#### bigmack69

Joined Oct 29, 2020
8
Hi Everybody,

I know that load are usually done with Mosfet but i'm trying to do this with Bjt instead.

Load current : 10mA -> 3A
Load Voltage : 1V -> 10V

I tried this schematic but something goes wrong when I try with 1V.

With a BJT, we must have at least 0.2V from Vce. On 3A and 1 Ohm resistor, the device must provide at least 3.2V. So i don't know how to make it works.
If someone have an idea don't hesitate !

Thank you !

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#### Alec_t

Joined Sep 17, 2013
13,221
Welcome to AAC!
You will have to reduce R1 if you want a 1V source to deliver 3A.
The dummy-load power will be shared by R1 and Q1, so you will need a meaty transistor. At 10V and 3A there is 30W to dissipate, so Q1 and/or R1 will need suitable large heatsinks.

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#### bigmack69

Joined Oct 29, 2020
8
Ok so for example with R1 = 0.1 Ohm and my input voltage V2 = 0.3V, it could be possible ?

#### Alec_t

Joined Sep 17, 2013
13,221
Yes.

#### bigmack69

Joined Oct 29, 2020
8
Ic and Ie will always be equal or not ?

#### bigmack69

Joined Oct 29, 2020
8
And there is also something i didn't understand, why do I have to reduce R1 ? with R1 = 1 and V2 = 3V, I still have 3A no ?

#### BobTPH

Joined Jun 5, 2013
6,035
But you are only supplying 1V! The 3V on the opamp input does not mean your 1V battery can suddenly output 3V.

A constant current source must have a higher input voltage than the load resistance times current.

Bob

#### bigmack69

Joined Oct 29, 2020
8
OK thank you i understand much better. However to comeback on that bjt, with that value of resistor 0.1 Ohm and V2 0.3V (maximum) there will be no problem to test for example a battery of 1V ?

#### BobTPH

Joined Jun 5, 2013
6,035
At least it does not violate Ohm’s law.

But you are only giving a.single example. Generally, a constant current source is specified over a range of output voltages..

Bob

#### bigmack69

Joined Oct 29, 2020
8
Load current : 10mA -> 3A
output Voltage : 1V -> 10V

This is my specifications.

The fact is I am a little bit lost around the transistor. I know i will have my current at the Emitter but what will be Ic (collector). Is the schematics enough with the specifications i'm trying to have

#### crutschow

Joined Mar 14, 2008
31,091
I know i will have my current at the Emitter but what will be Ic
The emitter current equals the collector current plus the base current (which is determined by the transistor current-gain (Beta).
This difference is why MOSFETs are usually used, since they have no appreciable gate current.
Is the schematics enough with the specifications i'm trying to have
Not completely.
Your op amp has to supply the base current for a 3A transistor collector current.
That current is determined by the Beta of the transistor you use.
If the op amp doesn't have sufficient output current for that (per its spec), then you may have to add a transistor (either to make a Darlington stage or a Sziklai Pair) to reduce the required base current, or go to a MOSFET.

#### bigmack69

Joined Oct 29, 2020
8
Ok i'll go for a darlington, but is there a problem with the voltage between the collector and the emitter is a little bit high like 1.2V.
Because the minimal voltage of the battery i want to test could be 1V

#### crutschow

Joined Mar 14, 2008
31,091
Ok i'll go for a darlington, but is there a problem with the voltage between the collector and the emitter is a little bit high like 1.2V.
Then run the collector of the first Darlington transistor to the op amp plus supply voltage, rather than the collector of the output transistor.
The minimum collector-emitter voltage of the output will then be the output transistor saturation voltage.

#### bigmack69

Joined Oct 29, 2020
8
Then run the collector of the first Darlington transistor to the op amp plus supply voltage, rather than the collector of the output transistor.
The minimum collector-emitter voltage of the output will then be the output transistor saturation voltage.
You mean like that ?

#### crutschow

Joined Mar 14, 2008
31,091
You mean like that ?
No.
Connect only the Q1 collector to the supply voltage Vcc, not the op amp input.
Connect the collector of Q2 to the DUT.
Remove the connection between the collector of Q1 and the collector of Q2.

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#### Ian0

Joined Aug 7, 2020
6,650
The connection between the test battery and the non-inverting input of the op-amp is wrong.