#### FuneralHomeJanitor

Joined Oct 12, 2019
54
In a darling array IC, such as the ULN2083, where the datasheet indicates the devices uses TTL 5V to output a 500 mA current, is there a linear current gain? For example, if I drive this with 5V from a microcontroller, should I expect a constant 500 mA? Or is there calculation needed beforehand? The diode clamps in the internal circuit diagram are confusing me a bit, and I tried to test one of the chips around 5V and my power supply started smoking. Thanks in advance, I am not working on any particular project, I am just trying to work with current sources, and am not fully grasping how this works.

#### ronsimpson

Joined Oct 7, 2019
3,133
is there a linear current gain?
It is about 1000. If you need more details I can supply.

#### crutschow

Joined Mar 14, 2008
34,680
For example, if I drive this with 5V from a microcontroller, should I expect a constant 500 mA?
The device is normally used as a switch, not as a current source, so the current depends upon the load.
I tried to test one of the chips around 5V and my power supply started smoking.
Then you apparently had the 5V connected to the output with no load, so it draws a high, short-circuit current limited by the transistor gain.
That's a big no-no.

#### FuneralHomeJanitor

Joined Oct 12, 2019
54
The device is normally used as a switch, not as a current source, so the current depends upon the load.
Then you apparently had the 5V connected to the output with no load, so it draws a high, short-circuit current limited by the transistor gain.
That's a big no-no.
I had it in series with my multimeter set to 10 A, attempting to test a no-load current

#### FuneralHomeJanitor

Joined Oct 12, 2019
54
It is about 1000. If you need more details I can supply.
View attachment 319647
For the 5V TTL as described in the bottom left, is a 5V value across the input suitable or does that need to be converted to current? How did you find 1000? Did you need to know each hfe for the circuit? I looked at the datasheet test circuits but thought they were a bit confusing. Thanks

#### WBahn

Joined Mar 31, 2012
30,239
I had it in series with my multimeter set to 10 A, attempting to test a no-load current
If there is no load, then the current would be zero.

You were attempting to test the short-circuit current, which is not a good idea.

Would you go out to your car and put a crowbar across your battery to see how much current it can deliver?

Your original post said ULN2083, but I'm guessing you meant ULN2803.

Here's the data sheet for the ST Micro part.

As others have said, these are intended to be used as switches; they are NOT current sources. As with all BJT transistors, the actual hFE is not well characterized and is quite variable, both under different operating conditions and from one device to another.

While the data sheet states indicates a minimum hFE of 1000, note that this is specifically for the ULN2801A, not the 2803. It's also for a specific set of operating conditions, namely a junction temperature of 25 °C, a Vce of 2 V, and an Ic of 350 mA.

#### WBahn

Joined Mar 31, 2012
30,239
For the 5V TTL as described in the bottom left, is a 5V value across the input suitable or does that need to be converted to current? How did you find 1000? Did you need to know each hfe for the circuit? I looked at the datasheet test circuits but thought they were a bit confusing. Thanks
The datasheet test circuits are not actual circuits, but just showing what measurements were made and what conditions were imposed. The actual circuitry that imposed those conditions are not shown.

For the 2803, the 5 V is the voltage that, when applied to the input, should result is adequate, but not excessive, base current to drive the output into saturation.

If we assume that both transistors have a Vbe of about 0.7 V, then with in input voltage of 5 V, there would be about 3.6 V across the 2.7 kΩ base resistor, yielding a current of 1.33 mA. This agrees quite well with the typical curve given in the data sheet:

Keep in mind that the notion of hFE being 1000 (or any high number) and the transistors being driven into hard saturation so that they behave like a good switch are mutually exclusive, since the former describes the operating in the linear region and the latter describes it in the saturation region. So don't try to use both relationships at the same time.

#### ronsimpson

Joined Oct 7, 2019
3,133
Most transistors, the current gain is measured when there are volts across the C-E. This part was designed to be driven full closed which takes more Base current.
This transistor cannot dissipate much heat. It should not be used in linear mode. Use it full on or full off.

#### crutschow

Joined Mar 14, 2008
34,680
I had it in series with my multimeter set to 10 A, attempting to test a no-load current
That's a test with a short-circuit to the supply voltage, not no-load.

#### Papabravo

Joined Feb 24, 2006
21,258
When a transistor is used as a switch it is not uncommon to guarantee the devices are in saturation by "forcing" the beta to be a modest fraction of what it is in the linear range. Typically for a single NPN configuration, a forced beta of 10 might be used. For the Darlington, a forced beta of 200-300 might be sufficient. Keep in mind that it is nearly impossible to drive Vce of the output stage to 0. You should expect a value in the vicinity of 1.0-1.3V depending on the load current.

Here is a simulation of a typical part. The load resistor is smaller than you would expect for a load current approaching 350 mA because the Vce of the output stage is rising with increasing current.

#### FuneralHomeJanitor

Joined Oct 12, 2019
54
That's a test with a short-circuit to the supply voltage, not no-load.
Understood, what a bonehead mistake lol but my goal was to see if I put 5V across the input terminals if 500 mA would flow out, and then use norton or thevenin theorem to possibly come up with a few resistor values to run a little pump I have that is rated at 250 mA if I liked the results, or do something using logic switching with different currents

#### crutschow

Joined Mar 14, 2008
34,680
I put 5V across the input terminals if 500 mA would flow out, and then use norton or thevenin theorem to possibly come up with a few resistor values to run a little pump I have that is rated at 250 mA
You don't adjust the current to a motor as it takes the current it needs for the load it is running.
You just apply the rated voltage to the motor, with the 2803 acting as a switch.

What is the motor's voltage rating?

#### FuneralHomeJanitor

Joined Oct 12, 2019
54
You don't adjust the current to a motor as it takes the current it needs for the load it is running.
You just apply the rated voltage to the motor, with the 2803 acting as a switch.

What is the motor's voltage rating?
12V. I was using the motor for another project where I switch it with a power MOSFET and was messing around trying to see if I could make a few different current sources for different speeds

#### WBahn

Joined Mar 31, 2012
30,239
12V. I was using the motor for another project where I switch it with a power MOSFET and was messing around trying to see if I could make a few different current sources for different speeds
Most DC motors can't be speed controlled by controlling the current. The motor current is proportional to the motor torque, not the speed, while the motor produced a back-EMF that is proportional to the speed. If you try to control the current, you will usually end up with extremely erratic speed variations and even runaway speeds. I found this out the hard way when I was an undergrad and got the bright idea to operate a small DC motor from a current-controlled supply in order to try to achieve a more stable speed, since the motor had a mirror mounted on the shaft that was used to bounce a laser off of to create Lissajous patterns on the wall (it was actually bounced of two or three mirrors in the path). The speed fluctuations made it hard to get a nice, stable pattern, so I thought that running constant-current might help. Wrong. What happened was that as I slowly increased the current to the point where the static friction was overcome, the lower dynamic friction that resulted once it started moving resulted in the motor immediately overspinning and the mirror grenading. Obvious in highsight.