To connect IC1 (DS3231M+ Real-Time Clock) to U3 (STM32F030C6T6 microcontroller), you need to make the following connections:

With these connections, the microcontroller (U3) can communicate with the Real-Time Clock (IC1) using the I2C protocol, control its reset, and make use of the interrupt/square wave output feature if needed.

Yes, the microcontroller (U3, STM32F030C6T6) appears to be wired correctly. Here are some of the connections:

Other control and monitoring pins like PA1, PA2, PA3, PA8, PA11, PA12, and PA13 are also connected appropriately. Overall, the microcontroller appears to be wired correctly based on the provided schematic.

I'm worried about BOOT0, what does that do on the MCU?

BOOT0 is a pin on the STM32F030C6T6 microcontroller used to select the boot mode during the startup process. By setting the BOOT0 pin state (either high or low), you can choose whether the MCU boots from its internal Flash memory, or an alternate boot source like system memory or bootloader for programming/firmware update purposes.

In the provided schematic, BOOT0 is connected to U3:PB8. For normal operation, you would generally connect BOOT0 to ground (low) so that the MCU boots from the internal Flash memory. However, if you ever need to update the firmware or use an alternate memory source, you can control the BOOT0 pin using the GPIO PB8 on the MCU. Keep in mind that this might require additional software configuration to ensure proper functionality.

thanks! I'll tie it low then

You're welcome! Tying BOOT0 low is a good choice for normal operation, as it ensures that the STM32F030C6T6 microcontroller boots from its internal Flash memory. If you need to update firmware or use an alternate memory source in the future, you can always modify the connections and software configuration to account for those changes. If you have any more questions or need further assistance, feel free to ask. Good luck with your project!

There could be several reasons for the -4V output on the op-amp cascade. Some possible causes include:

I recommend going through these steps to identify and resolve the issue.

i checked all of these and none are the problem. I'm seeing a 500kHz ripple on the output overlayed over the -4V output.

If you're seeing a 500kHz ripple on the output, it could be due to oscillation or noise coupling into the circuit. Here are some suggestions to address this issue:

Investigate these areas to identify and resolve the ripple issue.

what else could be causing the -4V on the output?

One possible cause of the -4V output could be output voltage phase-reversal, which occurs in some op-amps when the input common-mode voltage is exceeded. This issue is typically caused when one of the internal stages of the op-amp no longer has sufficient bias voltage across it and subsequently turns off, causing the output voltage to swing to the opposite rail until the input comes back within the common-mode range. This issue is most often associated with JFET and/or BiFET amplifiers, but some bipolar single-supply amplifiers are also susceptible to it. To address this issue, you may need to ensure that the input common-mode voltage is within the specified limits for the op-amp you are using. Source

To configure the OPA847 (U5) for a gain of 100 dB (~3162 V/V) at a bandwidth of 20 kHz, you can use a two-stage approach with the input biased at 4V. Here's how you can do it:

Note that this design assumes the OPA847 has sufficient bandwidth and can provide the required output voltage swing for your application. Be sure to check the datasheet for the specific device to ensure its performance characteristics meet your requirements.

would the OPA847 have enough bandwidth to achieve 50dB of gain at 20kHz?

The OPA847 has a gain-bandwidth product (GBP) of 3.9 GHz. To determine if the op-amp has sufficient bandwidth to provide 50 dB of gain at 20 kHz, you can calculate the required bandwidth for the desired gain and compare it to the GBP.

First, let's convert the gain from dB to a linear value:

Gain (linear) = 10^(Gain (dB) / 20) = 10^(50 / 20) = 316.2 V/V

Now, we can calculate the required bandwidth for the gain:

Required bandwidth = Gain * Signal bandwidth = 316.2 * 20,000 Hz = 6.324 MHz

The required bandwidth (6.324 MHz) is much lower than the OPA847's GBP (3.9 GHz), so the OPA847 should have sufficient bandwidth to achieve 50 dB of gain at 20 kHz. However, you should also ensure that the output voltage swing and other device characteristics meet your requirements by referring to the device's datasheet.

thanks copilot!

You're welcome, Mark! If you have any more questions or need further assistance, please don't hesitate to ask. I'm here to help!

@copilot, I just noticed my rail is 6V. How would I make a voltage divider to get 4V out?

To create a voltage divider that outputs 4V with a 6V input, you'll need to choose two resistors with the right ratio. The output voltage (Vout) for a voltage divider is determined by the following equation:

Vout = Vin * (R2 / (R1 + R2)),

where Vin is the input voltage (6V), R1 is the resistor connected to Vin, and R2 is the resistor connected to ground (GND). To find the ratio R2/(R1+R2) that will result in a 4V output:

R2 / (R1 + R2) = 4V / 6V = 2/3 ≈ 0.6667.

You can choose numerous resistor combinations to achieve this ratio, but some standard resistors to consider are R1 = 20 kΩ and R2 = 40 kΩ:

R2 / (R1 + R2) = 40 kΩ / (20 kΩ + 40 kΩ) = 40/60 ≈ 0.6667.

For this configuration, the output voltage will be:

Vout ≈ 6V * (40 kΩ / (20 kΩ + 40 kΩ)) ≈ 4V.

It's important to consider the input impedance of your op-amp stage so that the voltage divider output is not significantly affected. High impedance levels, typically in the MΩ range, are desirable. If necessary, use even larger resistor values for the voltage divider to achieve high input impedance while maintaining the desired ratio.

How much noise is at the output if I configure the op amp for 50dB of gain

To estimate the noise at the output of the op-amp configured for 50 dB of gain, you need to consider the various noise sources associated with the op-amp and the input network, such as input voltage noise, input current noise, and resistor thermal noise.

The OPA847 datasheet provides the following typical values for input voltage and current noise:

Assuming your input network consists of resistors RIN and RF, their thermal noise can also contribute to the overall noise. The thermal noise of a resistor can be calculated using:

Noise density = √(4 * k * T * R),

where k is Boltzmann's constant (1.38e-23 J/K), T is the temperature in Kelvin (assume room temperature, 298 K), and R is the resistance.

For RIN = 1 kΩ:

For RF = 316.2 kΩ:

The total input noise density can be estimated by adding the noise sources (op-amp and resistors) in quadrature:

Total input noise density ≈ √(1.1^2 + 4.07^2 + 22.6^2) ≈ 23.0 nV/√Hz

To get the total output noise, multiply the total input noise density by the gain:

Output noise density ≈ Gain * Total input noise density ≈ 316.2 * 23.0 nV/√Hz ≈ 7.26 µV/√Hz

To calculate the output noise over a specific bandwidth (e.g., 0 to 20 kHz), integrate the noise density over the frequency range:

Output noise (0 to 20 kHz) ≈ Output noise density * √(Bandwidth) ≈ 7.26 µV/√Hz * √(20e3) ≈ 324 µV

Keep in mind that these values are rough estimates based on typical noise densities provided in the datasheet. Actual values may vary depending on the specific device and operating conditions.