New EE Lab - HF Design

[Ian MacGregor]

Hello all,
I am back with more questions pertaining to a EE lab that I am currently taking. The link to our lab project is here: http://my.ece.ucsb.edu/bobsclass/137B/fiber_link.htm#_Toc37423576

In short, we are building an optical fiber link. This includes designing a random signal generator (for test) a Tx and a Rx. We are partially graded on how high we can get our frequency response in comparison to class mates. We have already successfully designed/built the random signal generator with a maximum operating speed of around 27MHz. Our questions involve the transmitter design.

Basically, the transmitter takes the 0-6V digital signal from the generator and turns the LED on/off. Our first question is why is the sample circuit so complicated??

We think we can design a circuit to perform this function with one or two transistors. Since all we are doing is converting an input voltage to an output current, wouldn't a single CE stage work as long as gm is designed to draw the desired current with our known input voltage??

Thanks, I'm sure we will have more questions coming.

Ian

 

[bcarso]

Have you covered the Miller effect yet in your study of simple amplifiers?

 

[Ian MacGregor]

Yes. We have covered the Miller effect briefly. Basically, all we know is that the intrinsic Ccb is multiplied by the gain the stage in a CE configuration. The multiplied Ccb is then across the the input to the stage as well. I think the hardest part about this transmitter design is that we are designing for a current output; we've always been doing voltage gain amps. I think most of the class is confused on how the circuit is supposed to actually work.

Ian

 

[bcarso]

OK. Well, the objective is to get a current flowing as quickly as possible in the LED, which as you know has the approximate circuit model shown (although I would like to know what your instructor means by an "ideal diode" in this context---especially as he later points out that the voltage drop is much larger than a typical diode---I guess ideal for the particular values of Is for the real part).

If you merely apply a voltage step to an LED, unless it is exquisitely fashioned to produce exactly the current you want it will under-drive or overdrive the device. This should be clear from the I V curves for the devices, where you see the very steep slope of current for an increase in voltage, and from the large temperature coefficient of this characteristic.

So, although you can look at the requirement as being to produce a specific voltage across the device, it is more apropos to think of producing a current---one more or less independent of voltage---as the drive signal. This means the LED foward voltage will have little effect on the current, which is what you want.

Take advantage of the suggestion to have a small current flowing in the LED in the logic zero condition too---this will help.

You and your classmates have correctly discerned that the transconductance provided by a transistor would do the job in principle. But the reason I mentioned the Miller effect is that this would be one mechanism preventing the highest speed from a given device. Also, bear in mind that the transconductance of a bipolar (especially) is also highly nonlinear and temperature sensitive, really only a small-signal number at a given bias point. So simply tying the emiiter to common and driving the base with a voltage step will not be robust.

On the other hand inserting a resistor in the emitter will make the circuit transconductance better controlled, but will also tend to slow it down, so there is a tradeoff there.

That should give you something to ponder and simulate.

 

[Ian MacGregor]

Ok, I've been pondering a bit here. I've come up with a CE circuit using emitter degeneration. I have not analyzed the HF response yet because I'm unsure how to do it. I know how to find Vo/Vin transfer functions, but I think I need to know Io/Vin. I'm a little confused on how to go about doing this. I've tried putting it all on paper, but I keep getting confused about how to figure out the nodal analysis for the ouptut half of the circuit. It seems like I keep trying to find Vo, when I don't think I need to (or want to). Also, at HF, do I model the diode as a Cap in series with a resistor of value 26mV/I?? I'm gonna post the schematic I've been working with in a sec.

Any ideas??

Ian

EDIT: Here is a link to a PDF of the schematic http://groupdiy.twin-x.com/albums/userpics/10021/CAM%20output%7E0.pdf

 

[bcarso]

That circuit is a good start in topology anyway. As far as the model, I believe you want the C across the current-dependent conductance of the diode. Remember to adjust the diode equation to give the right forward voltage characteristic. It's a little more complex than that I think because the C itself is a function of voltage, but a lumped model will get you close probably.

Add a means of setting the smaller logic zero current in the transistor (or just add an external i source---for the accuracy required a resistor to ground from the LED cathode is probably sufficient. Also be sure to inlcude the output impedance of the PRBS generator when considering the speed of the circuit.

What you want to do with this design is get the C charged as quickly as possible to the voltage across the LED corresponding to the settled value when driven by the desired current. That's what's going on when the guy talks about adding a zero to the network function, implemented by the parallel R-C network feeding the diode in the one schematic. In the LED datasheet sample circuit, the driver is more of a voltage source and the current is set by the series resistor. You are proposing a current source, so you have to get your initial charging current transient somewhere else---for example, a series R-C in parallel with the emitter resistor.

Another thing you don't want to do is to let the transistor saturate, or even get to too low a Vcb, but that shouldn't be hard with the 6V supply and a 1.9V forward voltage. But it means your 275 ohm R is too large, since that by itself would require 5.5V---the collector is running into the emitter. So you will have to do something to the output of your generator that attenutates the voltage swing and does so without significantly reducing circuit speed.

 

[Svart]

does it *have* to be a *BJT*? i didn't see where it had to be anything, just suggestions..

Is there a finite amount of current/voltage you can use?
(remember it has to be low impedance!)

Can you use both + AND - voltage for gate/base control?

if you absolutely positively have to use a BJT, how about a baker clamp?

MOSFETS can switch in the 10ns range.. is that fast enough?

How about an IGBT for lowside(ground side) switching of the LED? that way your LED is already biased via the source and you get the best of both BJT and MOSFET worlds.. fastass(not as fast as a true MOSFET but much faster than a BJT) switching without the conduction losses..


some other suggestions:

I would only consider MOSFETS or IGBTs for the switch if I were doing this.

drive it hard.. and you can put speed up diodes on the gates to..well.. to speed up the turn OFF. :green: Reversed biased, in parallel with the gate resistor. turn off is almost completely based on the gate drive circuit.

if you use an IGBT(or MOSFET), make sure it stays cold or you start to quickly loose any improvment in losses that you gained by using it..

Use a flywheel diode, but use the fastest one you can find.. look at IRF's FRED stuff for a start. the diode and the strength of your PSU limit your turn ON time.

keep the switching device close to the driving device.. the closer the better and faster.

I think if ye uses technology on thy side, ye shall win.

:guinness: :guinness:

 

[Svart]

"the zero-state LED current is to be less than 1 mA, while the on-state LED current is to be greater than 10 mA." interesting.. you could "warmup" the LED with a very high value R to ground from the cathode of the LED. that way the current is ready to flow through the MOSFET.. and give that damn LED all the current it can eat until it pops.. then get a new one and back the current off a bit.. :green:

" Do not exceed the maximum drive current to the LED, as replacing the LED is quite expensive. "

Digi-Key Part Number FB135-ND
Manufacturer Part Number IF-E98
Unit Price
14.00000

:thumb:

Oh and i took a look at the professor's background.. seems that he's really really really into highspeed data through various mediums.. Is that the class or is he secretly looking for breakthrough ideas from students to use in his research.. :shock:

 

[PRR]

> It seems like I keep trying to find Vo, when I don't think I need to (or want to).

Who cares about Vout?

You want the current in the intrinsic LED-diode. Not the LED, but the diode inside it. On the outside of the LED, the current is diode current plus parasitic capacitance current.

Say (made-up values) the diode is at zero current, the parasitic capacitance is 1,000pFd, the LED junction stays dark until it gets 2V drop. And you feed the minimum 10mA. We audio geeks call this "Slew Rate". 10mA into 1000pFd will slew 10mA/1000pFd= 10,000,000 volts per second, or 10V per uSec, or 2V in 0.2uSec. If your pulse rate is 2.5MHz, cycle time 0.4uSec, then you spend 0.2uSec slewing from zero voltage up to 2V, the LED just starts to light, and then you start slewing in the opposite direction.

And what is the slew rate on the discharge side of the cycle? Actually, this works in your favor. At the start, it is 10V/uSec. As diode current drops, discharge slows. This type of asymmetrical capacitor charge/discharge will tend to charge to just below the diode threshold. Your "off" state will be like 1.9V or 1.99V, so you don't have a large slew to get it on again. Note though, that this is a steady-state result: your first pulse(s) will be runts until the interbal cap charges to just below the diode threashold. With long meaningless data preables, error re-send, heartbeat pulses filling idle time, maybe not a problem. But if the prof asks you to send a short packet of data from a cold start, you fail because the first few bytes are mangled.

You can charge the cap faster if you throw big spikes of current at the LED. But big spikes in the active junction will burn-up the LED. (The LED may be cheap, but it is generally deep inside a box up on a pole; ladder-labor and downtime costs will kill your plan.) I don't know how the pros do it. There may be a clever plan. If I thought I knew the current and voltage in the LED active junction, and the parasitic capacitance was consistent unit to unit, I'd look at an R-C speed-up network. This could just be a few pFd bypassing your emitter resistor. No, that leads to "infinite" current in the spike, which is bad for bond-wires even if the spikes are short. This consideration leads to an R-C-R network; layout values left as a problem for the student.

And as bcarso says, you don't want your "collector.. running into the emitter". When that happens, the BJT junctions "flood" with electrons (or holes) and it will take a lot of time to drain them dry and back under control again.

I don't understand your input "bias" resistors. The input is not a linear amp, it takes what it gets, which should be huge swings. If you capacitor-couple a pulse-train, your average level depends on the pulse train duty cycle, which is usually a bad plan.

Don't overlook the trick of flowing full current through the LED, and shorting it to produce "dark". This is wasteful, but throws far less garbage into supply/ground rails. Pulling down to the 0.1V of a saturated BJT will force you to charge-up to threshold again, but you could short to a 1.5V reference and have much less far to rise.

 

[bcarso]

Ian, if you go to the website for the LED manufacturer you will see another faster diode they are selling, and with it another suggested drive circuit with a somewhat smaller series capacitor speedup circuit. But what is interesting is that for this diode, the IF-E99, there is actually a spec for capacitance: 10pF. So one could conjecture that the capacitance of the one you are to use is a bit larger and maybe its series R larger as well, to correspond to the ratio of rise/fall times (8ns for yours, 3ns for the faster one).

This might help your modeling a bit.

 

[Ian MacGregor]

[quote author="PRR"]>
Don't overlook the trick of flowing full current through the LED, and shorting it to produce "dark". This is wasteful, but throws far less garbage into supply/ground rails. Pulling down to the 0.1V of a saturated BJT will force you to charge-up to threshold again, but you could short to a 1.5V reference and have much less far to rise.[/quote]

I breadboarded up a quick version of this circuit using a standard LED, a 2N3904 and a 180 ohm resistor to limit current in the LED. As expected, it works just fine, but I am saturating the BJT. I tried putting a diode-connected 2N3904 from base to collector of the switching transistor, but that didn't change anything. I asked my professor about this idea and he said it was fine as long as the BJT doesn't saturate. He mentioned that fast BJT logic circuits have Schottky diodes from base to collector for this reason. I'm unfamiliar enough with Schottky diodes to not be able to design this right away. I'm gonna go read up about them in AOE tho.

Any help on how to keep the transistor from saturating?

Ian

 

[Svart]

I guess you *have* to use BJTs?

 

[Ian MacGregor]

We do not have to use BJTs, however, all I have on hand at the moment are BC184s (hehe.. Neve) and 2N3904/6s. If I understand correctly, MOSFETs do not have the saturation storage time problem that BJT's do?? So maybe we could just use a FET to short the LED??

Yeah, so to answer your questio, we can use anything as long as we can get ahold of it relatively soon (due date coming up!!)

Ian

 

[bcarso]

Ian, to prevent the 3904 from saturating you can use a small schottky from collector to base. The forward drop of the schottky is about 400mV so keps the collector of the BJT from getting much lower than 300mV (Vbe - schottky forward drop).

Actually, I like the current drive o.k. but I think I might use a voltage assist drive from a 3904/3906 complementary emitter follower, coupled in with a capacitor to the LED, in parallel with the i source drive. However, the polarity is conflicting with your transconductance stage so an inversion of the logic output (a fast one!) will be needed for one or the other drive signals.

The problem you are having with saturation is also a consequence of driving the base of your transconductance stage with the full 5 or 6 volt swing of your logic circuit. This needs to be cut down to size and the emitter resistor adjusted downward accordingly. A stiffish voltage divider should be adequate, maybe with a little speedup cap across the input resistor. A two-to-one division sounds about right. Then you will have for a 5V drive 2.5V at the base, about 1.85V across the emitter R, thus requiring about 180 ohms for 10mA. Then you will not need the schottky.

 

[Svart]

Ian, yes that is essentially correct for what you need here. the gate being voltage driven allows this to switch much faster than a BJT,in the nanosecond range on average, for many reasons. I haven't looked at the specs of your LED but I'm thinking this would be more than fast enough to be limited by the speed of the LED itself. As with my above posts, your turn ON speed is limited by your PSU and flywheel diode recovery time, again use a big PSU with plenty of capacitance and look at some of IRF's FRED type diodes for the fastest ones availiable( if you want a flywheel setup, you probably don't need it for this test). Your turn off speed is limited by the gate drive circuit. use a reversed biased diode in parallel with the gate resistor to speed that up a bit.

read up on some of the papers at IRF's website which go into much more detail than we can go into here.

I vote for a FET switch for many reasons, but do your research and decide which one is for you.

Check out the logic level FETs for some that can be driven directly from logic( although other ways might be a bit better)

good luck! :thumb:

 

[Ian MacGregor]

Svart,
I'm a little confused in regards to the IRF/flywheel diodes. Are these power supply elements?? For this lab, we are just using the bench supplies available in the lab. Also, are you suggesting that I use a FET in place of the BJT in the circuit I posted, or as PRR suggested, where the FET shorts out the LED?

Thanks,
Ian

 

[Svart]

The flywheel diode is a diode that allows reverse currents to flow through the FET. Most MOSFETs have them on-die. I doubt that will be a critical part of the design you are doing, but it's something to keep in mind for real-world design.

I would say to look at the MOSFET as the actual switching element. hook your LED anode to your current limit resistor which is hooked to your high rail, the MOSFET then hooks up to the cathode of the LED via the drain, the source to GND and control your gate with logic. if you control your gate with logic, make sure your logic pulls up AND down or we might need a 10k resistor to ground on the gate to keep it from floating up and turning the device on.

a quick look found this:
http://rocky.digikey.com/WebLib/Fairchild/Web%20Data/FDV301N.pdf

which ought to work nicely if the voltage is within specs.

:thumb:

 

[Ian MacGregor]

Like this: http://groupdiy.twin-x.com/albums/userpics/10021/FET_Switch.pdf

:?:

I think the symbol for the MOSFET is for a p-channel, I am using N-Channel. I did end up using a 100k resistor to tie the gate to ground. The logic input will switch from 0 to 6V.

I have breadboarded this circuit and it works when I touch 6V to the gate. Now I need to find out if it will work at high frequencies. Just to give an idea of the speed, our data generator works at a max of 27MHz, so we would like the transmitter to work up to that speed (if possible).

My main questions is: How do I analyze the transfer function of this circuit? More importantly, how do I model the I of the diode? Do I use the exponential diode equation as well as the intrinsic R and C of the diode??

Thanks,
Ian

 

[Svart]

that would be the circuit!

the FET you want to use is Nchannel, the one you have shown there is also N channel.

Depending on the FET you choose, It will likely work at those speeds.. what are the outputs of your logic? BJT or mosfet?

the MOSFET you have modelled does not have a flywheel diode. the specs of the LED might have the *real* info you need on that diode.

:thumb:

 

[Ian MacGregor]

The output of the logic is from a MM74HC74A D-flop, which is CMOS (I think??). We should be fine with our input impedence being around 100k, right?

We are gonna go into the lab tonight and try the circuit with the actual LED we must use.

Ian