Electronics > Repair
PV inverter - Voltage reading circuit
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fzabkar:
The AC voltage at the junction of the 10M and 10K resistors should be …

10K / (10K + 10M + 10M) x 240Vrms = 120mVrms
webgiorgio:
Sorry, it is ST, not Texas Instruments!
The marking has the ST logo, then "24I", and line below is "EZX6111". That's so strange that it does not come up anywhere. Attached photo is the most left chip.
The 8 pin is http://www.ti.com/lit/ds/symlink/opa2277-ep.pdf


My scope measurement on the L-N voltage between the 10k-10M node is showing 3 divisions times 250 mVpp (see earlier post).
250 mV * 3 /2 /sqrt(2) = 265 mV rms.
fzabkar:
The voltage at the junction of the two 10M resistors should be half the AC output. Is it?

An input voltage of 265mV rms (750mVpp) would drive the op amp output to each of its supply rails., ie it would clip. :-?

AISI, Vout and V+/V- should all be biased at 1.65VDC. I suspect that something is amiss at the op amp's V- pin.
webgiorgio:
Not half because of the 10 k resistors (and the op-amp need to be not saturated).

If the V+ equals V-, then it is like we have the 10 k connected together.
The voltage divider looks like 10M 10M 10k - 10k 10M 10M.

In theory the voltage I should see on the scope across the 20 k is:
V_LN*2*sqrt(2) * (20k / 20020 k) = 650 mVpp
I measure about 750 mVpp, so, it is correct (on the V_LN channel)

On the bad channel I see less, probably because the bad O-Amp is not producing an output that makes V- equal to V+. A bad op-amp or something else.
fzabkar:
I mistakenly thought that each of the L & N inputs were going to separate op amps. Now, IIUC, the two inputs are being sensed by a single op amp configured as a differential amp. If so, then the gain should be the same as I calculated previously, but you should be measuring the differential input between the two 10K-10M junctions and then comparing this against the L-to-N reading (Sorry, I see you have already done this). Am I right to assume that the op amp's 0V reference is floating midway between the L & N potentials?
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