Building a voltage-controlled current source with range and offset

[xieliwei]

Hello all!

I'm learning about and trying to build a VCCS for driving a laser diode at high speed (15MHz). Starting with this, https://en.wikipedia.org/wiki/File:Op-amp_current_source_with_pass_transistor.svg , I replaced the zener diode/resistor combination with my modulating signal and removed the transistor since the LM7171 I'm using is capable of producing the required current.

However, in both simulations and a prototype I built, the output current seems to peak off when my modulating voltage reaches 75% of Vcc (5V); why is that so? Other than increasing the voltage supply to the LM7171, is there a way to overcome this limitation?


Also, after reading up on laser diodes, it appears that the laser has a large excitation period when starting up. Hence, to achieve fast modulation, the laser must never become de-excited. That means that even if the modulation voltage is zero, the current source must still produce a current of 3mA.

So now the VCCS must put out 3mA at 0V and 10mA at 5V. How do I achieve the current offset? Is choosing the correct current sensing resistor a reasonable way to limit the current range?

Alternatively, is there already a driver out there that allows me to modulate the laser at 15MHz? A quick search picked up a few, but they seem excessively complicated since they're designed for SFP modules.

Thanks!

 

[MrAl]

Hi,

You'd need a fast op amp to use that circuit.

To get current offset, you simply make sure your control voltage is always above zero volts input. BTW you can not have 0 volts if you have 3ma, there will still be a small voltage.

 

[ronsimpson]

I am assuming you are using a 5 volt supply.
>The input of the LM7171 do not work if they are close to the supply voltage. Keep the inputs 1.65V away from the power supply.
>The output, driving 100 ohm load, does not work well if closer than 2.8V of the supply. (that explains your 75% of 5V)

I think you want a current source that outputs 3mA at 0V and 10mA at 5V.
>From input to (+) of op-amp connect a resistor. You will need a resistor form (+) of op-amp to ground.
>From (+) of op-amp connect a resistor to +5V.
>Scale the resistors so that when both resistors are at 5v there is 10mA and when one is at 0V and the other is at 5V the divider makes a voltage that gives you 3mA. That will take some math.

What diode are you using? What is its voltage?
-------------------------------------------------------
You have a good op-amp, 200mhz, but not overly fast. It will have problems with only 5 volts supply. You should keep the voltage across the current scene resistor low, 1v, not 5v because you only have a small supply voltage.

 

[xieliwei]

Thank you both for the reply!

MrAl:
Is the LM7171 fast enough? 200MHz at unity seems to be more than sufficient.

The problem is the signal I'm using to modulate is fixed and out of my control. I could probably add a DC offset myself, but I'm hoping there'd be a simpler method.

ronsimpson:
Yes, it was using a 5V supply. So I guess the only way is to power the op-amp with a higher voltage supply?

Interesting solution, I'll give it a try!

I'm using this diode: https://www.electro-tech-online.com/custompdfs/2012/02/OPV300.pdf Maximum Vf is 2.2V.

 

[ronsimpson]

The diode needs 2.2 volts and the current resistor might have 1 volt across it. (3.2 volt) That is much of your 5 volts and the LM7171 can not pull to the 5v rail. You could choose a resistor so the voltage is 0.5 or 0.1 volts.

If the voltage across the resistor is too small it will require a faster op-amp.

The op-amp data sheet shows in several places the output current and output voltage limits at certain supplies voltages.

 

[xieliwei]

Ah, so that's what the output swing of an op-amp means.

So I can try scaling both the op-amp (+) range (using the method you mentioned in the previous post) and the current sense resistor to obtain a solution that works for the output swing of the op-amp at 5V, did I understand correctly?

 

[xieliwei]

After some number crunching, this is what I came up with. Ignore the oscilloscope, CCVS, lone voltage divider and the general mess

View attachment 61388

It'll require a negative power supply though, or else it'll be impossible to obtain 3-10mA for the laser.

Is this a reasonable circuit? Simulation-wise it works, but I'll need to wait a while for those funky resistors before prototyping is possible.

 

[ronsimpson]

I don't see you need 0.1% resistors. Maybe 1%.
If your current detect resistor is +/-5% then no amount of 0.1% parts will help.
You 5 volt supply is not 0.1%. It is the reference everything is based from.
For now use 5% parts and see what happens.

 

[MrAl]

Hi again,


Oh so you are going to try to drive the laser diode with a constant current, is that a good idea?
Arent you supposed to use the built in feedback diode for the correct regulation?
Perhaps you have run some similar diodes like this before but...
The laser diode characteristics vary widely unlike a regular LED, unless you have a particular kind of laser diode that has some very narrow specs. I think what happens with a constant current is it is hard to adjust for the right laser current so that it might work once and then not work later, or work for a little while, then not, or you may accidentally go over the max power rating. Using the built in feedback sensor is supposed to eliminate the guess work. The only thing then is to minimize transients which could degrade the efficiency of the light output.

I have a feeling you should read up on laser diodes a bit more before you turn this thing on for the first time. A well designed circuit would most likely include feedback from the internal sensor.

As to the circuit itself, you'll need plus and minus 5v power supplies for that op amp selection. It looks like it may not work at all at just +5v.

 

[xieliwei]

ronsimpson:
Yes, 5% should suffice, but the distributor I'm getting my components from sell 0.1% resistors for a cent extra each. Doesn't hurt to buy them =)
I've placed the order, now to wait for them to arrive.

MrAl:
Yes, I should be driving at constant power but the diode I'm using (OPV300) does not have a built-in photodiode. The diode does have pretty tight specifications, so hopefully, by keeping a safe margin and some voltage clamps on the op-amp side, things won't go wrong.

Regarding the op-amp, I have switched to using a dual supply.