# Unity Gain Amplifier - How To Determine Appropriate Resistors For Op-Amp?

#### Mahonroy

Joined Oct 21, 2014
406
Hello, I have an 80 MHz oscillator connected up to 2x op-amps in a unity gain configuration. How do you determine the appropriate resistance values for the resistors? (green arrows below). From what I am reading, it seems people tend to put an arbitrary resistance here, e.g. 330 ohm, 1K, etc.?

#### crutschow

Joined Mar 14, 2008
34,842
The resistors are arbitrary and optional for a unity-gain follower circuit.
The circuit should work okay without them.

But why are you using an analog op amp to buffer a digital oscillator signal?

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#### dl324

Joined Mar 30, 2015
17,131
op-amps in a unity gain configuration. How do you determine the appropriate resistance values for the resistors?
The resistors are used to cancel effects from input bias currents. The one in the feedback path should equal the source impedance.

In many cases, they're not used.

#### crutschow

Joined Mar 14, 2008
34,842
The resistors are used to cancel effects from input bias currents.
But in a unity-gain follower that is not a factor as no resistors give the least output voltage offset effect from the bias currents.

#### dl324

Joined Mar 30, 2015
17,131
But in a unity-gain follower that is not a factor as no resistors give the least output voltage offset effect from the bias currents.
If you don't match the impedances on the inputs, input bias currents can affect the output.

This is from a 1979 Signetics Analog Applications manual.

#### crutschow

Joined Mar 14, 2008
34,842
If you don't match the impedances on the inputs, input bias currents can affect the output.
True.
But that means the feedback resistance should match the source resistance, not that you add a source resistance.

#### Audioguru again

Joined Oct 21, 2019
6,783
The OPA356 opamp is Cmos so its input bias current is very close to zero. It is a high frequency video opamp that almost oscillates at 300MHz if the feedback resistor is a piece of wire as shown in its datasheet.

#### Mahonroy

Joined Oct 21, 2014
406
But why are you using an analog op amp to buffer a digital oscillator signal?
I have 2 reasons for this, please let me know if I am incorrect here:

• The buffered signal needs needs to support 27.5 mA (its 3.3V, and I have a 120 ohm resistor, so a dead short would be 27.5mA). I wanted to use a buffer gate, but they tend to be just under this mA rating. E.g. a lot of them advertise 32mA output, but then in the datasheet, looking at 3.3V its actually closer to 24mA output. Plus it would be nice to have some headroom. So it seemed like maybe this wasn't the way to go.
• Also regarding the buffer gates, the fastest propagation delay I could find was between 2.5nS and 5.5nS. An 80MHz signal is 6.25nS high, and 6.25nS low. So I was worried I would get a distorted signal and that these buffer gates are not fast enough?

#### crutschow

Joined Mar 14, 2008
34,842
The buffered signal needs needs to support 27.5 mA (its 3.3V, and I have a 120 ohm resistor, so a dead short would be 27.5mA). I wanted to use a buffer gate, but they tend to be just under this mA rating.
Then use two buffer gates on the same chip in parallel.
Gates on the same chip should be well enough matched, so there's no significant conflict if their outputs are in parallel.
Also regarding the buffer gates, the fastest propagation delay I could find was between 2.5nS and 5.5nS. An 80MHz signal is 6.25nS high, and 6.25nS low. So I was worried I would get a distorted signal and that these buffer gates are not fast enough?
The OPA356 has a rise and fall time of about 10ns (the delay is not shown) which is longer than a fast gate.
And the propagation delay doesn't distort the signal, just delays it slightly.

#### tindel

Joined Sep 16, 2012
938
Lots of reasons to place those resistors or not. I'll add another.

A 80MHz clock will have 800MHz+ edges. At these frequencies I've often used resistors on the output of digital gates (50-300ohm) to reduce the signal bandwidth and ringing/overshoot. This has the opposite effect in the feedback of an op-amp circuit due to a pole in the feedback loop being shifted to a lower frequency and potentially destabilizing the amplifier. Tread cautiously.

I would also likely use a digital gate. The issue here isn't usually prop-delay, but gate-to-gate phase shift which will be low (sub-ns) if using a single chip. A phase shift from the source to the gate output usually doesn't matter.

If you haven't already realized, you'll need to use transmission lines if you need to keep 800MHz bandwidth and signals are traveling more than ~1".

A comparator may also be a good solution... not an opamp.

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#### Mahonroy

Joined Oct 21, 2014
406
Thanks again for the info, I decided to switch over to using 2x of these bus driver chips, and now no resistors:
https://www.digikey.com/en/products/detail/nexperia-usa-inc/74LVC1G125GV125/946686

Lots of reasons to place those resistors or not. I'll add another.

A 80MHz clock will have 800MHz+ edges. At these frequencies I've often used resistors on the output of digital gates (50-300ohm) to reduce the signal bandwidth and ringing/overshoot. This has the opposite effect in the feedback of an op-amp circuit due to a pole in the feedback loop being shifted to a lower frequency and potentially destabilizing the amplifier. Tread cautiously.

I would also likely use a digital gate. The issue here isn't usually prop-delay, but gate-to-gate phase shift which will be low (sub-ns) if using a single chip. A phase shift from the source to the gate output usually doesn't matter.

If you haven't already realized, you'll need to use transmission lines if you need to keep 800MHz bandwidth and signals are traveling more than ~1".

A comparator may also be a good solution... not an opamp.
Can you elaborate on "transmission lines" and signals that are traveling more than ~1"? What do you mean?

#### tindel

Joined Sep 16, 2012
938
Transmission lines are traces that have matched impedance over the entire length of the trace to maintain signal integrity. Physics dictates that as you approach 1/10 of the wavelength of the highest frequency of the signal that you start having reflections of the rising edge traversing back and fourth along the trace, not unlike splashing around in the bathtub and seeing waves hit one wall and reflecting and hitting the opposite wall of the tub. At 800MHz in FR4, your maximum length to travel is ~1" before violating this law of physics.

In electronics systems the driver usually has a 50ohm output, driving a 50ohm impedance transmission line, and is terminated by a 50ohm shunt to ground at the receiver.

Watch this #143: Transmission Line Terminations for Digital and RF signals - Intro/Tutorial for a much better explanation

#### crutschow

Joined Mar 14, 2008
34,842
Note that the PCB characteristic impedance of a controlled impedance trace over a ground plane is typically around 100 ohms.
You then need to terminate that in a 100Ω resistance to minimize reflections.