JFETs as a current source and a question.

[Herschel Peeler]

Getting to play with Junction FETs, discovering them as a current source. I had three on hand; 2N3819 N-JFET, PN5433 N-JFET and 2N5462 P-JET. I did the attached exercises in the configurations shown in the schematic. Each exercise means building it with all resistors, 10 ohm to 1 M ohm and noting Drain current, Source Voltage and calculating mho. (See the exercises.)
Doing the exercises and comparing my result to the data sheets was interesting. We discovered what makes one different from another.

Applying this theory ...
2N5462 was chosen. 5 mA was desired. Referencing the exercise that called for a 270 ohm resistor.
In the schematic, lower right drawing. I tried to set up a four LED constant current driver. My drive side had the expected 5 mA. I was expecting all four outputs to also be 5 mA. A four output current mirror.
Okay, not the first time my design didn't work as I expected. One output drew 4 mA, the other three about 3.8 to 3.9 mA. It didn't matter which LED was where.
I am running on a breadboard with a 5 Volt regulator, about a 250 mA current limit (LM78L05 with a 2N4403 booster).

So, any ideas on what went wrong? Why were they all not 5 mA out?

 

[Harald Kapp]

A few issues I can imagine:

• The ideal current mirror expects Vcb=0V whic is not the case here, This causes a bit of current mismatch, but not 20% as in your case.
• Strictly speaking the equations for the current mirror are for one mirrored output only. By having 4 output transistors, you add 4*Ib to the reference currrent (from Q5) instead of 1*Ib as in the standard equations for this type of current mirror.

Which measurements do you get when you connect only one mirror transistor (e.g. pins 3, 4 and 5) and disconnect the other three (pins 6...14)?

Read more here.

 

[Herschel Peeler]

A few issues I can imagine:

• The ideal current mirror expects Vcb=0V whic is not the case here, This causes a bit of current mismatch, but not 20% as in your case.
• Strictly speaking the equations for the current mirror are for one mirrored output only. By having 4 output transistors, you add 4*Ib to the reference currrent (from Q5) instead of 1*Ib as in the standard equations for this type of current mirror.

Which measurements do you get when you connect only one mirror transistor (e.g. pins 3, 4 and 5) and disconnect the other three (pins 6...14)?

Read more here.

Re: Vcb = 0 V. The driving side is as you say.

Re: Trying it with just one LED.
No change. And the driving side current stays at 5.5 mA no matter how many LEDs are connected, or none. ???? The transistor array was the closest thing I had to matched transistors. I thought they would be matched. ???? But yes, my symptoms would be accounted for by mismatched transistors, wouldn't it?

Why no change in drive current?
Is this the old question about collector current being related to Base Voltage, not Base Current? But what you say seems to make sense. It just doesn't happen. ???

Still lots for me to learn The more I test my knowledge the more I find out I really don't know.
Is anyone good with a simulator? I wonder what a simulator would show?

 

[Harald Kapp]

The transistor array was the closest thing I had to matched transistors.

Not a bad choice, but only the pair withinm the chip is really matched. However, being on the same chip and package, chances are that the Ca3046 is the best you can gaet (at an affordable price).

Is this the old question about collector current being related to Base Voltage, not Base Current?

Let's not start this discussion. It has been discused here in a lengthy thread.

Is anyone good with a simulator? I wonder what a simulator would show?

It shows what you expect:


I'm a bit at a loss concerning your observations. Are you sure input current is 5mA? in the breadboarded ciruit?

 

[Herschel Peeler]

Not a bad choice, but only the pair withinm the chip is really matched. However, being on the same chip and package, chances are that the Ca3046 is the best you can gaet (at an affordable price).


Let's not start this discussion. It has been discused here in a lengthy thread.


It shows what you expect:
View attachment 25380

I'm a bit at a loss concerning your observations. Are you sure input current is 5mA? in the breadboarded ciruit?

Super, thank you. Now add more transistors in parallel with Q2. Does the source current stay the same? On my breadboard it does. It does not increase as I add loads on the "Q2"s collectors.

Is that 5 mA split between the bases of all transistors, no matter how many there are?

I will try replacing my constant current source with a resistor tonight and see what difference it makes.

 

[Herschel Peeler]

Super, thank you. Now add more transistors in parallel with Q2. Does the source current stay the same? On my breadboard it does. It does not increase as I add loads on the "Q2"s collectors.

Is that 5 mA split between the bases of all transistors, no matter how many there are?

I will try replacing my constant current source with a resistor tonight and see what difference it makes.

Okay, replaced the JFET circuit with just a resistor (820 Ohms). Voltage and current through the resistor did not change very much with no LEDs, or 4 LEDs. Adding loads on the collector does not change base current proportionally, no surprise I guess. But this sheds no light on why my four outputs do not equal my input current.
I guess the next step is to measure the gain of each transistor and see if they are just not so well matched.

 

[(*steve*)]

If you add many transistors in parallel, you would expect their currents to be very similar to each other, however, as you add more transistors the load currents will drop from the current set by your constant current source.

 

[Herschel Peeler]

If you add many transistors in parallel, you would expect their currents to be very similar to each other, however, as you add more transistors the load currents will drop from the current set by your constant current source.

That makes sense, of course. As I add more transistors to the constant current source ... my current just divides between the transistor's bases??? Then my output currents should divide accordingly ??? They don't. They stay to about 4 mA or so.
On a gain check all my transistors match well on gain, but not under the application conditions. That check comes next. with 110 uA in I got 14.8 mA out with an LED as a load. Next I will test them with 5 mA in.

 

[(*steve*)]

Remember that your constant current supplies the collector current for one transistor, and also the base current for all of them. The base current is only small compared to the collector current, so whilst the collector currents are all approximately the same, the constant current source needs to provide a little more current than the collector current.

 

[GPG]

Put a degeneration resistor between the emitters and ground.10 to 100Ω should be enough depending on the match

 

[Herschel Peeler]

Remember that your constant current supplies the collector current for one transistor, and also the base current for all of them. The base current is only small compared to the collector current, so whilst the collector currents are all approximately the same, the constant current source needs to provide a little more current than the collector current.

Maybe part of my problem is not understanding current mirrors? Why does the output transistor have no gain? In a simple current mirror, one input,, one output, why is collector current the same as base current? Or is that the case? Is the input current split between the two bases? So the base current of the output transistor is half the input current??? But the output is twice that base current when the transistor has a gain of 100? The outputs don't go into saturation.

 

[Herschel Peeler]

Put a degeneration resistor between the emitters and ground.10 to 100Ω should be enough depending on the match

I'll try that. In op amps the current mirrors have emitter resistors on them. Used to trim offset!!! So I should be able to trim my outputs by trimming the emitter resistors. Sounds good.
Thanks.

 

[Harald Kapp]

In a simple current mirror, one input,, one output, why is collector current the same as base current?

It isn't. Collector current in the mirror transistor is the same as in the source transistor (leftmost) due to identical Vbe.

 

[Herschel Peeler]

I'll try that. In op amps the current mirrors have emitter resistors on them. Used to trim offset!!! So I should be able to trim my outputs by trimming the emitter resistors. Sounds good.
Thanks.

Something like this? No need for the odd ball transistor array.

 

[Herschel Peeler]

It isn't. Collector current in the mirror transistor is the same as in the source transistor (leftmost) due to identical Vbe.

But the transistor on the far left has no gain. The others should, or at least it would seem. The current design has 5.5 mA going in, but about 8 mA at the outputs. The transistors are not in saturation. About 2.7 V out. ??? Because of the constant voltage drop across the LED ???
Still using the CA3146.

 

[Harald Kapp]

But the transistor on the far left has no gain.

Why not? For the circuit I simulated the base current to Q1 is 25µA at 5mA collector current, a gain of 200.

The transistors are not in saturation. About 2.7 V out. ???

Of course not. In saturation Vce would be aroud 0.1V,leaving almost al or the 5V across the LED which in turn would burn out the LED (typical LED voltages range from 1.6V red to 3V white).

Super, thank you. Now add more transistors in parallel with Q2. Does the source current stay the same?

It does so in the simulation, too.

 

[Herschel Peeler]

Why not? For the circuit I simulated the base current to Q1 is 25µA at 5mA collector current, a gain of 200.


Of course not. In saturation Vce would be aroud 0.1V,leaving almost al or the 5V across the LED which in turn would burn out the LED (typical LED voltages range from 1.6V red to 3V white).


It does so in the simulation, too.

Re: simulated circuit

hanks for that. I gotta learn to use an simulator.

 

[GPG]

Something like this? No need for the odd ball transistor array.

The left transistor requires a resistor too. Not 10 AND 100, 10 to 100.

 

[Herschel Peeler]

The left transistor requires a resistor too. Not 10 AND 100, 10 to 100.

The two resistors in parallel were intended to be a 10 ohm resistor with a trimming resistor in parallel. After building it the second resistor was not needed. I got sufficient trimming with 5% values. As far as adding a resistor on the transistor on the left ... yes. Originally I figured it would be whatever it was and the others trimmed to match that. Not one of my better ideas.

My first attempt at simulation (Multisim Blue) I would not call a success. I tried to make a current mirror using 2N3904s and 1K resistors to +5 V.

Attached is the simulation and the same done up in ORCAD with measurements from an actual circuit. Current reports from the simulation do not agree with reality. The measurements shed some light on base currents. The bulk of the driving current goes through the collector of the first transistor. Adding more transistors detracts from the collector current???
Oops, the simulation file is too big to upload.

 

[Herschel Peeler]

Okay, current revision. Multi-column multi-row of white LEDs. Common constant current source. Dimmable with a 500 ohm pot. Up to 100 mA design. But the LEDs I used are rated at 30 mA so I wouldn't expect them to last long at 100 mA.
I had to get away from the 2N5642 and CA3146 to get 100 mA, but it works.