Need circuit advice - analog transistors

[mmagliaro]

This one is for nerdy electrical engineers and circuit theorists... (just warning you so you can happily scroll away if that ain't you)...

Below is a snippet of a typical transistor current amplifier (like I use in my throttle).  My main question is

"What are R3 and R4 for?"

Here's the circuit.  Questions and data are below it.



Observations.
Clearly, Q3 carries the lion's share of the current to the load.  That is by design.  But also note that
Q1, Q2 do conduct a small current through R3,R4 to the load.

D1 is just there to short out any accidental connection of another power source to the load.
Rload can be as low as 2 ohms in extreme load cases.

Why not just connect Q1's emitter to Q2's base and be done with it?  (And similarly, connect Q2's emitter
to Q3's base).   If R3, R4 are removed, the circuit still works, and in fact, since Q3 gets a little bit more drive
current to its base, I get a a bit more top end voltage to the load, which is always nice.

Are R3,R4 there to stabilize the circuit, so that Q1,Q2 are always saturated no matter how light
the load is, and so that they don't go in and out of saturation as the load changes?
R3,R4 put more load on Q1,Q2 than Q3 ever does, so it's almost like they are there to fully turn-on Q1,Q2
and not have them change their gain regardless of the load.  But I'm just guessing here.

[TrainCat2]

Q1 and Q2 are NPN transistors and R3 and R4 are there to set the operating bias current of the successive transistor. Basically, sets the operating range so there is the right voltage/current to drive the next stage within its operating range. So Q1/Q2 are basically amplifiers to be able to drive Q3 sufficiently from a much smaller input. Think of it as a 3-stage amp for a  low voltage/current input.

[mmagliaro]

Q1 and Q2 are NPN transistors and R3 and R4 are there to set the operating bias current of the successive transistor. Basically, sets the operating range so there is the right voltage/current to drive the next stage within its operating range. So Q1/Q2 are basically amplifiers to be able to drive Q3 sufficiently from a much smaller input. Think of it as a 3-stage amp for a  low voltage/current input.

Bob, yes, I think I understand the basics of it being a transistor current amp.  But I'm still confused on how R3,R4 get the transistors into
their operating range (I assume you mean we want them turned on well into their active region so they are in a good spot in their gain curve).
Let's focus on just one of them to keep it simple... R3.  So R3 is conducting a small current from Q1 from its emitter.  How does that have any effect on the current going into Q2's base?  I should point out that doing this in a circuit simulator, removing those resistors does not seem to have much effect on the current through Q1,Q2 or Q3, nor the current through the final load.  What condition are they trying to prevent by including those resistors?

[peteski]

Is a throttle circuit a real current amp?  Normally throttles control output voltage independent of the load current.  That is what makes them better than simple rheostat throttles.

[mmagliaro]

Is a throttle circuit a real current amp?  Normally throttles control output voltage independent of the load current.  That is what makes them better than simple rheostat throttles.

It is a current amplifier in that the speed voltage at the speed pot, and all the other effects, like pulse and momentum,
are all done in a relatively high resistance, low-current circuit, and then final voltage is passed through several stages of transistor
to amplify the current so that the current isn't drawn directly from the initial control pot like it is in a rheostat circuit.
As such, a higher load on the final transistor has very little effect on the voltage. 

Thinking of it another way, let's say you set the speed
pot to output 6v.  10uA goes into the 1st transistor, 1mA comes out, 50mA comes out of the second, and a full 1A comes out of the third to the track.
So we have gains of 100, 50, and 20 at each stage.  Assuming none of the transistors are "maxed out" on their gain, and assuming they are all fully turned on (so that the voltage drop through each transistor is as high as it is ever going to get),  if we suddenly draw 2 amps at the track, the transistors can still turn 10 uA into 2 amps, without the voltage at the speed pot dropping at all (now the gains might be 150, 50 and 27).   So you still get 6v.  The only "sag" you get is from the power transformer, which hopefully is beefy enough to deliver 2A in this case, and is being regulated so you see minimal sag from that as well.

Maybe the easiest way:
The emitter from each transistor has the same voltage as the one before it, except about 0.6v less due to the junction drop, but can deliver anywhere from 20-50 times more current, so it is a current amplifier.

[peteski]

While I'm rusty on transistor amplifier deigns, I'm wondering about the circuit simulator program.  Do you still have the prototype of the throttle?  WHat if you disconnect R3 and R4 in the actual circuit?  Will it behave like the simulated circuit (still work properly)?

The other question I have is what you based this circuit on?  Or did you design it from scratch?

[mmagliaro]

While I'm rusty on transistor amplifier deigns, I'm wondering about the circuit simulator program.  Do you still have the prototype of the throttle?  WHat if you disconnect R3 and R4 in the actual circuit?  Will it behave like the simulated circuit (still work properly)?

The other question I have is what you based this circuit on?  Or did you design it from scratch?

Short answers are yes and no.   Yes, I have actually built this section like this on a breadboard and it seems to work fine.
And no, I didn't design this part of the circuit myself.  The original design came from Peter Thorne's 1970s book.  About the only
original thing left in my throttle after all these years is this basic section of 3 transistors and the control pot.  The power supply, pulse generation, momentum, I redesigned and improved.

I am even using nearly similar transistors for Q1 and Q3.   He used a 2N3055, the big TO-3 metal cased one.  for Q3.   I used to use those.  But now
I used the TIP3055, a smaller form-factor with similar current and gain specs.  For Q2, I picked something with medium power capacity
like he did, but not exactly whatever his part was.

No matter how I ponder it, I see no reason to need those resistors.  They don't tie the bases to ground, so they can't be for making sure the transistors will all shut off at minimum throttle.   And in the event of a short circuit, with the original resistances Thorne had in there (470 and 56 ohm), quite a lot of power has to be dissipated by Q1,Q2 - maybe not enough to burn them out, but enough to send them close to 100 degrees C (I tested this, and it agrees with the wattage the simulator says and the deg C temp rise per watt that the data sheets say).  That's why I bumped them
up to 2.2k and 220 ohm.  Even R5 was originally a 330k, but I bumped it to 820k because otherwise, it draws just enough current through some resistor's that feed Q1's base that you lose nearly 1/2 volt.  All those little 0.1 volt here and 0.5 volt there caused Thorne's original design
to only hit about 10v maximum at full current load, whereas I am able to keep it above 12.  2v x 3A = 6 watts of wasted power.  That's a lot of heat I don't have.

[peteski]

Well, I did take electronics courses back in the day and while we covered some analog amp. circuits, I was more into digital electronics where transistors are used as switches.  For full enclosure, the electronic courses I took back in Poland also covered vacuum tube circuits, so when I look as a schematic I have an idea what it is.  I also was educated on telephone exchange equipment like rotary relays.  We had some fun with those when we actually practiced in live switching office.

One thing I'm curious about is why in your simulation you user relatively high load (1k) on the output transistor?  In real live that will be much lower resistance (locomotive's motor).  If you used lower resistance, would that affect the functionality of R3 and R4?

That circuit is reminiscent of what is called Darlington Transistor. That circuit is usually made up from 2 transistors, but a third is also mentioned in the article below.  And yes, there are no resistors used.
https://en.wikipedia.org/wiki/Darlington_transistor

[mmagliaro]

Well, I did take electronics courses back in the day and while we covered some analog amp. circuits, I was more into digital electronics where transistors are used as switches.  For full enclosure, the electronic courses I took back in Poland also covered vacuum tube circuits, so when I look as a schematic I have an idea what it is.  I also was educated on telephone exchange equipment like rotary relays.  We had some fun with those when we actually practiced in live switching office.

One thing I'm curious about is why in your simulation you user relatively high load (1k) on the output transistor?  In real live that will be much lower resistance (locomotive's motor).  If you used lower resistance, would that affect the functionality of R3 and R4?

That circuit is reminiscent of what is called Darlington Transistor. That circuit is usually made up from 2 transistors, but a third is also mentioned in the article below.  And yes, there are no resistors used.
https://en.wikipedia.org/wiki/Darlington_transistor

That's because a Darlington transistor really is just an emitter-follower current amp of 2 transistors inside a single package.

I don't always simulate with a 1k load.  That's just what I threw on there for this screen grab.  I certainly have simulated it with 4 ohm, 2 ohm,
and even ".01 ohm" (to see what happens in a short circuit condition).  The more current I draw, the better off I am if those resistors
are higher-valued (2.2k and 220, instead of the original 470 and 56).  With the lower values, and heavy load on the output,
Q1 and Q2 start having dissipate enought heat to get them hot (i.e. 60-70 C).  Not hot enough to destroy them, but unnecessarily hot.
And siphoning off current from Q1,Q2 that way only causes the input to Q3 to dip a little, getting me anywhere from 0.1 to 0.5v
lower top end under heavy load.

It could just be that Thorne selected "conservative" values (470 and 56)  for a widely varying selection
of transistors, and always be guaranteed to turn Q1 and Q2 on so the circuit would always work. 

(It's too bad he died many years ago, or I would try to hunt him down and ask him!)

Transistors do have an optimal gain region.  You need to draw enough collector current to get the transistor
into that region if you want the best gain out of them, but they are pretty forgiving.  The good "gain zone" is very wide. 
That may be all he was trying to do.

In my case, I know that the current I get through Q1,Q2 is definitely in a good zone based on their data sheet gain curves,
with the resistors I have chosen.

I guess I'll just have to be happy it works and let this drop.   

[mmagliaro]

Aha!  I FOUND IT.   After reading numerous articles on emitter-follower, I found something that finally
got this through my feeble brain.

The input resistance to any of the transistors is the emitter R x the gain of the transistor (and this makes intuitive sense, ta boot).
So for Q1, its input resistance is 2.2k x Q1's gain.  In my case, Q1, a 2n5088, has a gain of about 300.
So the input resistance to Q1 is 2.2k x 300 = 660k

The input resistance to Q2 is 220 x 60 = 13.2k
And the input R to Q3 = loadR x 20, so between .25A and 3A, this input R varies between 5 and 60 ohm

Because the load on each emitter from the following transistor is much less than the load from those resistors,
Q1 and Q2 see a nearly constant load regardless of what happens at Q3 and the final load resistance.

So.. going around in a big circle, those resistors are indeed there to stabilize the gain at each stage to make
the whole circuit keep a more constant voltage as the load goes up and down.  The lower I make those resistances,
the more stable Q1 and Q2 are, but at the expense of more wasted power and heat.

[peteski]

Thanks for doing the detective work (for yourself), and positing it here for others to educate themselves.

[mmagliaro]

Thanks for doing the detective work (for yourself), and positing it here for others to educate themselves.

You're most welcome.  I mostly did this in case somebody else in the group was a real transistor sharpie and would warn me,
"Whatever you do, don't increase those resistor values! "....

Here's some more, and very interesting (at least to me) data.
I tested the circuit with R3/R4 at  2.2k/220 ohm, and  then at 1.1k/110ohm.  So a 2x change in the R values.

2.2k/220 ohm
12 ohm load, 8v volts.  Change to a 4 ohm load, voltage dips to 7.61v

1.1k/110 ohm  SAME RESULT, exactly.  Dips from 8 to 7.61.
The resistors do not affect the stability at all.
That is what I was hoping I would see.  It means that my increased values are not messing up the transistor
gain performance.

It also shows that over a very wide swing in current, the output of the throttle is stable.
A 0.4 volt dip out of 8 volts (5%) for a 3x increase in current is quite good, way better than what would happen
with a rheostat, of course.

[peteski]

Now you got me thinking:  I wonder if you could come up with a circuit of similar stability but much lower power dissipation on the back side by using MOSFETs,  You could probably get by with just a single or two transistors.

Another thought I had was that I would have imagined that a transistorized throttle would be a voltage follower rather than a current amplifier.  That way the output voltage would follow the voltage at the wiper of the speed control pot.

[mmagliaro]

Now you got me thinking:  I wonder if you could come up with a circuit of similar stability but much lower power dissipation on the back side by using MOSFETs,  You could probably get by with just a single or two transistors.

Another thought I had was that I would have imagined that a transistorized throttle would be a voltage follower rather than a current amplifier.  That way the output voltage would follow the voltage at the wiper of the speed control pot.

Well, the output voltage does follow the voltage at the wiper of the speed control pot, just less the junction drops at the transistors.  It has to be both - it needs to follow the voltage, and amplify the current, since the pot itself can't conduct all that current.

I thought about MOSFETs, but to be honest, I'm more familiar with using them as power switches as opposed to linear gain devices.
I would have to read up to see if the gate on a MOSFET can be used to gradually increase the MOSFET's output, instead of just turning it on or off.  I'm not bothered by using 3 transistors.  They are small and cheap, relatively speaking, except for the final power device, and even those are only a buck or two.

[peteski]

Yes, while MOSFETs are usually associated with switching type applications (like the ubiquitous H-Bridge motor driver in DCC decoders) they can also be used in analog applications.  I know your design is finished - I was just curious whether you knew or considered using MOSFETs.

The 2N3055 power transistor (or its modern equivalents) are ancient technology. It has been around longer than N scale (since early '60s).  They are older than me! 
https://en.wikipedia.org/wiki/2N3055

I remember using them in the power stage of voltage regulator back in the '80s when I was building those circuits.