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."
Thanks!
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.
Thanks!
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