Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Introducing Easy Pulse: A DIY photoplethysmographic sensor for measuring heart rate

Status
Not open for further replies.
Hi Nishanth !!
It's a classic (non inverting) integrator. For high frequencies the capacitor acts as an open circuit, while the resistor provides the needed feedback in DC to make it stable.

Best regards
 

How does the capacitor acts as an open circuit at high frequency?

I didn't get.
 

How does the capacitor acts as an open circuit at high frequency?

I didn't get.

Hi Nishanth, the capacitor acts as an open circuits because it's capacitive reactance (Something like It's resistence in AC) depends of the frequency of the signal in AC: \[X_{c} = \frac{1}{w{\cdot}C}=\frac{1}{2{\cdot}{\pi}{\cdot}f{\cdot}C}\]. As the frequency gets high in the previous equation also the admitance of the capacitor gets high.

Best regards
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top