Spin-off topic - Constant Current Sources

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TomWaterman

Well-known member
Joined
Jun 4, 2004
Messages
1,151
Location
The Shire, UK
Hi peeps,

I've been looking at constant current sources since the little (but very helpful) discussion amongst the 'in another forum' thread.

I'm looking to CCS the outputs of a OPA627 for a headphone amp, and an OPA604 for the mono-summing circuit NYD kindly provided.

I've been reading H+H, the web and a paper from Siliconix AN103.

http://www.vishay.com/document/70596/70596.pdf

I have a few Q's - wondering if Brad, Paul, Feathery, whoever could shed some light.

Basically, its clear to me now that there are issues with the differing types of CCS.

I've narrowed it down to the JFET Cascode or a BJT one with the diodes as suggested in the other thread.

H+H seems to suggest that the BJT CCS will provide the most stable performance but I like the somewhat more simple JFET Cascode (fewer parts).

What would be the real advantage to the BJT type CCS as opposed to the JFET?

I was trying to calculate a source resistor for a 2N5457 NPN Jfet which is commonly available at Farnell and suitable for audio apps.

After going thru H+H, AN103 and 5457 datasheet I'm kind of at a loss.

Basically a couple of questions come up and I've realised I'm a retard!

Here's one circuits' scenario where I'd like to add a CCS.

Between the first stage OPA627 and the 2nd stage OPA627 in a phones amp. There's a 10k attenuator in between the place for the CCS and the 2nd stage.

The input Z of the next section is set to 100k at the mo.
So I calculated the input Z of the next stage + the attenuator, 1/(1/10k+1/100k)=9091ohms.

Lets say going by Pauls example of no lower than 0.5-0.75mA based upon max output (to be well covered on +/-18V rails) 9V pk....

I=9/9091
I=1mA+another 1mA to be safe so 2mA.

Something was mentioned about including the feedback network (of the 1st stage right?). My feedback resistor was going to be 820R....how do I calculate that as well? If I add it into the parallel load it brings it way down and doesn't seem right??

In the AN103 document it claims that the source resistor for the JFET is calculated by Vds/Id.

I'm lost as to where to get Vds from. So I'm kinda stuck knowing how to get the CCS to have an Id(?) of 2mA.

There is another formula which is:

-VGsoff/Id(1-root(Id/Idss))
which gave me an Rs of 275ohms but I had to guess a value for VGsoff as the datasheet says it varies from -0.5 to 6.0V. I chose -3V.

Ooooh another Q? - is the CCS actually more of a current sink than a source?

Thanks for any help explaining to a moron.
Cheers Tom
 
Tom

Here is the practical route from someone even more retarded. Grab a N-Channel JFET of your choice and hook it up as a current source. Power it with a 9V battery and start with a 100 ohm source resistor. Adjust the value of resistor until you get 2ma or whatever you want. Slap this little contraption onto the output of the opamp. Make sure to listen to it with and without the current source.

Cheers,
Tamas
 
The appeal of bipolar current sources for me is the predictability. Vbe is always reliably tight compared to Vgs no matter how much you pay for the manufacturer to select the FET.

If you really want to use the FETs then you can make an Idss measurement and a pinchoff voltage measurement for each part and then go back in to the basic FET equation, which works pretty well for long-channel FETs and not too well for short-channel ones. But like Tamas says, for getting moving just grab a part and play with it.

The application of the current source/sink is an issue here as well. If you are trying to do precise, temp-stable, and ultra high Z, then you will wind up using FETs somewhere in the end, in conjunction with a bunch of other parts. This is partially because bipolars run into a limitation on output Z due to base width modulation that sets an upper limit to output Z. If you use a darlington connection this can be made much higher, but at some point a FET is indicated as a final cascaded element. Then you can get output Z's in the 10^12 ohm range, so that the impedance will be strongly dominated even at low frequencies by the output capacitance. 3pF and a teraohm is a tau of 3 seconds!

But for just doing a job loading outputs of op amps to enforce class A operation over a wider voltage swing range without kicking the loop gain in the teeth, a simple or fairly simple circuit should be fine. The bipolar version will take a few extra parts but for the two-diode bias one, it will swing down to nearly just one diode drop above/below the particular rail, and in that sense will outperform even the lowest pinchoff voltage FET. Or for a more temp stable one use a GaAsP "standard red" LED, which has a pretty close to Si diode tempco but gives you a net 0.9V or so to work with for the emitter resistor that sets the current.

One of the things I like to use JFET current sources for is very low noise voltage references. There is a nearly zero tempco bias point for JFETs with pinchoff voltages more than about 0.7V. So if you put the resistor of the right value in the source and ground the gate you have a pretty stable current source with temperature. Use this to develop a voltage across a resistor and buffer that and you have a quieter voltage ref than anybody, if it's a good FET to begin with.

(although having said that I've been thinking about rolling my own discrete bandgap ref with these super low rbb bipolars....)
 
> including the feedback network (of the 1st stage right?). My feedback resistor was going to be 820R....how do I calculate that as well? If I add it into the parallel load it brings it way down and doesn't seem right??

That amp has to drive that feedback network; nobody else is going to do the job for it.

Actually the combination of both feedback resistors and the amp input impedance, but for gain of more than a little the shunt resistor is small so the total is roughly the series resistor.

Counting on thumbs, you must be driving 700-800 ohms. (BTW: why do you need such a low value feedback network?)

Taking your 9V peak as true, then you need over 10mA peak. If you are trying to force Class A bias, that's what your current source must supply. Oh, and you have over 180mW of heat in each channel; in a 2-fer opamp chip, that's hot.

> how to get the CCS to have an Id(?) of 2mA.

Get an FET and a lot of resistors (or a trim-pot). That AN103 rule applies, approximately, when desired current is much-much less than zero-bias current Idss. But 10mA is not much-much less than Idss of common small JFETs (it is more than many will pass with no resistor).

> There is another formula

This is more correct. The simple form is good-enuff when Id/Idss is very-small; i.e. when you want much less current than the FET will pass at zero-bias.

> I'm lost as to where to get Vds from

The specific part datasheet will show Vds, but only as a wide range like 2V-6V (or yes, 0.5V-6V). They can't control it except very roughly. They can sort it, but you pay more for that, and even so you can't get a tight spec on Vgs. If you actually care how much current the FET passes, you either have to trim a resistor or add a large voltage to the gate so Vgs variations seem small.

> is the CCS actually more of a current sink than a source?

No, no, no!!!! It is NEITHER! It is a Current Limiter. A current source is like a series-field generator made to run series arc-lamps, and you'll never see one of those. In modern electronics "current source/sink" is shorthand for "current limiter". And until you show me an electron I can see, "source" or "sink" is meaningless except to keep your math signs right in calculations.

> will provide the most stable performance

So what do you gain from stability? If you are talking a few mA, it hardly matters if it is 1mA or 3mA. Ideally you want to be sure it never drops so low that signal peaks shift to Class AB, but if the amp is even slightly audio-friendly then the occasional AB peak won't be audible.

You may however be in trouble at 10mA. Using your 9V ~900 ohm values, you need to never drop below 10mA to stay Class A on negative swing (assuming CCS to negative rail). But then the positive swing is 20mA: 10mA to the load, 10mA to the CSS. 20mA is hard work for some chips. If in addition your CCS varies from 10mA in January to 20mA in July, your opamp will need 30mA on positive peaks, and that can be a real strain. And if it can't do that, it clips which is grossly audible, unlike a shift to Class AB which should not be a gross event. So you want stable-enough that the amp is never worked too hard. If you have to compromise, let it shift to AB in worst-case rather than go into clipping.

I think most designers over-design their "current sinks". In many many cases, all you need is for dynamic resistance to be significantly higher than static resistance. Anything better than a resistor. Instead I find mirrored cascodes that give many-Meg dynamic resistance, at least up to a few KHz (everything has capacitance), in few-K circuits where such high resistances are just pointless. Even if parts are free, you are multiplying failure rate when you add parts. I really love (not) twin-triode long-tail diff-amps with hyper-elaborate tails: in a triode diff-amp the ideal tail impedance is actually much less than infinity, and if you have a modest negative rail a resistor works great. (Upsetting chip opamps does require something better than a resistor: either resistor to a high rail plus a protection diode, or a BJT/FET thing.)
 
Thanks a lot guys.
[quote author="PRR"]That amp has to drive that feedback network; nobody else is going to do the job for it.

Actually the combination of both feedback resistors and the amp input impedance....[/quote]
Both feedback resistors meaning the feedback of the first stage plus the amp input Z and feedback of the 2nd stage? So 1/820+1/820+1/10^13 etc etc?

It pretty obvious that the feedback network is much lower than it should be. My reasoning was as follows:

The datasheet for the 627 suggested low value feedback (2k) and mentioned it would also perform with a lower value such 820R. The headphone amps that I had been looking at for inspiration (per anders etc) seemed to use low feedback networks as well (1k etc).

I think after your nice explanation I'll have to increase it to at least 5k (more like 10k) to bring down the heat issues and loading on the amp?

[quote author="PRR"]Taking your 9V peak as true, then you need over 10mA peak. If you are trying to force Class A bias, that's what your current source must supply. Oh, and you have over 180mW of heat in each channel; in a 2-fer opamp chip, that's hot.[/quote]
Agreed that seems far too much, everything I have read, suggested that the 627 requires no more than about 2mA of bias. By upping that feedback network I can reduce that heat issue and only require 2-3mA. I chose 9V as it would be the max level the amp could produce, I would expect in reality for it to running at half that most of the time.

Thanks again PRR.

Right moving on.

It seems that when the circuit is built, the resistor is going to have to be trimmed on a 'per part' basis due to the variation in the FETs. So If i put a 5k trim pot in the schematic, I'd be good to go when breadboarding?

The way I see it, I need the CCS to be high input Z and low capacitance. How would an LED compare to the JFET? I've seen things on DIY Audio which mention good noise performance with LEDs, bypasssed with a capacitor but what about the input Z etc.

A 1k trim pot with the LED would give me the ability to trim between about 1 and 10mA, so could be used in all places, adjusted on test? Connected to the -ve rail at 18V, I would also need a dropper resistor for the LED?

I'm going to re-read through what I have got and try and make some sense of it.

Thanks again Brad, Tamas, PRR for all of your help.

One other thing, the 2nd stage OPA627 is running into a pair of paralleled BUF634s, providing a 40Meg ohm input z.

This makes it pull 0.225uA. In a situation where the OPA is running into such a high load, it hardly seems worth it to bias the output with a CCS.

Cheers Tom
 
Tom, you do understand that the LED is just in lieu of the two series diodes, to bias the base of a bipolar with an emitter resistor, right?

I have found the LEDs to be very quiet, although in a lengthy exchange among heavyweights (Jung, Curl etc.) in another forum I noticed someone said they got noisier at high currents (>>1mA).

Anyway the Z of the forward-biased LED is very similar to a standard Si diode at these currents. The circuit is so predictable that you can dispense with the potentiometer, unless you really want to vary the current a bunch.
 
Ahh right!

For some reason I have seen a CCS with just an LED - no BJT.

I think I'm still favouring the idea of the JFETs as I may just get my head around that!

Thanks again.
-Tom
 
Okay, here's how to calculate your total load. I'm assuming the opamp is noninverting; take Rinput (the resistor running from the - input to ground) and Rfeedback (the resistor running from the - input to the opamp output. Add them; that's the resistance of the feedback network. So if Rinput = 100R and Rfeedback = 820R then your total network resistance is 920R. Call that Rnetwork.

Now compute the total load: Rtotal = 1/(1/920 + 1/10,000 + 1/100000) = 835R. As others have noted, that's way low; it'll require at least 11.2mA of bias. I'd scale up the resistors around the opamp to the point where you can bias the amp at, say, 3mA. That'll keep the chip dissipation reasonable.

I generally use BJT currect sources because they're thoroughly predictable and cheap.

Peace,
Paul
 
Thanks a lot for the explanation Paul.

I will be increasing the feedback network. I wonder if Per Anders would like to explain why he uses such low values in his OPA/BUF QRV-04 headphone amp.

680R/220R+1 for a gain of 4. Can he get away with such low values as the BUF634 is inside the feedback loop, helping the amp to cope with the low loads?

Here's his amp:

http://home5.swipnet.se/~w-50719/hifi/qrv04/index.html

Do you have any suggestions for good through-hole BJTs to be used as a current source?

I'm scouring Farnell but am unsure of what to choose. Would like to consider the BJT CSS with LED as well as the JFET one.

Cheers Tom
 
"Can he get away with such low values as the BUF634 is inside the feedback loop, helping the amp to cope with the low loads?"

Yes, that is exactly correct---the BUF634 doesn't load the op amp preceding much at all.

However, if the equivalent load as calculated by Paul S. gets low enough, you will reach the same point with the BUF634, since it is also a class AB output part. It's so fast though that the transition through those regions is gong to be over in a hurry at any audio frequencies. But you can see where this stuff could get into an infinite regress situation.*

Someimtes I work my op amps pretty hard in order to trade off lower noise with potential crossover and other distortion. The gains in those situations are usually pretty low so I don't get into too much trouble. But many people don't realize how high the open-loop output Z of a typical o.k. op amp is. Often the structure is just a complementary emitter-follower fed from a fairly high-Z node, and the open-loop Z is roughly the high-Z node divided by the beta(s) of the output follower Q's. As an example, if I have a simple common-emitter stage and a current-source pullup driving a comp-sym e-follower output, the high-Z node might be 150k, and the output Z is then about 500 ohms for betas of 300. It's only with feedback that we get the circuit output Z's way lower.

In the case of the 990 otoh, the miller C around the second stage helps to lower the output Z of that node, so the open-loop output Z is lower at frequencies where the miller cap is working. You can also use a more complicated output buffer, like the diamond circuit, or even with FETs ahead of the bipolars so the high-Z node is not loaded much except by the capacitances.



*If anyone has a tape of the Conversations with Hoffnung he/she may recall his story of buying a penknife, which he was unable to open because he couldn't get his nail into the little slot. So the man sold him another penknife to open the first one with...
 
"Do you have any suggestions for good through-hole BJTs to be used as a current source? "

Virtually any small-signal device will do. Just make sure the breakdown voltage Vcb is reasonable, i.e. greater than your supply rails (hence >30V for +/-15V rails) and that you stay well within the dissipation limits. For an NPN this would occur at the maximum poisitve output swing---if you were able to swing all the way to the rails for this example the transistor would see almost 30V, and if pulling 10mA would dissipate almost 300mW at that point. So the lowly 2N3904 would be adequate, for example.
 
Thanks again guys.

I think I am getting somewhere.

Paul, how does the load calculation change when the opamp is inverting?
The phones amp is non-inv but the mono summing system is virtual earth and inverted amps are used.

Brad, number 10 on this CCS page is the LED set-up you mentioned. Can this just be reversed for connecting to -V, or is it best to connect to +V rail?

Test.gif

Cheers Tom
 
If what you want is to pull current to the negative rail, just use an NPN in a mirror image of the circuit shown. LED cathode (flat line) goes to the negative rail. Emitter resistor goes to the negative rail. LED anode/NPN base/other resistor tied together. The free end of that resistor goes to the positive rail (or ground if you like). Tie the collector to your load, in this case the op amp output. Run a mA or so through the LED, making the LED/base resistor about 27k for going to the +rail and 13k for going to ground, for +/- 15V rails. The difference between the LED voltage (for GaAsP red about 1.65 volts or so) and the base-emitter voltage of the transistor is what the emitter resistor sees, so the current is about (1.0V)/R. Note that the collector has to go somewhere more positive than the emitter for the circuit to work, so if you are testing it and adjusting the emitter R for the desired current, you might want to put a 100 ohm R in series with the collector to ground or +V, and read the voltage across it to see what current you are getting.

You could use the current measuring function of your DMM as such, but I avoid this when I can because people forget that they plugged the test leads into the amps input, and then go to measure a voltage later and blow the amps fuse. I'd say half or more of the DMMs in labs have blown fuses.
 
Thanks Brad.

Sorry to be asking so many questions!!

Would there be any advantage to running the base resistor to the +Ve rail as opposed to ground?

-Tom
 
A slight one: the ratio of the resistor to the LED impedance will be larger and thus the modulation of current by the difference between the + rail and the - rail might be smaller.

However, if the variations in rail voltages are equal in magnitude but opposite in sign, as might be the case if they are due to mains voltage variations and similar voltage regulator performance on each rail, then the variation is nearly doubled and the advantage almost evaporates.

And of course you're dissipating twice the power, although this is still small.

There's a second order effect associated with the LED temp and forward voltage but it's really small.

For your purposes for biasing output stages, none of these are of any practical significance.
 
Cool!

Great info to know for future reference.

You really are a great Lab member Brad, very helpful.

I think I'll go with the resistor to ground.

-Tom
 
Hehe, well it might be something to consider for sure.

When I get around to building this sucker I'll try it out! :green:
What would you suggest to cover it, paint?

-Tom
 

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