Constant current Opamp circuit issues

Question

I have implemented a constant current source for an high power LED. I am using the following topology.

I have the following problems:

• Even a small noise on IN+ causes some current to pass through Rsense and make the LED turn on, although it is very weak, it consumes power and not ideal. How can I make this work only above certain voltage on IN+, for example 1V.

• I couldn't push the current through LED to very high numbers. I use a 5A capable BJT and MCP6021 as opamp. I can see up to 1.5A but cannot get pass that. (1.5A measured on Rsense's voltage drop). The capacitors are 4x 420uF Electrolytic. (I didn't have a 1mF handy so I put in 4 parallel caps) Do I need a special cap?

Answer 1

It's a wonder you get 1.5A at all.

edit
Since you mention 3.3V as the input for your boost regulator I'll presume that this is also the power supply for your 555 and opamp. The LM555 doesn't work at this low voltage, so you'll have to use the CMOS version, TLC555.

\$H_{FE}\$ for the 2N5191 at 1.5A is minimum 25, so you'll need 60mA of base current. An \$H_{FE}\$ of 25 makes that the opamp sees the 0.5\$\Omega\$ emitter resistor as 12.5\$\Omega\$. That's still low, so it won't be the limiting factor for the base current: \$\dfrac{3.3V - 0.7V}{12.5\Omega}\$ = 208mA.

The problem is the opamp. At 2.5V power supply the MCP6021's short circuit current is typically 30mA, which in a worst case situation will only give you 750mA collector current. The reason you do get 1.5A is that the 25 is the transistor's minimum \$H_{FE}\$, it can go up to 100.

The \$H_{FE}\$ you get is typical for a power transistor, so swapping it for another type probably won't help much. You can use a Darlington, or make your own by adding a small signal transistor to drive the power transistor's base. The BC817 will be suitable.

A MOSFET is also a possibility, but this should be chosen with care. Even logic MOSFETs don't always give their maximum current at 5V \$V_{GS}\$, and the sense resistor subtracts also a certain voltage from it. The 555's output divider gives 1.65V for the opamp's non-inverting input, and given enough drive current the sense resistor will also settle to 1.65V, giving 3.3A to the LEDs. That means that the MOSFET should be able to give 3.3A of drain current at only 3.3V - 1.65V = 1.65V \$V_{GS}\$. That's very low! But by a lucky coincidence an answer to another recent question showed this graph:

That damned AO6408 delivers 8A at only 1.5V \$V_{GS}\$! Tailor-made for the job!

Answer 2

• Even a small noise on IN+ causes some current to pass through Rsense and make the LED turn on, although it is very weak, it consumes power and not ideal. How can I make this work only above certain voltage on IN+, for example 1V.

Simply take your feedback from the output of the op-amp (the base of the BJT). Now if the input signal is less than about 0.6 V, no IR output.

If you still want to have the same maximum output you can adjust the two resistors (not labelled in your schematic) to increase the "on" voltage input to the op-amp by 0.7 V or so.

EDIT

As Nick Alexeev points out the above is a bit of a "quick-and-dirty" solution. It gives about 100-200 mA of uncertainty of the output current. It might also restrict your options with solving the issue of the op-amp not having enough "oomph" to provide the base current in the BJT that you need to get the output current you want (as detailed in Nick's answer).

Here's a solution that solves both problems (this replaces the op-amp, BJT, and sense resistor in your circuit):

With this circuit you don't need to worry about the drive capability of the op-amp. You also reduce the output current uncertainty. The original circuit had possible errors due to the resistor variation (0.5 - 5%, depending what resistor you buy) and the BJT beta (1 - 2%).

With this you are left with just the resistor variation and an error due to the variation in the diode forward voltage.

If the 47 kOhm resistor makes this circuit respond too slowly for your needs, you may need to reduce the values of your voltage divider resistors to enable you to reduce the value of the new resistor.

EDIT 2

To reduce component count, you could build the diode in to your resistor divider. R1 here replaces the lower resistor in the divider. The value of R1 would be reduced to 10 kOhms. An additional resistor is added in series with the diode to be the upper resistor in the divider. The input then comes straight from the 555.

The value of the upper resistor should be chosen to give the output current you want when the input is high.

Answer 3

• I couldn't push the current through LED to very high numbers. I use a 5A capable BJT and MCP6021 as opamp. I can see up to 1.5A but cannot get pass that. (1.5A measured on Rsense's voltage drop).

The max output current of MCP6021 is 30mA, which is the gate base current \$I_b\$. The LED current is the collector current \$I_c=I_b h_{FE}\$. We don't know what model Q1 is, but the values of \$h_{FE}\$ (sometimes it's also called beta) are usually between 10 and few hundred. It agrees with 1.5A max, which you're getting. A Darlington pair or a MOSFET could give you more current.