Another Way to Measure Transistor Gain

We've talked about hfe, which represents current gain, but there is another way to measure gain in a transistor. Some devices, like vacuum tubes or FETS, don't draw any input current, so the concept of current gain makes no sense. For those devices, engineers use the term "tranconductance," which is a fancy way of saying how much the output current changes when you change the input voltage. BJTs also exhibit transconductance, which gives us another way to evaluate gain. Remember we talked about how hfe has pretty large manufacturing tolerances and varies from device to device? Transconductance is just the opposite. It's extremely predictable and is the same for every part number, silicon and germanium! That's because transconductance in BJTs is a fundamental property of solid-state physics. Transconductance is represented by the symbol gm. gm is easily calculated; it is collector current divided by the Thermal Voltage (vt), which is about 26mV at room temp: gm = ic / vt. For example, a transistor with a collector current of 520uA would have the transconductance 520uA / 26mV = 20mS (milliSiemens). Conductance is measured in units called Siemens (go ahead, get your snickering out of the way). Siemens is the reciprocal of Ohms. A 1K resistor has a conductance of 1/ 1K = 1mS. To calculate the voltage gain of a transistor stage, we multiply the transconductance by the load resistance. Suppose the transistor in the previous example has a 15K load resistor. Its voltage gain would be 20mS * 15K = 300. Ok, so it's not quite that simple because BJTs have output impedance which is effectively in parallel with the load resistance. We can get a pretty good estimate of a transistor's output impedance by dividing 50V by the collector current. It's just a rule of thumb, but it's plenty accurate for what we're doing. The transistor in the previous example has a collector current = 520uA, so the output impedance is approx. 50V / 520uA = 96K. We put that in parallel with the 15K load resistor and get 13K, so the voltage gain is actually 260. Notice that hfe plays no part in calculating transconductance or voltage gain. A germanium transistor with hfe = 50 has the same transconductance as a silicon transistor with hfe = 800, as long as their collector currents are the same.

Two more quick thoughts related to transconductance and we'll call it a day.

If we know hfe & gm, we can calculated the input impedance (Zin) of a transistor. Zin = hfe / gm. Let's say the transistor in the previous examples has hfe = 200. Then Zin = 200 / 20mS = 10KΩ.

The output impedance of an emitter follower is low, but how low? That's easy to calculate. The output impedance (Zout) of an emitter follower is Zout = 1 / gm. Using that same transistor as an emitter follower: Zout = 1 / 20mS = 50Ω.

Next time: What's All This Impedance Stuff About?
 

Chuck D. Bones

Circuit Wizard
It certainly could, depends on the surrounding circuitry and whether the designer gives a shit. The input impedance of the first stage usually makes a big different because that's what's loading the guitar or pedal that's feeding the circuit in question. Most pickups and some pedals are sensitive to loading. Some examples:
A Fuzz Face has the input jack connected to the base of the first transistor via a large capacitor. The emitter is grounded. Therefore, the guitar is looking into the emitter-base junction of a transistor. Let's say that transistor has an hfe = 75 and the collector current is 180uA. The input impedance is Zin = hfe * vt / ic = 75 * 26mV / 0.18mA = 11K. This will load the pickups a bit, reducing brightness. Not really a problem with a distortion pedal.

Fuzz Face - Arbiter PNP Germanium Re-issue.jpg

Now let's consider the Range Master.
Dallas Rangemaster 2.jpg

The emitter is grounded in an AC sense via the 25uF cap. The collector current will be around 500uA. We'll assume hfe = 60. Zin = 60 * 26mV / .5mA = 3.1K. That and the 6.8nF cap will load the pickups which reduces the height of the pickup's resonant peak and moves it down an octave or so. It must sound good because everyone loves a Range Master, right?

How transistor stages interact with each other is as much a matter of current drive capability as it is impedance. Couple that with the fact that the input and output impedance is non-linear (it changes with signal amplitude) and frequency dependent and it gets complicated.

I use LTSpice, a breadboard, a scope and my ears to see what's going on in a circuit because the simple calculations I did here only get us in the ballpark. Professional pedal designers like Brian Wampler do the same thing. Some pedal builders fly by the seat of their pants and it shows in their designs.

Kind of a wordy response. The short answer to your question is: "yes, but it ain't that simple."
 
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BurntFingers

Well-known member
Do you have an "if in doubt" approach? We've all heard the "high impedance in, low impedance out" commandment many times. Is that a good rule to follow in most circumstances? Whack a 2.2m pull down on the input and job done sorta thing?

Also, if you don't mind explaining candidly: what are the negatives of the 'wrong' impedance? Is it strictly a loss of treble/volume or are there any other things going on that affect what we hear through the speakers?
 

Chuck D. Bones

Circuit Wizard
I prefer the high impedance in, low impedance out approach because it simplifies stacking. But, ease of stacking is not the only consideration. Sometimes we want something else. I have yet to run into a situation when a high output impedance is better. Not saying it couldn't happen, but I don't expect it will happen, at least not to me. If the pedal is first in the chain, then the input impedance affects how it interacts with the guitar (pickups, controls and most important, fingers). High input impedance will let all of the signal thru without altering the pickup's voicing. A high input impedance pedal will be less sensitive to the guitar's volume control. A few pedals contain internal trimmers that let you choose an input impedance. Then there's the question of how high is high enough? If you've ever tried change the 250K volume pot in a Strat to 500K to see what happens, then you kinda know the answer. Somewhere around 500K it stops making much difference. Friedman sets the input impedance on his pedals at 330K. So let's call that "high enough."

Now about that 2.2M pull-down resistor... That's not the only component influencing the input impedance. You have to take everything in parallel with it into account. I know I've got an example here somewhere...

OK, here's the front-end of the Marshall Blues Breaker:
1616545247185.png
What's the input impedance? It's not 2.2M. It's 2.2M in parallel with 1M in parallel with the input impedance of IC1.1. The impedance of the + input of a JFET opamp is so high we can treat it as infinite. So we're down to 2.2M in parallel with 1M = 688K. Still plenty high.

I like to pick on EQD, so let's look at one of their pedals, the Acapulco Gold. Pretty simple 386-based overdrive.
There's a 1M pot up front, but the LM386's input impedance is 50K. So if we dime the GAIN control, the input impedance is just under 50K. As we turn GAIN down, the input impedance rises because now part of the GAIN control is in series with the input. With GAIN at noon, the input impedance is around 200K (it's a C-taper pot).

EQD Acapulco Gold.png

Make sense?
 

BurntFingers

Well-known member
Yep. What about at the other end? Output impedance and volume pots.

Is that determined by the pot in the EQD example above?

I've often wondered why 1 circuit will use a 10k pot and another a 100k. I get that sometimes that's because the pot is working as part of a filter with the output cap, but is that all there is to the decision making?
 

fig

Village Idiot
the input and output impedance is non-linear (it changes with signal amplitude) and frequency dependent and it gets complicated.
multiple interdependent variables are always a treat to anticipate or determine compensation.

I appreciate your time and attention. You don't simply answer 'yes, no, or f*-off', which is both impressive and helpful.

I glean from this that some of these variables (vt for example) can *sometimes be estimated with little to no noticeable effect.

*this also is dependent of any influences from all other components in the circuit. If, for example there are components that influence the vt such as, but not limited to the thermal dissipation of a voltage regulator."

I also picked up a vibe that you take exception to some circuit designers who "grip it and rip it" (as John Daly would say). In the industry I worked for 35+ years, we operated under a very strict set of protocols. They were developed to meet government standards and guidelines (government-speak for 'or else'). Guitar pedal circuit design standards are seemingly written [or not] by the individual designer, but enforced by the end-user.
 

Chuck D. Bones

Circuit Wizard
vt is known as "thermal voltage." It's equal to k * T / q where k = Boltzmann's constant (in Joules/degree Kelvin), T = absolute temperature (in Kelvins), and q is the charge of an electron (in coulombs). k and q are defined as constants and T = 300K at room temp. A 30 deg C change in T corresponds to a 10% change in vt, so it's safe to pretend T is constant in this case and not have to bother figuring out the junction temp of each transistor. We can assume vt = is always 26mV and not be that far off. I say "in this case" because the leakage and hFE of Germanium transistors are very temperature sensitive. If we use those parts, we have to be concerned about the effects of temperature changes.

In essentially all pedal circuits, a 10% change in any one parameter will have negligible effect on overall performance. Because variations like initial tolerance, aging and thermal drift are generally uncorrelated, the vast majority of the combinations of variations in all parameters also has negligible effect on performance.

Part of my job was presenting my designs at design reviews and being a reviewer at other peoples' design reviews. I got to see it all from innovative, carefully designed circuits to sloppy "git that bullshit outta here" designs. I welcomed critique from other engineers because I wanted my stuff to be as good as it could be. If there was a mistake or opportunity for improvement, I wanted to know. I bring that same mentality when I look at pedal circuits. Some are very clever and well thought out, while others fall into the "git that bullshit outta here" category. Our customers were very demanding and it was on us to convince them they were getting what they paid for. I do not hold commercial designs to the same standards as hi-rel electronics, but I do have some standards. If you build 100 pedals of the same design, I expect at least 95 will work right out of the chute. A well-designed, well-built pedal will last decades if it's not abused.
 

Chuck D. Bones

Circuit Wizard
Yep. What about at the other end? Output impedance and volume pots.

Is that determined by the pot in the EQD example above?

I've often wondered why 1 circuit will use a 10k pot and another a 100k. I get that sometimes that's because the pot is working as part of a filter with the output cap, but is that all there is to the decision making?

I think that the value of the volume control is frequently determined by what the builder has in stock. A100K appear to be the most common value for Gain and Volume controls. In the case of the Acapulco Gold, the Volume control could have been A1K because the LM386 is designed to drive an 8Ω speaker. Driving a 1K pot with that circuit is easy.

Going back the Blues Breaker, the Volume pot loads the tone circuit. A100K is a good value but lower values will work. The downside to using a lower value Volume pot is the volume will change when we turn the Tone knob. Not a big deal, but avoidable.

1616601089313.png

The Volume control is not always at the tail end of the circuit. When it is, it determines the pedal's output impedance and the output impedance will change when you turn the Volume knob.

If you're driving a long cable, lower output impedance is desirable because there is less high-freq loss in the cable. The EQD Hoof has a 1M Volume pot at the end and that could have issues with cable loading. I put A100K in my Hoof.

I chose A50K Volume pots for the Biggus Dickus and Mojito because that was a good compromise between low output impedance and loading the tone controls (BD) or output stage (Mojito). A100K would have worked too.
 

fig

Village Idiot
It's always encouraging and refreshing to encounter individuals who have an attitude of ownership in their work. Most often I've found those same individuals extend the same to most areas of their lives.

the leakage and hFE of Germanium transistors are very temperature sensitive. If we use those parts, we have to be concerned about the effects of temperature changes.

Would you go a bit deeper into under what conditions that concern might translate to something actionable and what that action might be? In other words, "whatcha gonna do bout it?"
 

Chuck D. Bones

Circuit Wizard
I've been meaning to write an article on the practical aspects of BJT biasing. I guess it's time...

The short answer to Fig's question "whatcha gonna do bout it?" is you either
  • live with bias drift - most players will find this unacceptable because your pedal can crap out in the middle of a gig
  • have a bias pot on the front panel so you can manually compensate for bias drift - the Sunface has this
  • design a bias circuit that uses negative feedback to stabilize the bias - my preferred method
 
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