TIA Optimal Bandwidth
f−3dB = √(fGBW / (2π · Rf · Cin))
Quick answer
This calculator solves f−3dB = √(fGBW / (2π × Rf × Cin)) for Bandwidth. With Gain-Bandwidth 10 MHz, Feedback Resistor 2.2 MΩ, Input Capacitance 100 pF, the result is 85.05 kHz. It solves in 4 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
When a transimpedance amplifier is compensated for maximum flatness, its achievable closed-loop bandwidth is the geometric mean of the amplifier’s gain-bandwidth product and the feedback-network pole. This sets the speed limit of a photodiode front end: higher gain (larger Rf) or more input capacitance reduces bandwidth, while a faster op-amp raises it.
Variables
- f−3dB — Closed-loop bandwidth (Hz)
- fGBW — Amplifier gain-bandwidth product (Hz)
- Rf — Feedback resistor (Ω)
- Cin — Total input capacitance (F)
Practical Notes
Bandwidth scales only as the square root of GBW, so doubling speed needs 4× the GBW. There is a direct gain-bandwidth trade: raising Rf for more transimpedance gain lowers the bandwidth.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| f3dB | Bandwidth | kHz | Hz, kHz, MHz, GHz |
| fGBW | Gain-Bandwidth | MHz | Hz, kHz, MHz, GHz |
| Rf | Feedback Resistor | MΩ | mΩ, Ω, kΩ, MΩ |
| Cin | Input Capacitance | pF | pF, nF, µF, mF, F |
Step by step
- 1. What you know: Gain-Bandwidth 10 MHz, Feedback Resistor 2.2 MΩ, Input Capacitance 100 pF
- 2. The relation: f−3dB = √(fGBW / (2π × Rf × Cin))
- 3. Substitute: fGBW = 10 MHz, Rf = 2.2 MΩ, Cin = 100 pF
- 4. Result: Bandwidth = 85.05 kHz
Worked examples
Minimum range
Given: Gain-Bandwidth 100 kHz, Feedback Resistor 100 kΩ, Input Capacitance 1 pF
Result: Bandwidth = 0.3989 MHz
Low range
Given: Gain-Bandwidth 1 MHz, Feedback Resistor 1 MΩ, Input Capacitance 10 pF
Result: Bandwidth = 126.2 kHz
Typical
Given: Gain-Bandwidth 10 MHz, Feedback Resistor 2.2 MΩ, Input Capacitance 100 pF
Result: Bandwidth = 85.05 kHz
High range
Given: Gain-Bandwidth 100 MHz, Feedback Resistor 10 MΩ, Input Capacitance 1 nF
Result: Bandwidth = 39.89 kHz
Maximum range
Given: Gain-Bandwidth 1 GHz, Feedback Resistor 100 MΩ, Input Capacitance 10 nF
Result: Bandwidth = 12.62 kHz
Solving for each variable
Solve for Bandwidth
Gain-Bandwidth 100 kHz, Feedback Resistor 100 kΩ, Input Capacitance 1 pF → 0.3989 MHz
Solve for Gain-Bandwidth
Bandwidth 100 Hz, Feedback Resistor 100 kΩ, Input Capacitance 1 pF → 0.006283 Hz
Solve for Feedback Resistor
Bandwidth 100 Hz, Gain-Bandwidth 100 kHz, Input Capacitance 1 pF → 1.59 × 10⁶ MΩ
Solve for Input Capacitance
Bandwidth 100 Hz, Gain-Bandwidth 100 kHz, Feedback Resistor 100 kΩ → 15.92 µF
How Bandwidth changes with Gain-Bandwidth
| Gain-Bandwidth (input) | Bandwidth (result) |
|---|---|
| 1 MHz | 26.9 kHz |
| 2.5 MHz | 42.53 kHz |
| 5 MHz | 60.14 kHz |
| 7.5 MHz | 73.66 kHz |
| 10 MHz | 85.05 kHz |
| 15 MHz | 104.2 kHz |
| 20 MHz | 120.3 kHz |
| 40 MHz | 0.1701 MHz |
| 70 MHz | 0.225 MHz |
| 100 MHz | 0.269 MHz |
The same result in other units
| Unit | Value |
|---|---|
| Hz | 85050 |
| kHz | 85.05 |
| MHz | 0.08505 |
Embed this calculator
Paste this into any page. It keeps working when this page updates, and credits the source.
<iframe src="https://vbatpower.com/embed/tia-bandwidth-optimal/" width="100%" height="460" style="border:1px solid #1f2a37;border-radius:12px" loading="lazy" title="TIA Optimal Bandwidth"></iframe>
<p><a href="https://vbatpower.com/formulas/tia-bandwidth-optimal/">TIA Optimal Bandwidth</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate TIA Optimal Bandwidth?
Use f−3dB = √(fGBW / (2π × Rf × Cin)). With Gain-Bandwidth 10 MHz, Feedback Resistor 2.2 MΩ, Input Capacitance 100 pF, Bandwidth works out to 85.05 kHz.
What is the formula for TIA Optimal Bandwidth?
The formula is f−3dB = √(fGBW / (2π × Rf × Cin)). When a transimpedance amplifier is compensated for maximum flatness, its achievable closed-loop bandwidth is the geometric mean of the amplifier’s gain-bandwidth product and the feedback-network pole.
How do you find Gain-Bandwidth when you know the other values?
Rearrange for Gain-Bandwidth: with Bandwidth 100 Hz, Feedback Resistor 100 kΩ, Input Capacitance 1 pF, Gain-Bandwidth is 0.006283 Hz.
How do you find Feedback Resistor when you know the other values?
Rearrange for Feedback Resistor: with Bandwidth 100 Hz, Gain-Bandwidth 100 kHz, Input Capacitance 1 pF, Feedback Resistor is 1.59 × 10⁶ MΩ.
What units does TIA Optimal Bandwidth use?
Bandwidth is shown in Hz, kHz, MHz. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
Related Concepts
All Op-Amps formulas →Need more features?
Save calculations, import telemetry data, simulate battery discharge, and collaborate with your team.
Try the App