2
\$\begingroup\$

I come from this post Guitar Pre-Amp stage/filtering and STM32F4. I found this circuit below in the answer by Dave Branton.

Why does it include 2 clamps, one at the input and one at the output?

schematic

Moreover, I have simulated the circuit in PSPICE and even it suits perfectly fine with my purpose. I don't understand the use of the diodes. At the input we have the voltage divider along with the diodes, and I understand that adds 4.5VDC to the input signal. But, if I remove D4, the simulation is the same, and what I learned is that the main diode in a clamp is that diode connected to ground, the one connected to VCC is only to protect the first one.

Can anyone clarify this for me?

\$\endgroup\$
0

2 Answers 2

Reset to default
6
\$\begingroup\$

The diodes are over/under-voltage protection devices. Typical op-amp inputs can't be driven more than 0.5 V greater than the positive supply rail or more than 0.5 V below the negative supply rail. This includes when the device is powered down.

Guitars generally output signals of a 100 mV to a volt or so but people do crazy stuff and could, for example, plug an amplifier's output into the input of this device. As you have noticed, the op-amp input is biased to 4.5 V so that means that in input signal of > ±5 V would drive the op-amp input into the danger region. D3 and D4 clamp the signal to prevent damage to the op-amp.

Meanwhile the pre-amplifier has a gain of about 10 but is feeding an ADC with, effecively, a range of ±1.65 V. Therefore an input of ±165 mV would cause the ADC to go into clipping. D1 and D2 limit the over/under voltage with the help of R8.

You should be able to prove this by increasing the magnitude of the input signal. The clipping should be visible on the ADC output in particular.

\$\endgroup\$
4
\$\begingroup\$

Most circuits have limits you can't exceed.

Since the op-amp input can come from anywhere, it must be limited to be within supply voltage, even when supply is off.

But the 10k series resistor effectively prevents that, although it takes some effort for the signal to charge the 10uF cap first.

The output is from 0V to 9V, and the MCU input needs to be limited to 0V and 3.3V. So it needs another set of diodes to clamp to MCU supplies.

\$\endgroup\$
4
  • \$\begingroup\$ Thank you for you answer! So what is clamping here it is actually the voltage divider and D3?? If I remove D4 in PSpice the result is the same. \$\endgroup\$
    – jjavi66
    Commented Dec 27, 2023 at 22:40
  • \$\begingroup\$ @jjavi66, try overloading the input with a large signal - maybe 12 V peak-to-peak. \$\endgroup\$
    – Transistor
    Commented Dec 27, 2023 at 22:42
  • \$\begingroup\$ thank you for you comment @Transistor with 5v amplitude it behaves as I am used to see a clamp, now I undestand the voltage resistor is meant for clamping low inputs and the diodes start to act when the input is higher, am I right?? \$\endgroup\$
    – jjavi66
    Commented Dec 27, 2023 at 22:47
  • \$\begingroup\$ The resistors set the bias voltage. The upper diodes conduct if the voltage is too high, the lower diodes conduct if the voltage is too low. \$\endgroup\$
    – evildemonic
    Commented Dec 28, 2023 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.