Author Topic: Q for RG ... hfe/gain in buffers  (Read 2830 times)

RedHouse

Q for RG ... hfe/gain in buffers
« on: April 26, 2004, 11:53:47 AM »
RG,

Reading recently one of your posts you mention the MPS-A18 as being the lowest noise transistor (US made) and so looking at the transistor data sheet it says it hfe is like 500->1500, that's some range.

Anyway my question is regarding buffers and the hfe of the transistor used, does the hfe set the gain of the bufer?

For instance take JD's circuit here:


if I used the 220K/220k/3.3k resistors and an MPS-A18 for Q1 would it be a buffer or would I be getting the huge 500->1500 gains the hfe spec indicates from it?

My goal is finding a minimal gain buffer to use as a building block when isolating two circuits, say a output buffer for a classic Wah or FF or something, hence the need for low noise.
(I'm not understanding the hfe/gain relationship I guess)

puretube

Q for RG ... hfe/gain in buffers
« Reply #1 on: April 26, 2004, 01:03:04 PM »
that emitter-follower circuit always has a gain of ~1,
regardless of trans.-type.
(someone else may explain more into detail...)

petemoore

trying different Q's
« Reply #2 on: April 26, 2004, 01:08:15 PM »
I don't even socket buffers transistors anymore. One can stick just about anything in there that's NPN and it will work, some have lower noise though.
  I usually just grab a 2N5089 for buffers, and solder it right in./
  Could be I'm missing something, but they drive just fine.
Convention creates following, following creates convention.

R.G.

Q for RG ... hfe/gain in buffers
« Reply #3 on: April 26, 2004, 01:49:52 PM »
Follower circuits by their nature try to but never achieve a gain of 1.

Emitter-, source-, and cathode-followers all employ 100% feedback to do their work. They do this by means of their transconductance - the amount of current that flows in the output as as result of a change in the input voltage. The output current is converted to the output voltage by flowing through a resistor.

The transconductance is literally the change in output current with a change in input voltage. For FETs and tubes, the transconductance is a specified quantity on the data sheet, for bipolars, it is usually not.

Tubes have the lowest transconductance of the bunch; a tube cathode follower will have a "gain" of about 0.90 to 0.95. This is because the output current changes little (on an absolute basis) for each volt of input voltage change.

JFETs are better, and will get into the higher 0.9's. MOSFETs have a big transconductance, usually about 1A/V, and so they do a much better job.

Bipolars are usually the winner in this contest, because they have a transconductance typically 10x that of MOSFETs. Only a tiny change in base voltage produces major changes in output current. In fact, the change needed is so small that to a first approximation we think of bipolar base-emitter voltage as constant at a fraction of a volt. The gain of a bipolar emitter follower is typically 0.99 for low gain devices, and it can get much better. It's best thought of as a gain of one minus an eyelash.

Even within bipolars, the higher the current gain, the closer the output gain is to one.

But your question is about noise.

Noise comes from a couple of places in a follower. It comes from thermal noise in the follower resistor, it comes from the thermal and excess noise in the transistor, and it comes from the thermal noise in the biasing arrangement. The good news is that since the gain is so low, you usually can ignore these sources. The only time that is not true is when the transistor is noisy. Most modern transistors are very good, and the 2N5088, 2N5089, and MPSA18 are exceptionally good. Just don't ever reverse-break the base-emitter junction and they'll stay that way.

So use any one of those with impunity as your buffer. They will not be where the noise comes from.
R.G.

Quick IQ Test: If anyone in a governmental position suspected that YOU had top-secret information on YOUR computer, how many minutes would you remain outside a jail cell?

RedHouse

Q for RG ... hfe/gain in buffers
« Reply #4 on: April 27, 2004, 11:11:01 AM »
Thanks RG, always the gentleman!

Also thank you puretube and petemoore for posting on this.

RedHouse

Would Darlington get a gain of 1?
« Reply #5 on: April 27, 2004, 11:17:27 AM »
Would the use of a darlington in this circuit (MPS-A13) get the gain up to 1?

Dunlop in the newer Crybaby circuits, use an MPSA13 as an input buffer, could this possibly be why.

puretube

Q for RG ... hfe/gain in buffers
« Reply #6 on: April 27, 2004, 11:30:35 AM »
the MPSA is probably rather there for the high input-impedance (?)

Joe Davisson

  • Guest
Q for RG ... hfe/gain in buffers
« Reply #7 on: April 27, 2004, 08:17:05 PM »
For non-inverting voltage gain you can use a common-base amp, but the input sensitivity is kinda low. Otherwise it will work for audio, as long as the desired gain is pretty low (like 1x-2x).

Example (2x gain):

Base: 10k/10k divider.
Collector: 20k to 9v.
Emitter: 20k to ground.

Input into emitter, output from collector. You can get some (unpleasent?) sounds by using high resistances. Driving it directly with an emitter-follower corrects the impedance, giving you a non-inverting buffer with voltage gain =) Great until it clips =(

-Joe

R.G.

Q for RG ... hfe/gain in buffers
« Reply #8 on: April 27, 2004, 08:20:28 PM »
Quote
Would the use of a darlington in this circuit (MPS-A13) get the gain up to 1?

No. The math works out that the gain of a simple emitter follower approaches 1 asymptotically - that is, it gets infinitely close to 1 as current gain increases, but never actually gets there. A darlington gets insignificantly nearer 1 than a for instance 500hfe ordinary BJT.

Quote
the MPSA is probably rather there for the high input-impedance (?)

Yes. Using a very high gain transistor (MPSA18 at 500 - 1200) or darlington (1000 to 50,000) lets you get follower action with a vanishingly small input base current, so you can make the biasing resistors much larger. A darlington lets you get into the 1M region with biasing resistors, so the input of the buffer is much higher impedance, and sucks tone less there.

However, you can easily add noise with darlingtons if you're not careful. Darlingtons invite the casual designer to bias them with a simple resistor string from + to ground connected to the base. This happens to *maximize* the resistors' noise into the base, especially with high value resistors. Using two lower value resistors, say about 10K, to set the voltage and a third high value resistor to the base puts a very small voltage across the impedance setting resistor at the base and lets you use a capacitor from the junction of the three resistors to ground to make the bias voltage quiet. This is known in some circles as "noiseless" biasing. It's worth remembering any time you use high gain bipolars or darlingtons, or any time you feel the urge to reach for a 1M.
R.G.

Quick IQ Test: If anyone in a governmental position suspected that YOU had top-secret information on YOUR computer, how many minutes would you remain outside a jail cell?

RedHouse

Q for RG ... hfe/gain in buffers
« Reply #9 on: April 28, 2004, 10:19:11 PM »
Thanks RG, and thank you too Joe.

This is great stuff, you (I) just don't get this depth of info from books.