# Question about LED modulation circuit, OP-AMP

#### Col John Matrix

Joined Apr 16, 2009
14
Hi,

i'm looking at an LED modulation circuit I found posted online, and I'm a bit puzzled about some of it. I'll post the circuit and description from the site first...circuit diagram is attached:

"circuit is useful up to about 50 MHz and uses a high frequency analog op amp, a Linear
Technology LT1363 with a GBW of 70 MHz, a slewrate of 1000 V/us and an output current drive
capability of at least 50 mA.

The circuit is a transconductance or voltage to current converter configuration. The output is DC
biased using a noninverting input voltage divider network at a quiescent current of Vcc/2/Ri or about
15 mA using the component values shown with a supply voltage of +/ 10V. The modulation input is
capacitatively coupled to the non inverting put via the RC lead network with a low frequency
modulation cutoff at about 3 Hz. Since dynamic resistance of the LED is low compared to Ri the
fe
edback fraction is roughly unity, and therefore the expected modulation bandwidth will be on the
order of 70 MHz. To achieve a high LED current modulation index in this configuration, a large
output voltage swing across Ri is required and for operation at highfrequency, a very high slewrate
op amp is needed. For example, to achieve a peak modulation current of 10 mA (total is 25 mA)
requires a modulation input of Vp ~ 3V. For modulation up to 50 MHz, the required slewrate is
2*Pi*Vp*f or about 950 V/us which is just within the bandwidth and slewrate specifications of the
LT1363.
"

My first questions are about this sentence: "The output is DC biased using a noninverting input voltage divider network at a quiescent current of Vcc/2/Ri or about 15 mA using the component values shown with a supply voltage of +/ 10V."

• Does this mean that the LED would be biased with a 15mA DC offset, and modulated around this value?
• Why is the quiescent current given by Vcc/2/Ri? Ri does not appear to be connected to the Op Amp supply voltage.
The next question I have is what is the purpose of V3 and the R2/R1 network it's connected to? V3 is not mentioned in the description at all.

Thanks!

#### Attachments

• 136.7 KB Views: 23

#### ErnieM

Joined Apr 24, 2011
8,363
1. Yes.

2. Because the op amp will attempt to regulate the voltage at the + terminal across Ri.

3. V3 is a source of bias to offset the zero input current up from ground. Loks like the author assumes it is the same voltage as the op amp runs at.

The led needs to be running at some level of brightness so the input can vary the brightness around that level. If you don't do that you will only modulate the positive halves of the input wave.

#### Col John Matrix

Joined Apr 16, 2009
14
Hi Ernie, thanks for the reply.

I'm not sure I understand 2)...the circuit looks like a voltage follower to me, like the output should follow Vm rather than Vcc.
I'm not grasping what you mean when you say " the op amp will attempt to regulate the voltage at the + terminal across Ri." As a result I also don't see where the divide by two comes from either.

Sorry if I'm being really slow here, but any further explanation would be much appreciated!

#### hp1729

Joined Nov 23, 2015
2,304
Hi,

i'm looking at an LED modulation circuit I found posted online, and I'm a bit puzzled about some of it. I'll post the circuit and description from the site first...circuit diagram is attached:

"circuit is useful up to about 50 MHz and uses a high frequency analog op amp, a Linear
Technology LT1363 with a GBW of 70 MHz, a slewrate of 1000 V/us and an output current drive
capability of at least 50 mA.

The circuit is a transconductance or voltage to current converter configuration. The output is DC
biased using a noninverting input voltage divider network at a quiescent current of Vcc/2/Ri or about
15 mA using the component values shown with a supply voltage of +/ 10V. The modulation input is
capacitatively coupled to the non inverting put via the RC lead network with a low frequency
modulation cutoff at about 3 Hz. Since dynamic resistance of the LED is low compared to Ri the
fe
edback fraction is roughly unity, and therefore the expected modulation bandwidth will be on the
order of 70 MHz. To achieve a high LED current modulation index in this configuration, a large
output voltage swing across Ri is required and for operation at highfrequency, a very high slewrate
op amp is needed. For example, to achieve a peak modulation current of 10 mA (total is 25 mA)
requires a modulation input of Vp ~ 3V. For modulation up to 50 MHz, the required slewrate is
2*Pi*Vp*f or about 950 V/us which is just within the bandwidth and slewrate specifications of the
LT1363.
"

My first questions are about this sentence: "The output is DC biased using a noninverting input voltage divider network at a quiescent current of Vcc/2/Ri or about 15 mA using the component values shown with a supply voltage of +/ 10V."

• Does this mean that the LED would be biased with a 15mA DC offset, and modulated around this value?
• Why is the quiescent current given by Vcc/2/Ri? Ri does not appear to be connected to the Op Amp supply voltage.
The next question I have is what is the purpose of V3 and the R2/R1 network it's connected to? V3 is not mentioned in the description at all.

Thanks!
Not homework, right?

Not knowing for sure what V3 is it is kind of hard to say much about R2/R1. Effectively it would set the 15 mA level in the LED with no input signal. So I would say adjust V3 for 15 mA at the output.
15 mA across 330 ohms sounds like 5 V so the junction of R2 and R1 must be 5 V also. V3 = 10 V?

Last edited:

#### hp1729

Joined Nov 23, 2015
2,304
Hi,

i'm looking at an LED modulation circuit I found posted online, and I'm a bit puzzled about some of it. I'll post the circuit and description from the site first...circuit diagram is attached:

"circuit is useful up to about 50 MHz and uses a high frequency analog op amp, a Linear
Technology LT1363 with a GBW of 70 MHz, a slewrate of 1000 V/us and an output current drive
capability of at least 50 mA.

The circuit is a transconductance or voltage to current converter configuration. The output is DC
biased using a noninverting input voltage divider network at a quiescent current of Vcc/2/Ri or about
15 mA using the component values shown with a supply voltage of +/ 10V. The modulation input is
capacitatively coupled to the non inverting put via the RC lead network with a low frequency
modulation cutoff at about 3 Hz. Since dynamic resistance of the LED is low compared to Ri the
fe
edback fraction is roughly unity, and therefore the expected modulation bandwidth will be on the
order of 70 MHz. To achieve a high LED current modulation index in this configuration, a large
output voltage swing across Ri is required and for operation at highfrequency, a very high slewrate
op amp is needed. For example, to achieve a peak modulation current of 10 mA (total is 25 mA)
requires a modulation input of Vp ~ 3V. For modulation up to 50 MHz, the required slewrate is
2*Pi*Vp*f or about 950 V/us which is just within the bandwidth and slewrate specifications of the
LT1363.
"

My first questions are about this sentence: "The output is DC biased using a noninverting input voltage divider network at a quiescent current of Vcc/2/Ri or about 15 mA using the component values shown with a supply voltage of +/ 10V."

• Does this mean that the LED would be biased with a 15mA DC offset, and modulated around this value?
• Why is the quiescent current given by Vcc/2/Ri? Ri does not appear to be connected to the Op Amp supply voltage.
The next question I have is what is the purpose of V3 and the R2/R1 network it's connected to? V3 is not mentioned in the description at all.

Thanks!
Well I built it and the circuit works as designed. Changes in intensity for a few volts in isn't real apparent to the eye but through fiber optics to a sensitive receiver it may make a difference. I used a variable voltage source for V3. Yes, set at 10 V it works as expected.

#### #12

Joined Nov 30, 2010
18,224
The first thing to know about op-amps (operated in a linear fashion) is that, "For any input voltage, the amplifier output will do whatever it takes to make its other input pin the same voltage as the first input pin."

Yeah. Right. It took me a few days to understand that statement. The implications open a whole new world, compared to stage-by-stage amplification that most people are used to.

In this case, if the + input is 3 volts, the op-amp will shove as much current as necessary through the LED to make 3 volts happen at the top of Ri.
Change the input voltage to 2 volts or 4 volts and the op-amp responds by driving enough current through Ri to make the other input 2 volts or 4 volts.
The voltage across the LED is only changing by a few tenths of a volt, but the current through Ri changes in exact proportion to the input voltage, therefore the current through the LED changes exactly the same as the current through Ri.

#### AnalogKid

Joined Aug 1, 2013
10,798
In your schematic, the instantaneous voltage at the + input to the opamp is V3 divided by two (R1 and R2 form a 2:1 voltage divider) *plus* the instantaneous value of the input voltage Vm. In a negative feedback circuit (which this is), the output of the opamp does whatever it takes to move the - input so it is exactly the same voltage as the + input. The - input is connected to the top of Ri, so the opamp forces the input voltage to appear at Ri. Thus, the current through Ri is E ((V3/2) + Vm) / Ri. Adjusting R1 and R2 wil change the steady-state bias current through the LED.

This looks like an amplitude modulator for an optical audio link. Note that an LED's brightness is not a linear function of the current through it, so expect some harmonic distortion at the receiving end.

ak

#### Col John Matrix

Joined Apr 16, 2009
14
Thanks everyone for your replies, they're very helpful. I think it seems that the original article assumes V3 = 10 V, though this is not explicitly stated, and this is is largely where my confusion cam from.

I think I've got a good grasp of how it works now, I'm going to put it together and have a play.

Not homework, right?
Well, not formal set homework, rather some homework I'd set myself! I had a quick scan (too quick?) of the forums and decided this was the a fairly suitable place to post my question (or at least, not the worst place...). Will use more consideration in future!

Thanks again

#### dannyf

Joined Sep 13, 2015
2,197
There is very little change this thing will work to 50Mhz, or even 1Mhz.

You should be careful about any reactance, like small capacitance, on Ri.