I have a wire with current I1 going thru it. As the voltage and I1 rise there is a transistor Q which turns on and it in turn stops the current from flowing. I1 and the voltage referenced is tied to the base of the transistor.
My question is:
When I1=.90amps Vbe = .63 volts
When I1=1.3 amps Vbe = .58 volts
When I1=1.5 amps Vbe=.38 volts

Why does the turn on voltage of the transistor keep dropping as the current increases?
(q1 is 2N3904)

Added picture upon request. Making it a bit more specific to my circuit

 

As the transistor saturates, Vbe drops and Collector current increases.

 

Vbe is the potential barrier due to the electrons and holes combined at the base emitter junction.
A very strong charge needs to be applied to reduce this barrier and free the electrons so recombination cannot occur as readily.

So now itsw like overdrive less potential difference at the junction to overcome, once the transistor shuts down agfain, electron hole recombination takes place, producing the potential barrier, which again needs that extra voltage Vbe to reduce the barrier.

 

Wait. it doesn't drop over time, when I is steady.
When I set the current to 1.5 amps I find Vbe turns on at .38 volts
when I set the current to <.9 amps I find Vbe turns on and is steady at .65 volts

How can I fix it so it turns on around .65 volts(or a constant voltage), with an I from 700mA to 2.5 amps?

.

 

I have a wire with current I1 going thru it. As the voltage and I1 rise there is a transistor Q which turns on and it turn stops the current from flowing. I1 and the voltage referenced is tied to the base of the transistor.
My question is:
When I1=.90amps Vbe = .63 volts
When I1=1.3 amps Vbe = .58 volts
When I1=1.5 amps Vbe=.38 volts

Why does the turn on voltage of the transistor keep dropping as the current increases?
(q1 is 2N3904)

Because a VBE drops 2mV for every degree C increase. If you are putting 1 - 1.5A of current through a 2N3904, you are cooking the living snot out of it. I'm surprised it didn't melt.

BTW: the absolute maximum current rating for 2N3904 is 200 mA.

http://www.fairchildsemi.com/ds/2N/2N3904.pdf

 

I have a wire with current I1 going thru it. As the voltage and I1 rise there is a transistor Q which turns on and it turn stops the current from flowing. I1 and the voltage referenced is tied to the base of the transistor.

If you can provide a circuit diagram for it ,it may help others to help you.

I think you are not passing that much of current through the transistor as the transistor itself will not allow it ,what I understood is that you are passing the current through the wire and you want to control this current using a transistor by turning the current source on and off.

Is that voltage reference a sensing resistor ????

Without a circuit diagram I can't say much...

 

Ok here's a schematic of what I'm testing.
The npn transistor stays cool, I'm assuming it doesn't take 3amps b-e.
How can I keep the turn on voltage near .65?

 

R3 is a sense resistor. As the voltage across it tries to increase Q1 conducts, bringing ground to the gate of Q2, which tends to turn it off. As Q2 turns off the voltage across R3 also drops, which causes Q1 to conduct less, raising the voltage at the collector, causing the gate voltage on Q2 to increase, which in turn causes the voltage across R3 to increase.

I am not a fan of this circuit, I don't like using MOSFETs in analog modes. The circuit is dynamic, the setpoint current is where the desired current causes 0.6VDC on the sense resistor R3 if Q1 were not there. Q2 is the component that will get hot, this is the norm for almost all current regulators, and depending on R1, Q1 will stay cool.

To set the current setpoint, you vary R3's resistance. I = R3 / 0.6V

The circuit will be somewhat temperature sensitive, though I believe it can be lived with. This is because the characteristics of Q1 will change a little over temperature.

A lot of power supply circuits use a similar scheme for current foldback on their regulators, especially the LM723.

Current mirrors can eliminate the temperature sensitive issues. I didn't much like them either when I first looked at them, but I have since come around.

LED Drivers

I'll probably add your schematic to the list of circuits on this album page.

 

You can probably minimize the change in Vbe with load current by making R1 larger.

 

I have a wire with current I1 going thru it. As the voltage and I1 rise there is a transistor Q which turns on and it in turn stops the current from flowing. I1 and the voltage referenced is tied to the base of the transistor.
My question is:
When I1=.90amps Vbe = .63 volts
When I1=1.3 amps Vbe = .58 volts
When I1=1.5 amps Vbe=.38 volts

Why does the turn on voltage of the transistor keep dropping as the current increases?
(q1 is 2N3904)

Added picture upon request. Making it a bit more specific to my circuit

How are you changing the current?

 

R3 is a sense resistor. As the voltage across it tries to increase Q1 conducts, bringing ground to the gate of Q2, which tends to turn it off. As Q2 turns off the voltage across R3 also drops, which causes Q1 to conduct less, raising the voltage at the collector, causing the gate voltage on Q2 to increase, which in turn causes the voltage across R3 to increase.

I am not a fan of this circuit, I don't like using MOSFETs in analog modes. The circuit is dynamic, the setpoint current is where the desired current causes 0.6VDC on the sense resistor R3 if Q1 were not there. Q2 is the component that will get hot, this is the norm for almost all current regulators, and depending on R1, Q1 will stay cool.

To set the current setpoint, you vary R3's resistance. I = R3 / 0.6V

The circuit will be somewhat temperature sensitive, though I believe it can be lived with. This is because the characteristics of Q1 will change a little over temperature.

A lot of power supply circuits use a similar scheme for current foldback on their regulators, especially the LM723.

Current mirrors can eliminate the temperature sensitive issues. I didn't much like them either when I first looked at them, but I have since come around.

LED Drivers

I'll probably add your schematic to the list of circuits on this album page.

But as I try to calculate the current for higher current I find the turn on voltage drops from .6 volts down. I tested all the way down to .2 volts. I can't draw more than 1.45 amps since now anytime I decrease the sense resistor the turn on voltage decreases. Example: I put .2 ohm sense resistor then the transIstir turns on around .29v. Current I find is still around 1.45 a.
Will increasing resistance to the collector help?
Will adding a resistor to the base help?

 

What is your supply voltage and R1 = ?? and what type of a MOPSFET you are using.
Are your power supply is capable of to deliver such a large current??

 

At 0.2V on R3 Q1 is effectively off, which in turn would turn Q2 fully on. You would see the maximum voltage on the gate at this point.

Now we get to the point I am not a fan of this circuit. It works, but to be fully on a standard MOSFET must have at least 10V on the gate as measure from the source. Otherwise it is likely in the linear region. Since MOSFETs are not very linear it becomes hard to predict at this point.

However, as I mentioned the circuit is dynamic, it will find its own balance. It absolutely requires more than 10V on the power supply though to give the GS on Q2 what it needs.

When the current regulator is working R3 will have around 0.6V across it. It is core to how this circuit works.

 

What is your supply voltage and R1 = ?? and what type of a MOPSFET you are using.
Are your power supply is capable of to deliver such a large current??

Supply voltage is around 7.5-8.5 volts (2 lithium cells)
R1 Im using = 1k
mosfet, currently = IRF510
power supply has delivered more than 2.6 amps before.

 

At 0.2V on R3 Q1 is effectively off, which in turn would turn Q2 fully on. You would see the maximum voltage on the gate at this point.

Now we get to the point I am not a fan of this circuit. It works, but to be fully on a standard MOSFET must have at least 10V on the gate as measure from the source. Otherwise it is likely in the linear region. Since MOSFETs are not very linear it becomes hard to predict at this point.

However, as I mentioned the circuit is dynamic, it will find its own balance. It absolutely requires more than 10V on the power supply though to give the GS on Q2 what it needs.

When the current regulator is working R3 will have around 0.6V across it. It is core to how this circuit works.

I built the circuit, as I test the circuit and change out the sense resistor to smaller values I find that the voltage across Vbe drops. I find Q1 is on at 0.2V.
If I increase the sense resistor a little from .15 to .23 Ω, I measure Vbe to be 0.3 V.

What can I do to increase the current?
I have always found the current measured and Vbe measured accurately predict the resistor value i've used
(since I have found decreasing the sense resistor decreases the turn on voltage of Q1, what else can I change in the circuit to keep Q1 at .6 volts, or draw more current)(the circuit Ive tested works fine under 1 amp, with minimal Q1 Vbe change)

 

You likely have oscillation then, because if Q1 is a silicon device, it will not be on at 0.2V. The rules on this are pretty simple with BJTs, and 0.6V is actually a little low. For any current to flow through the BE junction you have to reach the break over voltage. What is the voltage on the gate of the MOSFET?

The reason I suspect oscillation is it would provide the spikes needed to cross over the voltage of the BE junction and turn the transistor on, but it would be invisible to a DVM.

Just curious, are you building this on a protoboard? This is what I call a protoboard...

 

I agree with the Bill that there might be oscillation. It would explain some of the odd measurements.

In the future, I would suggest for each data set, you make a table that includes supply voltage, R1, R3, collector voltage, and base voltage. That would help a lot to see a "snapshot" of what's happening.

 

I have it soldered up.
I'll make a table too mlog, good idea

How can I get rid of the oscillation?

 

Voltage __R1__ __R3__ __Vbe__ __I__
7.6 v 1k 0.69 0.627 0.91
7.6 v 1k 0.33 0.468 1.37
7.6 v 1k 0.5 0.57 1.34
7.6 v 1k 0.25 0.388 1.47
7.6 v 1k 0.22 0.33 1.44

These were some test results, I know I dropped R3(sense) down to .12 homs and the current draw was still 1.5 ish

 

Do not know for sure, since I've never used this circuit.

A capacitor in the right place might damp things down, but you have to be careful, since it could also cause a surge of current.

Just as a test (I would replace the LEDs with regular diodes) I might put a capacitor across R3 to see if it damps things down. The reason this is a risk is while the cap is charging there will be a massive surge of current through Q2.

Putting a cap on the gate of Q2 would actually cause a soft start, a good thing. I think it would be less effective at damping oscillation though.