R.G.]

Don't do it that way.

Change the 220Ks to 47Ks. Put a resistor between the 47K/47K divider network and the base. Make this about 470K to 1M. Signal capacitor still goes to base. More on that later. Bypass the 47K/47K to ground with a 10uF cap. This (1) sets a bias voltage (2) makes the bias voltage very quiet (3) isolates the bias for DC (4) ensures a high input impedance.

If you want to go to the trouble of using a bootstrap, make the resistor between divider and base about 100K and insert another 100K in series with a 1uF cap to the emitter.

[fuzzy645]

Don't do it that way.

Change the 220Ks to 47Ks. Put a resistor between the 47K/47K divider network and the base. Make this about 470K to 1M. Signal capacitor still goes to base. More on that later. Bypass the 47K/47K to ground with a 10uF cap. This (1) sets a bias voltage (2) makes the bias voltage very quiet (3) isolates the bias for DC (4) ensures a high input impedance.

If you want to go to the trouble of using a bootstrap, make the resistor between divider and base about 100K and insert another 100K in series with a 1uF cap to the emitter.

Thanks all for your assistance with this.

RG - thanks again for all your help with this, and my other thread too.  I have drawn out what you just recommended (below).

2 questions:

1.  Did I interpret your recommendations correctly in the drawing?
2. The easy drive also has an input/coupling cap (.1 uf).  Is it OK that I have the .1 uf output cap on this buffer that feeds directly into the .1 uf input cap of the easy drive, or is that redundant?  I guess what I'm asking is can I leave off the last cap C3 since the easy drive has an input coupling cap.  Is there any harm in leaving it in?
3. Lastly, is any particular bipolar transistor suited for this or can I use the 2N5089 since I have em' laying around

Thanks!

[R.G.]

1.  Did I interpret your recommendations correctly in the drawing?

Mostly. There is one thing which should be different. The input capacitor C1 should go to the transistor base, not the bias divider. It should connect to the junction of R3 and the base.

2. The easy drive also has an input/coupling cap (.1 uf).  Is it OK that I have the .1 uf output cap on this buffer that feeds directly into the .1 uf input cap of the easy drive, or is that redundant?  I guess what I'm asking is can I leave off the last cap C3 since the easy drive has an input coupling cap.  Is there any harm in leaving it in?

If this is going to be a permanent addition to the easy drive, leave off one of the two capacitors. The only harm in leaving it in is that the output cap on this buffer and the input cap on the easy drive causes a reduction in bass response because two capacitors in series is the same as one smaller-value capacitor.

3. Lastly, is any particular bipolar transistor suited for this or can I use the 2N5089 since I have em' laying around

2N5089 is a very good transistor for this.

Thanks!

[fuzzy645]

Got it RG., thanks.  So like this. I will try this tonight after work

[lopsided]

Got it RG., thanks.  So like this. I will try this tonight after work

this way DC would leak through the input. What, I believe, R.G. meant was connecting one side of the cap directly to input and the other directly to the base (where also R3 connects).

J.

[ubersam]

its like this:



just ignore the values I used and omit C1, C4, & C5.

[Earthscum]

Here... relevant info for you to bookmark... actually, just bookmark http://www.geofex.com/

But, relevant to the biasing, R.G. explains it in full in the "Mod your Mu-amp/Minibooster" (link) article. Good read, even if you aren't building a mu amp.

ETA: Thinking about it, I believe that is why I started to bias my jfet buffers with a single resistor to ground. As long as the source resistor is close to the range that the Jfet likes, it biases up with plenty of room to swing either way, and I don't have to worry about a biasing network.

[fuzzy645]

OK, it sounds like I interpreted the diagram incorrectly. Is the version I have on the top correct (labeled V2), as compared with the version on the bottom?  I moved C1 (although not labeled) to the junction of R3 and the base.

         

[R.G.]

OK, it sounds like I interpreted the diagram incorrectly. Is the version I have on the top correct (labeled V2), as compared with the version on the bottom?  I moved C1 (although not labeled) to the junction of R3 and the base.

V2 is the correct way to get input into it.

I noticed one more thing I should have seen earlier. The bottom side of R2 goes to ground. The bottom side of C2 also goes to ground, and the top side of C2 goes to the junction of R1, R2, and   R3.

[fuzzy645]

I noticed one more thing I should have seen earlier. The bottom side of R2 goes to ground. The bottom side of C2 also goes to ground, and the top side of C2 goes to the junction of R1, R2, and   R3.

Ahhh, got it.   Thanks!!   Like this, correct??

[PRR]

> 2N5089 ...a pair of 220K resistors + 3.3K and measured the input impedance and got about 25K....

That can't be right.

Take emitter R as 3K and your 7K number for EZdrive (which also seems low) in parallel. That's ~~2K. Multiply by hFE of 2N5089, at least 300. 600K looking into the base.

This is in parallel with the two 220K which are in parallel for AC, 110K.

600K||110K = 93K.

It "can't" be ~~25K input unless the external load is near 100 ohms.

> this method I found on the net ...Here is the link

Show your raw numbers. Maybe we can figure out why your measurement is so low.

[fuzzy645]

> 2N5089 ...a pair of 220K resistors + 3.3K and measured the input impedance and got about 25K....

That can't be right.

Take emitter R as 3K and your 7K number for EZdrive (which also seems low) in parallel. That's ~~2K. Multiply by hFE of 2N5089, at least 300. 600K looking into the base.

This is in parallel with the two 220K which are in parallel for AC, 110K.

600K||110K = 93K.

It "can't" be ~~25K input unless the external load is near 100 ohms.

> this method I found on the net ...Here is the link

Show your raw numbers. Maybe we can figure out why your measurement is so low.

I used methods described here: http://www.zen22142.zen.co.uk/Theory/inzoz.htm



I used a 10K load resistor for the test with a function generator set to a fixed frequency of 1kHz.  As per the sites recommendation, I measured the AC voltage at points V1 and V2.  I'm sure the method isn't perfect, but perhaps this approximation can be useful.

FYI - for comparison I tried measuring the input impedance of my Fulltone Fulldrive with this exact same method, and the test demonstrated an input impedance in the 500K range.

Anyway, here is the  measurement I took of a slightly modified version of the EZ Drive, but with no buffers of any kind.  

RL (the fixed load resistor) = 10K
Voltage measured (V2) was 16.6 mv (therefore 0.0166 volts)
Voltage with 10K load resistor (V1) was 6.6mV, (therefore 0.0066 volts)

Input_Current = (V2 - V1) / RL= (.0166 - .0066) / 10000 = 0.000001
Z (input Impedance) = V1 / Input_Current = .0066 / 0.000001 = 6.6K

This is super low, and one might assume I made a wiring mistake.  That is certainly possible, but I must add the pedal sounds fantastic as-is, although I have not tried it with high output humbuckers as of yet.

Anyway, when I added the ORMAN buffer out front, here is what I got as far as impedance measurements (using the same parameters as above). FYI - I have NOT yet tried RG's buffer (ran out of time last night), so there numbers below are from the ORMAN buffer I hooked up yesterday):

Voltage measured (V2) was 16.4 mv (therefore 0.0164 volts)
Voltage with 10K load resistor (V1) was 11.7mV, (therefore 0.0117 volts)

Input_Current = (V2 - V1) / RL= (.0164 - .0117) / 10000 = 0.00000047
Z (input Impedance) = V1 / Input_Current = .0117 / 0.00000047 = 24.89K

As a form of "control group" (as they call it in statistics) I then measured my Fulltone Fulldrive, using this same method and came up with V2 = 16.1 mV and V1 = 15.8 mV, and running through the same math as above will give you an Z input impedance of = 526.67K.  I probably would have expected closer to 1M for the Fulltone, but at least we are in the right ballpark there.

In general I have noticed that if V1 and V2 are really tightly close together, the impedance # is very high. If there is a "gap" between them the impedance turns out low.  

[PRR]

> a slightly modified version of the EZ Drive

Uh,,,,, not like this?



Yes, this plan is wonky.

The input may already be over-loaded at 16mV. Try changing your input level. If it never gets over some tens of mV, then you can't use the linear assumptions of simple input impedance. The input impedance may be somewhat higher at 1mV and drop severely at higher levels.

The Orman BJT buffer should be quite linear at 16mV even 1V. Be sure power is on, and roughly 3VDC at the emitter. Be sure the output is not shorted. Re-check for wiring errors.

> I have noticed that if V1 and V2 are really tightly close together, the impedance # is very high

Well yeah; specifically "high compared to 10K".

And when not-high and not needing precision, you can eye-ball:

V2 16.4 mv
V1 11.7mV

17 to 12 is almost like 3 to 2, or we are dropping 1/3rd in 10K and 2/3rd in unknown load, so the unknown must be about twice of 10K. That's heavily rounded, but for most purposes "20K" is as good as "24.89K".

Or if your signal source is stuck on 16.5mV, you can apply dummy-loads of 1K, 3K3, 10K, 33K, 100K, 330K, note the resulting mV, then compare your unknown for a quick no-math approximation.

[fuzzy645]

Any chance I have something wrong in this buffer circuit as depicted?  I tried wiring it up independently of the easy drive initially and something is wrong. Here are my voltages:

9.4 V at collector
1.7 V at base
1.1 V at emitter
4.6 V at the junction of R1, R2, R3 and C2.

FYI - for purposes of testing I did add an output .1 output cap although not shown in the diagram.  I even plugged in a guitar and it is sounding bad so something is wrong. Let me know if it seems like I have a major mistake in the diagram below as I believe I have hooked it up as drawn.   If not, I'll keep checking my soldering work but it really seems like its hooked up as drawn.

Thoughts???

Thanks in advance