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What is the output voltage of the circuit(L7987) below? Please tell me how I can set the output voltage of L7987.
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The data sheet says about the function of the feedback in section 4:
The L7987 device is based on a voltage mode, constant frequency control loop. The output voltage VOUT, sensed by the feedback pin (FB), is compared to an internal reference (0.8 V) providing an error signal on the COMP pin. The COMP voltage level is then compared to a fixed frequency sawtooth ramp, which finally controls the on- and off-time of the power switch.
In the circuit, the voltage divider to derive VFB from the output voltage VDCDC consists of resistors R19 (51k) and R21 (4.02k). Please do not be confused by the components C19/R20, which together with R14, C40/44 form a Type III Compensation Network (data sheet, section 5.4.2).
The voltage VFB at FB is 0.8V in stable state (data sheet, table 5, Error Amplifier).
This results in the voltage VDCDC = VFB * (R19/R21+1) = 10.95V, i.e. about 11V.
You can adjust the output voltage (here VDCDC) accordingly, e.g. by calculating the resistor R19 according to a simple voltage divider, which divides VDCDC down to the 800mV. So if you want to generate e.g. 15V you would have to set R19 = (VDCDC/VFB-1) * R21, so with R21=4.02k, R19 should be set to 71.36k.
Does it answer your question?
Regards
/Peter
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