What do you propose to deconvolve in this situation? If you lowpass the data, you just kill bandwidth, which is hard to come by here. If you try to extend the bandwidth, you magnify the noise... a linear transfer function can't create precision where it doesn't exist.
Deconvolution is one of the family of "ill-posed problems", in the sense that, given an imperfect signal A, and an ideal signal B, what is the transfer function T that satisfies
B = A ** T
where ** is convolution.
Forests have died making papers about this problem.
The obvious Fourier solution (f.transform A and B, divide fB/fA, reverse transform) tends to have singularities (ie, zero lines in A) or bad noise problems (near-zero ditto). This ain't turf for the faint of heart, or wikipedia bs artists, either.
I've developed an iterative time-domain method, and it's fun to play with. A little iterating is pretty good, but if you get greedy and try to equalize too well, namely insist on making B perfect, it blows up.
Hi Winfield - my response is so slow that you may miss it altogeteher
- but I finally had time to work through your circuits. (getting stuck in the airport for a day and a half really frees you up!). To me, it looks like the 200 ohm resistor on the emitter of the transistor should be a 210 ohm resistor. Otherwise, by my maths, you get a constant (assuming the HV supply is constant) offset as well as a vout not quite equal to 500*ILoad. It looks to me like you are assuming that the current through the shunt is almost entirely the load current, which at the low currents we're talking about is not entirely accurate, as the resistive divider on the non-inverting terminal of the LT1783 draws about a quarter of a ma through the shunt.
By the way, if I wanted much higher bandwidth (say 100MHz) - do you think this circuit might still be able to work? Certainly the the LT1783 would need to be swapped out. Can you tell me why exactly you chose that part? I don't see why an over-the-top op-amp is needed.
Yes, you are a little slow at reading the postings! You are correct, or at least partially so. In earlier postings, I caught my error and changed the other resistor to 190 ohms. That's the way to preserve a 1/20 mirror gain while keeping a zero dc-offset from the bias current.
It was a convenient sot-23 low-current over-1MHz rail-rail part. Hopefully there are better ones.
As for higher bandwidths, yes, certainly you can, with lower resistor values, and at higher mirror currents. But at frequencies well below 100MHz the sense resistor's inductance, etc., will be a big problem. For moderately- high RF-frequencies I suggest trying a current transformer.
Hello again Winfield - This circuit was a little tricky for me to understand at first, so then I just started making assumptions to simplify things and I think I figured out what you were going for, but I was hoping you might be able to address a couple questions/worries I have about it:
It looks like you are assuming that the base to emitter voltages of the two matched transistors are the same. This would require the base currents to be the same, and the collector to emitter voltages to be the same. This clearly isn't the case, however. Am I wrong in assuming that you are treating the VBEs of the two matched transistors to be the same? Or is the error small enough that the approximation will work?
I think the 2N3906 is being used to protect the left matched transistor from the early effect by providing a fairly constant VCE across that transistor. It would also provide a path for current to flow from the matched transistors to ground. Is that right?
Another assumption I think you made was that the base currents were all small enough to ignore. This seems fairly safe to me, but I just want to be sure that I'm seeing things correctly.
The MPSA92 just serves to drop the voltage down to ground, correct? How is it able to regulate the current through it? It seems to me that it'll have a relatively constant VBE and VCE. This really confuses me.
How did you choose the values of the 270K, 1K, and 4.02K resistors? I understand that the 1K and 4.02K resistors needed to be chosen with a specific ratio so as to provide a specific current divider, but why the magnitudes that you chose? It looks like the left matched transistor will have .2-2.2ma running through it, while the right matched transistor will have about 1ma through it, so did you just aim to place the current of the right matched transistor to be in the middle of the range of currents going through the left matched transistor, so as to keep their VBEs as close as possible?
I'm getting the final output to look like: (2.49K/4.99) * ILoad - (1/4.99-1/5.02)*IQ2 (where IQ2 is the current through the right matched transistor, and should be about 1ma, thus making that second term approximately 1uV and thus negligible). This is all from the assumptions mentioned above, thus this is why I think my assumption about your assumptions is probably right.
Sorry for bugging you with so many questions - this circuit just left me a bit more confused than the other.
Yes, the circuit suffers from errors from the Vbe dependence on transistor current. Let's see. At 10mA full-scale load current, we have 11mA through the 10 ohms and 2.2mA though the 50 ohms, with 0.2mA worth subtracted out by the second opamp. The Q1 and Q2 transistors are operating at 2.2mA and 1mA, for an offset error of about 20mV or -20%.
At zero current load transistor Q1 would nominally operate at 1mA 10/50 = 0.2mA, ignoring the Vbe issue, which adds an offset of about 40mV = +40% of full scale. Hmm, doesn't look too good.
Yes and yes.
That's the common-base connection, where Ic = Ie, minus 1 to 2% for the base current.
That's a 1/5 ratio, matching the 10/50 ohms up top, to subtract out the effect of the standing 1mA bias current.
Yes, wishful thinking apparently.
Well, I think you were chipping away at the weak foundation of the matched-voltage emitter-input differential amplifier, and my only defense is the good performance of the opamp circuit I posted later, which you replied to in your post.
Ah, I missed that! The 190 ohm change is a much better change.
Current transormers are more for measuring AC, not DC, correct? I should have mentioned more about my application: I'm looking to be able to analyze the current through an avalanche photodiode that is receiving really sharp pulses of light. Specifically, my plan is to have a very high speed comparator on the output of the current sensor circuit, as well as an integrator and a peak detector. Thus I think a current transormer would not work very well, right? Is there a good way to handle these really high speeds? I'm planning on the pulses having rise and fall times in the single digit (double digit worst case) picoseonds, and lengths of probably a nanosecond or two. Thanks,
How did you calculate these values? Wouldn't that value depend on the Is of the transistor chosen?
Interesting - I was not familiar with that type of amplifier. Looks like I need to do a bit of reading!
To be honest, I prefer the idea behind this circuit. I've always preferred circuits that you can do without op-amps, and I really liked how you connected the two bases together. I think it's because there's this tendency these days just to connect black boxes, and I'm always trying to avoid that tendency. It's unfortunate that it won't work perfectly, though.
To follow up on my post from a few minutes ago - is the idea of a common base amplifier that VBE is high enough that the transistor will push as much current through itself as possible, but VCE is kept high enough to keep it active so that the base current stays minimal? Looks like hFE is around 80 at 25=B0C with an Ic from 0-10ma, so 1-2% current loss to the base would make sense. Thanks,
No. It's easy with two identical transistors, you can take ratios and get a simple equation, dV = kT/q ln Ix/Iy. See AoE page 91. Figure 2.53 plots dV vs current ratios.
[ snip ]
It's not hard to fix the circuit's Vbe-vs-Ic problem, if we allow ourselves one more transistor, in order to force both Ic's equal, as I should have done in the first place. You think about it, and I'll try to find time to post the answer later today or tomorrow.
The goals for our circuit are to present a 0-5V output at ground, for a 0-10mA current on a +250V rail, with a 1MHz or greater bandwidth, to an accuracy of say 2%.
Here's my original poor circuit, with very low gain and a 60mV input error required to get its full-scale output, With its high errors, this circuit met only two goals.
To reduce the input error below say 1%, or 1mV, both of the input transistors must always operate at the same current, to within e^dV/25mV = 1.04, or 4% change for a full-scale output swing. So we need more gain in the differential stage. As before, it must operate from signals at the positive rail, and it must be pre-biased, to have more-than 1MHz bandwidth at zero input signal.
We'll split our bias current into two equal parts, using new transistors Q3 Q4, to feed input transistors Q1 Q2. Except for the base current of Q5, the high-voltage PNP output transistor, Q1 and Q2 operate at equal collector currents of about 450uA each (assuming similar betas). The Q5 current ranges from 450uA, to insure high speed, to 1.45mA at full scale. A 100k collector resistor holds its power dissipation down to 150mW.
. +250V Rs 10.0 Iout . IN ----+---/\\/\\----+------o ---> 0 to 10mA load . | | . R1 R2 . 100 190 all resistors 1% . | | R1 = 10Rs . ,-------+ | R2 = 2R1 - Rs . | | | . | e e . | Q1 b --+-- b Q2 Q1 Q2 matched pnp . | c | c pnp Q3 Q4 matched npn . e | | | low-voltage types . b ----+ +-----+ . c Q5 | | | . | mps c Q3 | c Q4 . | A92 b --+-- b npn . | e e . | | | . | R3 R4 10k 1% . | |___________| . | | . \\ Ib/2 | 270k 4.99k . / Ib '--/\\/\\---+--/\\/\\--- gnd . \\ 100k 0.5W | . | 0.5W 5% \\ 4.99k . | / . | 4.99k \\ . | ,--/\\/\\---, | 4.99k . | | __ | +--/\\/\\---, 0 to +5V . '---+--|- \\ | 4.99k | __ | out, for . | >--+---/\\/\\--+--|- \\ | 0 to 10mA . gnd ---|+_/ | >--+---- . gnd ---|+_/
If Q5's beta is more than 100 (the datasheet plot says 150), then Q1's change in Ic is under 10uA, for 0 to full scale, or half of our 2% budget. So this circuit should fix the problem.
Hi Winfield - you tripped me up when you said it could be done with one more transistor. I was trying to figure out a way to add a transistor to your original circuit, instead of making a fairly different one!
Anyways - I just ran through the numbers and I think I understand most everything that is going on in your circuit, but I'm still confused about a couple things:
First of all, R3 would be 10K, correct? I didn't see it marked, but otherwise things wouldn't make sense.
Q5's operation is a bit strange to me. It really confused me until I decided to just assume that the emitter currents in Q1 and Q2 were the same and that Q5 would get the remainder of the current flowing through R1. So that tells me that Q5 is acting as a current buffer, so once again the MPSA92 is in the common base configuration. The common base configuration really confuses me... Is there a section in AOE that goes over it? I couldn't find anything about it. As far as I can tell, however, the idea of the common base configuration is to keep VBE large enough that the transistor attempts to push as much current through the collector as possible. Is that at all accurate?
Lastly, could you please explain the feedback loop that you are using? I must admit that I'm having trouble seeing why the circuit you posted earlier today would not work while this one would.
Well, imagine a current source feeding an emitter. If the compliance voltage is high enough to create more than Vbe across the base-emitter junction, it'll carry all of the current-source's current. But we know the transistor only needs 1/beta of that for the base; the rest, almost equal to the original emitter current, comes from the collector. That fraction, called alpha, is nearly equal to 1.
One way: you can follow each element in the feedback path, looking for an extra inverted element after adding them up. Another way, imagine the current in the output transistor Q5 is too high. Voltage across R1 too high. What happens to the rest of the circuit, does it increase or decrease Q5's current?
Imagine, if Q5's current is too high, the voltage across R1 is larger (more negative) than the R2 - Q2e node, and Q2's emitter voltage is more negative than it should be, and Q2 takes more current from R2 than it should, and this pulls the Q4-Q2 collector node toward Q2, tending to shutoff Q5, and reverse the process. That's negative feedback. You can see this quickly, by observing that every stage is a follower, or noninverting -- except the Q2 inverting stage that gives us negative feedback.
So it looks to me like the base of Q5 can go from about .2V below it's emitter to about 1V below it's emitter, assuming Q1-Q4 always stay active. So then the feedback mechanism is in place to regulate that and keep the proper amount of current flowing through Q5 by changing Q1's VBE. Your description of the feedback mechanism makes sense, but I'm still trying to figure out this common base configuration. So for a common base amplifier you have to have some sort of feedback system in place to keep VBE within the proper range, right? I mean without feedback it seems like you could easily have VBE slightly too large and have all the current flow from E to B.
I guess this all makes sense in a hand wavey way... but going from understanding it to designing it... I still need alot of work there.
Ebers-Moll tells us that Vbe relates not to Ie directly, but instead to Ic. So what happens in a common-base amplifier is that we present a current Ie to the emitter, and the transistor adjusts its own Vbe, so that Ic = Ie - Ic/beta, or Ie minus the base current.
So to use the common-base configuration, all we have to do is present the emitter current, and collect the collector current. :-)
But my emitter-coupled differential amplifier works a little differently, as you can see. Q1 sets up the base voltage for itself and Q2, and forces Q2's emitter to follow it. In the original "defective" version, I collected the resulting collector current, ignoring the big change in Vbe over the amplifier's range. In the new version, we collect the current that's left over, after the differential amplifier's feedback loop responds to the forcing currents from Q3 and Q4.
But you also have to present some way for VBE to change, right? You can't have the base and the emitter held at specific voltages, correct? Because otherwise, say you presented the transistor with a VBE of .8V, it seems to me that all current would flow from B to E and probably just bypass the collector.
Why would Q2's emitter have to follow the base of Q2? Maybe that is where I am getting confused... To me it looks like the base of Q5 is not set at a specific voltage - more it looks like it can be anywhere from about .2V below Q5's emitter to a couple VBEs below Q5's emitter.
By the way - do you have any idea what sort of frequency this circuit could operate to? My goal is to measure and analyze the current through an avalanche photodiode. It will have pulses of current though it that should have rise and fall times in the single digit pico seconds, and durations in the single digit nanoseconds. Ideally I'd like to have this circuit's output follow that very closely. If there is a set delay in the output that is fine as I can fix that elsewhere. Do you think that a circuit like this would be a good fit for my application?
Thanks, and sorry I'm being so thick skulled about this. Some days I wonder if I'd be better off just sticking to soldering all day.
I copied the drawing below, for reference. The current through Q3 (which is equal to Q4 and is half the bias current), goes through Q1 and sets its Vbe voltage. Q2's base is tied to Q1's so, assuming they have equal current, Q2's emitter voltage has to equal Q1's emitter voltage. Once we know that we can start evaluating the currents, to see what they do. Q2's collector is especially interesting because if Q1 and Q2 don't have the same current, as programmed by Q3 and Q4, then Q5's base changes to control the current Q5 takes from R1, and bring Q1 and Q2 in balance.
. +250V Rs 10.0 Iout . IN ----+---/\\/\\----+------o ---> 0 to 10mA load . | | . R1 R2 . 100 190 all resistors 1% . | | R1 = 10Rs . ,-------+ | R2 = 2R1 - Rs . | | | . | e e . | Q1 b --+-- b Q2 Q1 Q2 matched pnp . | c | c pnp Q3 Q4 matched npn . | | | | low-voltagehigh-gain . e +-----' | . b --- | ----------+ . c Q5 +-----, | . | mps | | | . | A92 c Q3 | c Q4 . | b --+-- b npn . | e e . | | | . | R3 R4 10k 1% . | |___________| . | | . \\ Ib/2 | 270k 4.99k . / Ib '--/\\/\\---+--/\\/\\--- gnd . \\ 100k 0.5W | . | 0.5W 5% \\ 4.99k . | / . | 4.99k \\ . | ,--/\\/\\---, | 4.99k . | | __ | +--/\\/\\---, 0 to +5V . '---+--|- \\ | 4.99k | __ | out, for . | >--+---/\\/\\--+--|- \\ | 0 to 10mA . gnd ---|+_/ | >--+---- . gnd ---|+_/
OK, I'm curious too. I'll run a spice model.
No. I don't think you're allowed to let the APD voltage change much, once you've found the magic value. I allowed 100mV fs, which might be too much. You could use a transimpedance amplifier, sitting at 250V, but what you really want is a Tektronix CT-1 AC Current Probe, like this one,
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You just run the wire through the little hole in the 3rd photo, and you're all set! Measure away, with your 2GHz 50-ohm scope! Ahem!
Excellent! Pretty soon we can enroll you as an apprentice IC designer... you put your finger right on it... balance... balance... balance... match... match... match ;-)
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
America: Land of the Free, Because of the Brave
Could you send me the spice file? (or post it here - but not on ABSE as I don't have access) I'd love to play around with it: viewing circuits in spice really helps me to understand them.
Well, the current gain of the APD decreases as you decrease the voltage across it. Take for example:
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- figure 2-4 and 2-5 show gain vs. voltage. I guess at the higher reverse voltages (where I will probably be running it) the gain is quite sensitive to changes in that voltage. Then again, the main thing I'm looking for is just an on/off pulse, so it's not the end of the world if the gain fluctuates some. Just how much gain fluctuation is allowable is just impossible to estimate at this point, though.
Ah phooie - here all I have is a 60GHz scope. Err, hmmmm... make that MHz. My trusty old Tek 2215A, to be exact :)
So - allow me to go into my plan in a bit more detail: My plan is to power everything off of a single 5V supply. The analog supply (probably +-12V) will be generated with normal boost and inverting DC/ DC converters. The 250V supply will be generated with a non-isolated flyback supply. It'll be held at something like 25V above the APD supply. An op-amp driving a couple FETs will actively regulate the voltage across the APD, with the idea of eliminating, or at least reducing, the noise from the flyback supply. In theory, that could keep the voltage across the APD at a constant, no matter what current was flowing. In reality, of course, I'm not sure if it's possible to get the bandwidth of that circuit high enough. The design I'm working on looks alot like Figure 6.47 "High-voltage regulated supply" from AOE, but I'm seeing bandwidth in possibly the single digit MHz region at best. It still needs alot of work to be sure - I'm still at a pretty early stage on it.
So the Tek current probe is just a current transformer, right? Are there plain current transformers on the market with similar performance? I'm aiming to keep parts cost for this entire project at around the cost of a single one of those guys (~$500 new), as I'd like to make a bunch of them. Also, what would the output of such a circuit look like when looking at the current through my APD? I had thought that current transformers were more for periodic signals, while the signals through the APD will not be periodic.
To me what might make sense is to float a high speed instrumentation amplifier with a ground at say 250-10V using a zener. Use that to watch the voltage across a very small shunt, one that might produce
50mV or so at peak current. I don't believe a drop that small will affect the APD in any significant manner. Run the output of the instrumentation amp into an op-amp acting as a transconductance amp that puts the current into say a current buffer like what you've done with the MPSA92, then take that current and run it through a transimpedance amp to generate a voltage.
Does that make sense at all? That's alot of black boxes, which always scares me... Also, it seems most instrumentation amplifiers top out in the single or double digit MHz region, which probably isn't enough.
The reason I wanted current sensing to be high side was so that I could use the APD side of the shunt resistor as the feedback for my high voltage regulator circuit. But if I could use a small enough shunt, I think it'd be just fine for it to be low side, which would make things easier I would think as I could just use a non-inverting amplifier to get the voltage up to something reasonable. Maybe that would be a more reasonable path to take?
I'm going to cut myself off now before my post becomes novel length. Hopefully it isn't all completely insane.
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