Low voltage op amp to control high voltage series pass transistor

[d.arbitman]

I am in the middle of designing a low drift and low ripple power supply for myself. To control the output voltage, which ranges from 0 to 20V, I am using an op amp whose output is fed into a Darlington pair of power bjts. The output of the Darlington pair is both fed back to the op amp's inverting input (i.e. the op amp is configured as a voltage follower) as well as to the load. The desired output voltage is present on the non-inverting input to the op amp as a reference. I have a transformer that produces +/- 40V rails so I am using those to power the op amp. I have a high voltage op amp but recently, the op amp was fried even though it can be powered from +/- 45V. As a result, I have decided to use a low voltage op amp to control the +/- 40V rails. The problem is, I have no idea how to go about doing this. I have tried using single transistor amplifiers to translate 0 -> 5V to 0 -> 20V but have had no luck. So the question, how do I use a low voltage op amp that is powered from +/- 5V to control +/- 40V rails so that my output ranges from 0 to 20V?

 

[pullmanwa]

I am in the middle of designing a low drift and low ripple power supply for myself. To control the output voltage, which ranges from 0 to 20V, I am using an op amp whose output is fed into a Darlington pair of power bjts. The output of the Darlington pair is both fed back to the op amp's inverting input (i.e. the op amp is configured as a voltage follower) as well as to the load. The desired output voltage is present on the non-inverting input to the op amp as a reference. I have a transformer that produces +/- 40V rails so I am using those to power the op amp. I have a high voltage op amp but recently, the op amp was fried even though it can be powered from +/- 45V. As a result, I have decided to use a low voltage op amp to control the +/- 40V rails. The problem is, I have no idea how to go about doing this. I have tried using single transistor amplifiers to translate 0 -> 5V to 0 -> 20V but have had no luck. So the question, how do I use a low voltage op amp that is powered from +/- 5V to control +/- 40V rails so that my output ranges from 0 to 20V?

Show us your schematic, please.

 

[d.arbitman]

Show us your schematic, please.

Uploaded. As you can see, the op amp is powered from +/- 40V but I don't want to use a high voltage op amp like I'm doing at the moment.

 

[pullmanwa]

Uploaded. As you can see, the op amp is powered from +/- 40V but I don't want to use a high voltage op amp like I'm doing at the moment.

I have a question if you don't mind.

1. where is the output voltage taken across from? I'm asking where is your ground reference for the output voltage?

** The + input to the opamp is missing --- that will cause a problem

 

[meBigGuy]

That's a pretty primitive schematic, but I see what you are trying to do.

You need an amplifier with gain since you want to go from say a 1V swing to a 40V swing.

Here is a circuit that does what you want and does it well.

http://circuits.linear.com/High_Speed_Amplifiers--all-439

Once you understand it, you can simplify it.

The simplest approach is to just drive into a single stage common emitter amp. But that dosn't have much drive. Next simplest is to follow that with an emitter follower. Don't forget to deal with the common-emitter inversion and divide down the emitter follower output for the feedback.

 

[d.arbitman]

I have a question if you don't mind.

1. where is the output voltage taken across from? I'm asking where is your ground reference for the output voltage?

** The + input to the opamp is missing --- that will cause a problem

I know. That's the input voltage and assumed to be in the range of 0 to 20V. The ground reference is labeled on the schematic (for lack of a better term...the triangle connected to the power supplies is ground).

I have configured the op amp to behave as a voltage follower.

 

[d.arbitman]

That's a pretty primitive schematic, but I see what you are trying to do.

You need an amplifier with gain since you want to go from say a 1V swing to a 40V swing.

Here is a circuit that does what you want and does it well.

http://circuits.linear.com/High_Speed_Amplifiers--all-439

Once you understand it, you can simplify it.

The simplest approach is to just drive into a single stage common emitter amp. But that dosn't have much drive.

That op amp is supplied from +/- 125V which is what I want to stay away from. I want to use a low voltage op amp to do what I want. I not only want the input to be scaled down but also the rails to be, say +/- 5V but I still want it to be able to drive a series pass transistor whose rail is at +40V

I have attached a complete schematic of what I have.

R1 is the load resistance. U1 and its associated resistors mimic a 50V/V current sense amplifier.

 

[meBigGuy]

No it isn't. Notice the zener diodes that drop the voltage for the op amp. Or supply the opamp from different supplies.

 

[d.arbitman]

No it isn't. Notice the zener diodes that drop the voltage for the op amp. Or supply the opamp from different supplies.

Oh you're absolutely right.

 

[meBigGuy]

OK --- You just need to drive the opamp into a common emitter stage with a pullup to 40V. Follow the common emitter stage with your darlington. Don't forget the inversion of the common emitter stage, which changes things a bit (notice the feedback to the + input in the 125V circuit). You can make U2 inverting, or whatever.

 

[d.arbitman]

I have no idea what's going on in that circuit. I figure the 2n3440 and 2n5415 furthest to the right are the series pass transistors and they are the ones that supply the current. What's the point of the 27R resistors and the 1k that are connected to them? What's the point of the 2n222 and 2n2907 transistors? Why are the diodes 1n4148 connected across the base of the 2n3440 series pass transistor to the output?

How do I go about understanding what that circuit does?

 

[meBigGuy]

The circuit is a bit fancy for what you want to do. It provides high power symmetrical drive with current limiting.

The 27R resistors are for current limiting sense. When the voltage across them reaches 0.7V the 2907 and 2222 transistors turn on, reducing the drive to the output transistor.

See if you can decipher what I said in post 10.

 

[meBigGuy]

describing the circuit a bit more for you. Think of it as two common emitter amps followed by 2 emitter followers.

If you look at the circuit as a big X it is easier to conceptualize what it basically does. The bottom left 3440 is a common emitter amplifier, and the top right 3440 is sort of its emitter follower. And the 2 5415's operate the same way. In combination they are very symmetrical, which you really don't need.

Look at T1 and T5 in this design. It has a double inversion so maybe that will make things easier. Just drive into T1. Read the wikipeda page on common emitter amplifiers to get it biased right.
http://www.bogobit.de/bogobox/

 

[d.arbitman]

The circuit is a bit fancy for what you want to do. It provides high power symmetrical drive with current limiting.

The 27R resistors are for current limiting sense. When the voltage across them reaches 0.7V the 2907 and 2222 transistors turn on, reducing the drive to the output transistor.

See if you can decipher what I said in post 10.

What do the 1k resistors do? What is the purpose of the first transistor stage whose base is connected to 1M resistors? How does that first stage work? I figure the diodes prevent crossover distortion, but what about everything else?

As far as post 10 is concerned I don't know which circuit you are referring to when you say drive the op amp into a common emitter stage.

What do you mean by pullup to 40V?

In a common emitter configuration, the output is taken at the collector and the output is inverted. As a result I should make U2 in my schematic inverting? It seems like I would be introducing more parts than I already have to do the same thing.

 

[meBigGuy]

Explaining basic transistor circuits is a bit beyond what can be done here. I can point you to a topology that can do what you want and you have to take it from there. I suggested a common emitter followed by a darlington, but two common emitters is a better idea.

Again, look at T1 and T5 in http://www.bogobit.de/bogobox/ That's a basic topolgy for what you want to do.

Don't forget that you need a voltage divider in your feedback loop to keep high voltage from frying the opamp.

I'm not sure whether you will run into stability problems with the additional gain. You may want to rolloff U2.

 

[jim hardy]

here's an approach

http://circuits.linear.com/Operational_Amplifiers_(Op_Amps)--all-439

±120V Output Precision Op Amp

http://circuits.linear.com/img/439_circuit_1.jpg

 

[meBigGuy]

That's what I posted. But for his circuit the T1/T5 combo in http://www.bogobit.de/bogobox/ is probably adequate. What do you think? I need to get a schematic tool. Too hard to describe circuits.

 

[d.arbitman]

Explaining basic transistor circuits is a bit beyond what can be done here. I can point you to a topology that can do what you want and you have to take it from there. I suggested a common emitter followed by a darlington, but two common emitters is a better idea.

Again, look at T1 and T5 in http://www.bogobit.de/bogobox/ That's a basic topolgy for what you want to do.

Don't forget that you need a voltage divider in your feedback loop to keep high voltage from frying the opamp.

I'm not sure whether you will run into stability problems with the additional gain. You may want to rolloff U2.

I rebuilt the circuit in that same way...op amp drives an NPN which sinks the base current from a PNP via its collector. I ran some simulations and boy that PNP will dissipate a ton of power.

Thank you for helping by the way.

I have attached my latest schematic.

P.S. I tried loading the circuit that you showed me from LT, the +/- 120V output, with a 10\Omega load and it failed miserably.

 

[jim hardy]

That's what I posted. But for his circuit the T1/T5 combo in http://www.bogobit.de/bogobox/ is probably adequate. What do you think? I need to get a schematic tool. Too hard to describe circuits.

Well so it is exact same link ! Sorry - I only saw bogobox...

in his schematic
he needs a current source to turn on his Q1, and a pull-down transistor controlled by the opamp to turn Q1 off.

Establish some base drive to Q1, enough to make rated current
and steal that base drive away to control output voltage.


As you said he should figure out why that Linear Technology circuit works.

Also as you said, it is difficult to paint a good mental picture with just words.

old jim

 

[jim hardy]

P.S. I tried loading the circuit that you showed me from LT, the +/- 120V output, with a 10\Omega load and it failed miserably.

Well

120 volts across ten ohms is twelve amps.
Did you notice that the 27 ohm resistor and 1N4148 diode form a ~25 milliamp current limiter on the output stage?
How's it do below 1/4 volt on that ten ohm load?


When you can describe in words how a circuit works you are coming to understand it.

 

[d.arbitman]

Well

120 volts across ten ohms is twelve amps.
Did you notice that the 27 ohm resistor and 1N4148 diode form a ~25 milliamp current limiter on the output stage?
How's it do below 1/4 volt on that ten ohm load?


When you can describe in words how a circuit works you are coming to understand it.

Wait a minute. If the 1n4148 AND THE 27ohm form the current limiter, then what do the 2n2222 and the 2n2907 do?

 

[jim hardy]

Wait a minute. If the 1n4148 AND THE 27ohm form the current limiter, then what do the 2n2222 and the 2n2907 do?

You got me good !

They make a better one.

 

[d.arbitman]

What do the 510 and 330 ohm resistors do, why are they chosen so small?

Is the smaller of the two values connected to the emitter of the PNP rather than the larger in order to forward bias the transistor?

 

[jim hardy]

What do the 510 and 330 ohm resistors do, why are they chosen so small?

Is the smaller of the two values connected to the emitter of the PNP rather than the larger in order to forward bias the transistor?

They're to make the transistor mimic the current flowing through the 50K ohm resistor. That transistor is a very simple(if admittedly imprecise) controlled current source.
I did arithmetic assuming opamp is at zero out and get two milliamps through that transistor.
That is the base drive current available to the final transistors, but some of it is diverted by your current limiters..

The basis of this thing is a balancing act with two controlled current sources driven by opamp through those 50K resistors.

See how cleverly they've separated the high voltage from opamp?
50K resistors and high voltage transistors.
The output is "felt" through the 100K resistor and balanced against input via its 10K

Also observe the driver stages invert polarity, that's why the LT 1055's + and - inputs are backward from your normal inverting opamp configuration.

I'm still trying to figure out the biasing at full output.
Perhaps you or me-big-guy will beat me to it ...

Here's as far as I got;
---------------------------------------------------------------------------------------------------

http://circuits.linear.com/img/439_circuit_1.jpg

The objective is to control ~ 120 volts with only a ~12 volt "handle".
We have to do that because
The opamp is powered by those zener sources, look at their datasheet they're 15 volts.
So the LT1055opamp can only move his pin6 up and down about 12 volts.
We'll call that point in the circuit 'our 12 volt handle'...

Let us look at top half of circuit when handle is at zero volts.
The majority of the voltage drop between power supply and our "handle" is across the 50K resistors.
Current through the 50K resistor is around ~(125 ) / 50.5K = 2.475ma.
Voltage across the 510 = 2.475ma X 510 ohms = 1.262 volts
If Veb for the 2N5415 is 0.6V, the there must be (1.262 - 0.6) = 0.662V across the 330 ohm resistor. That'd be 2 milliamps through it.

That 2 ma multiplied by 2N3440's probable h_fe of ~ 30, would cause 60 ma to flow out of its emitter but your current limiters limit it to ~25 ma.

By symmetry, the bottom half of the circuit is doing the exact same thing, except of course current is in other direction - leavingthe output terminal instead of entering it...
So the two 25ma currents cancel and output is zero. Now let us push our 'handle' down to -12 volts.
Current through the 50K resistor is now ~(137) / 50.5K = 2.712 ma
Voltage across the 510 = 2.673ma X 510 ohms = 1.383 volt
voltage across 330 = (1.383-0.6) = 0.783 v
that'd be 2.374 ma through it

So - lowering our 'handle ' twelve volts increased the base drive current for top 2N3440 by 374microamps.

By symmetry, bottom output transistor should see close to same decrease in base drive.---------------------------------------------------------------------------------------------------

I've not yet considered the 1 meg resistors because I think they're negative feedback,,it's late and I'm pooped. I hope you guys have it explained in the morning !

old jim

 

[d.arbitman]

http://cds.linear.com/docs/en/application-note/an18f.pdf

See page 6.

It seems that the 50k and 1M allow the op amp to control a large voltage swing on the output from a small swing on the op amp's ouput. I just can't see it. How exactly does that inner loop gain work?

 

[meBigGuy]

Think of the voltage on the base of Q1 as effectively being across the 330 ohm emitter resistor thereby modulating the Q1-Q2 current. The same thing is happening with Q2. If Q2 is "more off" the the voltage at the collector of Q1 will be high, and vice versa. Note that there is no dot between the diodes. The diodes just sort of parallel the two Vbe diodes on the output devices. The 1M feedback resistors kind of clean things up.

 

[d.arbitman]

Think of the voltage on the base of Q1 as effectively being across the 330 ohm emitter resistor thereby modulating the Q1-Q2 current. The same thing is happening with Q2. If Q2 is "more off" the the voltage a the collector of Q1 will be high, and vice versa. Note that there is no dot between the diodes. The diodes just sort of parallel the two Vbe diodes on the output devices. The 1M feedback resistors kind of clean things up.

"The local 1M-50k feedback pairs set stage gain at 20, allowing ±10V LT1055 drives to cause full 120V output swing." Page 6 of that PDF.

What I noticed through simulation is that changing the 1M to 100k lowers the output when the input voltage remains the same. The question is, why and how. Notice that the voltage at the base didn't change much. The voltage at the collector changed by an order of magnitude.

 

[meBigGuy]

OK. I had it wrong (well, the 1M is critical to setting the gain --- the rest explained the mechanism).

The 50K - 1M combination is exactly like the source and feedback resistors in an inverting opamp. Current into the Q1 base node through the 50K must be pulled out through the 1M. Reducing the 1M to 100K reduces the gain by 10 since a 10 times lower voltage will create the current.

I learned something too.

 

[d.arbitman]

OK. I had it wrong (well, the 1M is critical to setting the gain --- the rest explained the mechanism).

The 50K - 1M combination is exactly like the source and feedback resistors in an inverting opamp. Current into the Q1 base node through the 50K must be pulled out through the 1M. Reducing the 1M to 100K reduces the gain by 10 since a 10 times lower voltage will create the current.

I learned something too.

I drew that analog with an inverting op amp too, but still doesn't explain it. Notice in my attached simulation pictures that the current draw increase is through the 510ohm resistor. The current out of the base increased by 1uA when the 1M resistor was changed to 100k

Current into the Q1 base node through the 50K must be pulled out through the 1M. Reducing the 1M to 100K reduces the gain by 10 since a 10 times lower voltage will create the current.

I don't understand what that means.

 

[meBigGuy]

The opamp model explains it, but there are other bias currents and finite open loop gain. Notice that the total gain didn't change by exactly 10. Basically the current through the 50K and the current through the 1M will be comparable. Take the 1M out and measure the open loop gain. I'm guessing it will be low enough to account for the discrepency. (model an opamp with an open loop gain of 10 and see what happens).

If there was infinite open loop gain and no input bias currents the current through the 50K and the current through the 1M would be equal and opposite.

 

[d.arbitman]

The opamp model explains it, but there are other bias currents and finite open loop gain. Notice that the total gain didn't change by exactly 10. Basically the current through the 50K and the current through the 1M will be comparable. Take the 1M out and measure the open loop gain. I'm guessing it will be low enough to account for the discrepency. (model an opamp with an open loop gain of 10 and see what happens).

I noticed that the total gain did not change by the same ratio that the resistance changed. I am assuming that is due to the nonzero current change in the base of Q2 (Q1 in LT's schematic) in my schematic as well as the non-linearity of transistors. (Q1 in LT's schematic).

How are the currents through the 50k and 1M comparable?

I attached a picture of a schematic that ought to simulate the behavior of the 1M-50k gain stage. Hopefully it's an accurate model of the gain stage.

So if we set Rg = 50k and Rf = 1M, we know that the output voltage must be higher than if Rf was less 1M due to KCL.

I am having trouble applying this to the actual circuit because of the increase in current through the 510ohm resistor.

 

[jim hardy]

Thanks Guys ! I was thinking about this while installing a new front door on the house today.

You have verified what I had come to think - the 1meg::50K set gain of output stages to 20, so we have two amplifiers in cascade.

And your simulation confirmed something I was trying to reason out:
Observe that of the ~2ma flowing down through R8, hardly any goes into Q5's base.
None should go into Q3 since it's only got 31mv Vbe
which means it has to flow down through D3 and D4 where it robs base drive from Q6.
So as top half of circuit pulls output up, bottom half pushes up due to mismatched base drive currents .
That makes gain of the booster stage quite high, so that wrapping R6 around the booster loop gives a nicely controlled voltage gain. Same way a self-saturating magamp from the 1940's works.

That is what I was trying to reason out when I gave up last night.
So D3 and D4 are NOT current limiters like I initially thought, instead they are part of the biasing.
What REALLY had me going was the dot between D3 & D4 that appeared to tie that point to output.
I couldn't make the circuit go with that dot there.
But it is NOT there in the appnote page 7 !

Ahhh so my alleged brain isn't totally fried. Whew !

I learned a couple things today too -
1. How to hang a door and true up the jamb (it's sure nice having retired carpenter friends)
2. How a nice voltage booster really works (it's sure nice having bright young friends who like electronics)

Thanks Again Guys ! You really made my day !

And congratulations on your victory figuring this thing out...

old jim

 

[d.arbitman]

And congratulations on your victory figuring this thing out...

I still can't figure out how that inner gain loop works. Mainly how the ratio of 1M:50k controls the ratio of total output swing to the op amp output swing.

 

[meBigGuy]

I think I am out of my element here. I don't fully understand the 510 ohm resistor and its sizing. I'll have to think about it. It's more confusing because there are two sides playing against each other. Have you played with changing the 510R to see what that does.

 

[d.arbitman]

I think I am out of my element here. I don't fully understand the 510 ohm resistor and its sizing. I'll have to think about it. It's more confusing because there are two sides playing against each other. Have you played with changing the 510R to see what that does.

Changing the 510 changes the bias voltage at the base of Q1/Q2, which changes the voltage drop across the 330 and as a result changes base drive to the output transistors. Although I tried and nothing major happened.

I am mainly interested how changing the 1M to 100k changes the maximum output voltage swing, specifically reduces it. In general, how changing the 1M resistor change the maximum output voltage swing.

 

[meBigGuy]

Good, I'm glad the 510 has minimal impact. The 1M is the negative feedback resistor that sucks or adds current to the base node to control the gain. It does that for sure. Take it out to see the open loop gain.

Normalize the currents to 0V input and look at the deltas. It may make more sense then. The currents for 0 output voltage are the basic bias currents, and they swing relative to that. The open loop gain affects the gain change vs. resistor size.

 

[d.arbitman]

I simplified the circuit. I took out the op amp and feedback resistors because they were unnecessary for understanding what is happening. I also took out the current limiting transistors because there was no load.

So I applied 0V to the input and -4V with a 1M resistor. I attached the results.

 

[meBigGuy]

For example the 0 level current for the 1M resistor is 123 uA, so when it changes to 23uA when the output goes to 100V, that's actually a -100uA change in current. Then look at the delta through the 50K.

 

[meBigGuy]

It seems there is significant offset since you get 43V for 0 input. Not sure where that is coming from.

 

[d.arbitman]

It seems there is significant offset since you get 43V for 0 input. Not sure where that is coming from.

It's using real models as opposed to ideal transistors. NPN and PNP transistors are very different.

For example the 0 level current for the 1M resistor is 123 uA, so when it changes to 23uA when the output goes to 100V, that's actually a -100uA change in current. Then look at the delta through the 50K.

Where do you see 123uA?

 

[meBigGuy]

If the output voltage was 0V (and all should be relative to an OUTPUT of 0V) then there would be 123V or so across the 1M resistor. When the output goes to 100V, there will be 23V or so across the 1M. That means the 1M current changed by -100uA. What do you think the 50K current will change by? Maybe not exactly 100uA, since there is finite open loop gain and input bias currents, but it should be consistant with the gain change you are seeing.

 

[d.arbitman]

If the output voltage was 0V (and all should be relative to an OUTPUT of 0V) then there would be 123V or so across the 1M resistor. When the output goes to 100V, there will be 23V or so across the 1M. That means the 1M current changed by -100uA. What do you think the 50K current will change by? Maybe not exactly 100uA, since there is finite open loop gain and input bias currents, but it should be consistant with the gain change you are seeing.

Oh, you're considering an ideal case; I thought you were talking about my simulation and I couldn't find 123uA anywhere. You were starting to scare me.

According to the simulation pictures I posted. The current decreases by ~60uA through the 1M while the current increases by ~80uA through the 50k which makes sense, since the voltage decreases by 4V.

4V/50k = 80uA


The difference of 20uA is compensated for via the base drive and the current through the 510ohm resistor. I still can't explain the inner loop gain.

 

[d.arbitman]

So if we assume that the input can swing between -10V and +10V, I have simulated the output swing using 1M and 100k resistors. I only changed the resistor values and I want to know why/how the output swing is larger using the 1M resistors.

 

[meBigGuy]

The difference of 20uA is compensated for via the base drive and the current through the 510ohm resistor. I still can't explain the inner loop gain.

You just explained it. The input voltage goes negative causing an increased current to be pulled out of the 50K. That current comes from the 1M resistor. (well, the 1M resistor draws less current because the output voltage went up). The output had to go up enough to change the current by the same amount as the 50K current did, therefore a 1M/50K gain (almost).

 

[d.arbitman]

I finally figured it out. I forgot to take into account the quiescent current and the associated change in said current when the input drops. Because 510 is 1% of 50k, the increase in current through the 510 is negligible, so when the input drops, the current through the 1M has to decrease by roughly the same amount. When you use a lower value resistor the voltage change is not as significant because more quiescent current flows and a small change in current doesn't produce a large change in voltage. However when the resistor is large, the quiescent current is small and a small change in current changes the output voltage by a larger value.

 

[meBigGuy]

Not sure I totally agree with your use of quiescent, but I'm glad you see the 50K/1M currents vs the resistor ratio as setting the gain.

Think of the base of Q1 as a contant voltage point since the Q1 collector current does not change significantly. Or, think of the Q1 current as constant since the voltage at Q1's base doesn't change significantly.

 

[d.arbitman]

Not sure I totally agree with your use of quiescent, but I'm glad you see the 50K/1M currents vs the resistor ratio as setting the gain.

Think of the base of Q1 as a contant voltage point since the Q1 collector current does not change significantly. Or, think of the Q1 current as constant since the voltage at Q1's base doesn't change significantly.

By quiescent, I meant when the input is 0V.

 

[meBigGuy]

As I reread what you said, I think you almost got it. Basically the current change in the 50K needs to be matched to the current change in the 1M. The output goes to whatever voltage is required to do that. The gain is not ideal because the effective impedance at the Q1 base is not infinite, and the open loop gain is not infinite.

The quiescent current has nothing to do with it. It is just quiescent. Maybe it helps you think about it, but it isn't part of the equation.

 

[d.arbitman]

As I reread what you said, I think you almost got it. Basically the current change in the 50K needs to be matched to the current change in the 1M. The output goes to whatever voltage is required to do that. The gain is not ideal because the effective impedance at the Q1 base is not infinite, and the open loop gain is not infinite.

The quiescent current has nothing to do with it. It is just quiescent. Maybe it helps you think about it, but it isn't part of the equation.

It's the change in current from the state when no input is applied. When the in put is -4V, then change in current is 80uA. The drop in voltage at the base is almost negligible (and the additional current through the 510ohm is on the order of 2uA) which is why the additional change in current out of the base is 1uA or so. The rest of the current must be decreased through the 1M which causes less of a voltage drop across it. Since ~77uA through 1M is much greater than through 100k, the output can swing much more.

 

[meBigGuy]

The output "MUST" swing much more as opposed to "can".

Is this the implementation you will use? Now that you understand this circuit, look at the simpler two transistor solution using T1 and T5 in http://www.bogobit.de/bogobox/