questions about constant-current source

Winfield Hill wrote: > Innocent men are being held at Guantanamo Bay as our prisoners, > according to our Military's tribunals. Our right to know this fact, > and the men's right to tell it, both guaranteed through habeas > corpus, is being taken away by the Bush administration, through a > last minute amendment to the Military Authorization Bill, brought up > on the floor of the Senate without committee deliberations, and > virtually no advance warning. Sheesh! I am so ashamed of our > government. It's extremely distressing.

Habeus Corpus, which derives from the tyranny of the English kings' and their propensity for indefinite detention of dissidents, means a writ, issued by a Court of *jurisdiction*, requiring the government to demonstrate legal cause to "hold the body." Since Guatanamo is offshore, government activities there are not under the *jurisdiction* of any US court, which means among other things, no writ can be issued and no legal cause for the detention is needed. You have no right to publication of information about government activity that is guaranteed proprietary status under much higher Constitutional authority, and the accused have no rights whatsoever.

Reply to
eehinjor
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Something is wrong, your listed MOSFETs are p-channel devices, so the gate voltage should be a volt or so more negative than the source.

Same story, something wrong.

That's the right direction, closer to a good gate-voltage value. If Vs says it's at 7.00V, this means your 5k programming resistor has 5.00V across it, and must have 1.00mA through it, if your circuit is exactly as shown above.

I wonder if you have the right MOSFET installed, if it's wired in correctly, or of it's damaged? I'd also wonder about opamp oscillation, but your 470pF should make that nice and stable. What kind of opamp are you using?

What's the story with Q2?

--
 Thanks,
    - Win
Reply to
Winfield Hill

That's an interesting idea, adding a pair of opamp input resistors. They shouldn't be necessary in a simple circuit, but we don't know how eehinjor changes his resistor values, what kind of wiring and panel switch, etc., he's using. These need to be isolated from the opamp's high-frequency feedback capabilities to prevent oscillation.

--
 Thanks,
    - Win
Reply to
Winfield Hill

Agreed, but not a fundamental reason why the reed approach won't work if manufactured properly.

One issue might be that the need for more spacing increases the size and inductance of the reed coil, but this ought to be manageable with suitably increased drive current.

Reply to
Fred Bloggs

Thank you,Winfield Hill. I know there is something wrong with it.But the chip is all right.At first I doubted this problem,and replaced the mosfet and OP.It works as before.When the output is 100uA/10uA/1uA,it works all right.

My opamp is OPA602.

Q2 is to limit the voltage on RL below +5.5V.

Reply to
eehinjor

Because my output is 1mA/100uA/10uA/1uA,the resistor is

5K/50K/500K/5M.The multipluxer is DG409.

Fled Bloggs is right,there should be input resistors.In this ciruit Vg should be constant. The data: When Rl=0,Vg=7.01940V. When Rl=240,Vg=7.02626V. When Rl=2.5K,Vg=7.03293V. When Rl=5K,Vg=7.03323. from the data above,we can know the input voltage is influenced by the opamp.

I don't know what's wront with this circuit,the 100,470pf or others?

Reply to
eehinjor

Thank you,Robert. I am afraid I can't accept your explain. If Vds is 2V,this circuit can not work.Why it can work when the outpu is 100uA?

Reply to
eehinjor

It may be delivering the output you want, but when you take your meter off the output and look at the gate voltage, it becomes defective at that point, showing an incorrect value.

--
 Thanks,
    - Win
Reply to
Winfield Hill

The FET will be happy with 2V, hey at 1mA it'll "work" with anything more than Id*Rds(on), which is less than 10mV or so. But how did you get the 2V value, by assuming a high-value load? Or an open circuit? Presumable eehinjor tests his 1mA output with a small load resistor?

--
 Thanks,
    - Win
Reply to
Winfield Hill

Seems to me that *if* 1mA was flowing thru Rl of 5K, then the drain of the FET would be at 5V; continuing with ASSuME-tions, let the opamp be working "properly" so that the source of the FET would be at 7V. That would make for only 2V across the FET; will it actually work there? The various part numbers given do not help; they are different parts and some are obsolets.

Reply to
Robert Baer

What should I do next? So confused.

Reply to
eehinjor

Backup.

Think.

Try to understand what it's supposed to be doing.

I've been busy picking on Win ;-) and haven't kept close track of this thread... what's your problem/misunderstanding?

...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 | |

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Reply to
Jim Thompson

Thank you,Jim. My question was posted.The output of this circuit is not constant.

Reply to
eehinjor

Hi,Tony. Yes,you are right.Later I removed Q2 from my board.The output is better than before.But if there is no Q2,how can I restrict the voltage on RL below 5.5V?

Bur the error is more than I want,especially when RL changed from 0 to

5K,the output can change from 1.0024mA to 0.9998mA.My goal is 1mA,0.01%.

Would you give me more advice?

Reply to
eehinjor

Well, then there is less drop making for larger Vds, putting any FEt into its operational range (what i call "linear" range - where the Id VS Vds is flat). Note that my explaination said "if".. so if the FET cannot work at Vds=2V, then the ASSuME-ptions would be false, and your measurements should show what is happening or not happening.

Reply to
Robert Baer

In article ,

The drop to 0.87mA in RL could be explained if the balance was going through Q2. Possible causes could be..... 1. Effective RL, (+ ammeter shunt-R?), is greater than 5k. 2. The base voltage of Q2 is lower than 5.5V.

Put a temporary 1k in series with Q2 collector to check for Q2 current.

Doesn't make sense.

BTW: For this circuit all voltages should be referenced to the +12V. eg, the 7V ref should be a -5V ref hanging down from the +12V, and most DVM measurements taken with respect to the +12V.

BTW2: When the 5k is switched in, the -ve input of the opamp might get temporarily pulsed above the rated CMV. For some opamps this has been known to reverse the polarity of the inputs and cause a lockup. The solution is a diode between -ve and +ve, (-ve --|>|-- +ve).

--
Tony Williams.
Reply to
Tony Williams

Ok, there is something wrong with the Q2 circuit. Forget it for now, get the C-C circuit working first, then re-visit the voltage clamp.

Note though that Q2 should have affected the lower current ranges even worse. A puzzler to ponder.

That's a 2.6uA change for a 0-5v swing. Almost as if you have a 2meg shunt across the 5k+Ammeter.

Question1. Do you have a component in there that you did not fit for the other ranges. For example, (noticing that you are reading 1mA to 4 decimal places), have you dropped a capacitor across the output to cut noise on the Ammeter?

Question2. Do you have a scope, permanently on Vg and set to AC, 10mV, so that you have a continuous indication of possible oscillation?

Question3. Can you (re)confirm that the lower ranges worked perfectly. In fact publish the current outputs for the same 0V and 5V outputs.

Thanks,

--
Tony Williams.
Reply to
Tony Williams
[snip]

Using my own post to suggest some experiments.

Expt1. ~~~~~ Move the ammeter so that it is directly in series with the Source of the Mosfet. See what the 1mA does when RL is changed from 0 to 5k.

The purpose of the experiment is to ensure that all the current coming down R1 goes into the source. ie, No leakage path around the opamp.

Expt2. ~~~~~ Put a DVM (10uV-resolution) between opamp -ve and +ve. Look at the change in opamp input voltage when RL is changed from 0 to 5k..... it should be zero.

Purpose is to check the faint chance that the opamp gain is very low (for some unknown reason).

The scope on Vg to look for oscillation is probably essential for this experiment.

--
Tony Williams.
Reply to
Tony Williams

...and add in the meter loading when making measurements. Perhaps adding in opamp voltage followers will give sufficent isolation - but will add in their own errors (input offset voltage, input current loading).

Reply to
Robert Baer

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