# How to choose resistor values in op-amps  Ok so my question is what resistor values should I choose for this amplifier. Vin is 0.5Vpp, 50kHz and the desired Vout is 25Vpp, so the overall voltage gain is 50. I know that the voltage gain for an inverting amplifier is -Rf/Rin. What's the difference between choosing resistor values like 50ohms and 1ohm or like let's say 50kohm and 1kohm. Also, what should my +-Vcc supply voltage to the amplifier be? The 2nd picture below is the op-amp's(LF351) datasheet Wondering if that could affect anything. Any help is much appreciated, thanks.

• Welcome to the forum! Inverting amplification pulls current from the signal line. It would be a good idea to buffer it with unity gain op amp and then amplify it with high value resistors (buffer will still have to output a bit of current, but it won't distort the signal itself anymore). Check op amp vide of eevblog on the youtube, it's one of the most helpful op amp vids ever
– Ilya
Jul 3, 2020 at 10:31
• Interesting, thanks for the suggestion will check it out! Jul 3, 2020 at 12:27
• @Ilya EE is not a forum. Jul 3, 2020 at 13:22
• I’d recommend a newer op amp with superior specifications, such as OPA197 Jul 3, 2020 at 14:21 And this graph just about permits a gain of 50 with a load of 2 kohm: - Regards resistor values, I'd be thinking of 22 kohm for the feedback resistor and about 440 ohms for the input resistor. You might be able to go a little higher and still get the 50 kHz bandwidth you need. Maybe as high as 220 kohm for the feedback resistor.

What's the difference between choosing resistor values like 50ohms and 1ohm or like let's say 50kohm and 1kohm.

The op-amp is inverting hence the inverting input is at 0 volts hence the output load IS the feedback resistor and you can't have this too low or you won't get the output voltage amplitude. On the other hand, you can't go too big because the parasitic capacitances of the op-amp will start to reduce gain too much at higher frequencies.

Regards numbers in the data sheet (rather than graphs), this is a summary on page 4: - The supply is listed as +/- 15 volts and the output swing is typically +/- 13.5 volts. This means that you can't push the output to typically within 1.5 volts of the positive or negative power rail but, this could be as bad as 3 volts if you take the minimum figure. The 1.5 volts and 3 volts numbers are saturation voltages and the load is 10 kohm.

• Interesting, how does the value of the load resistor affect my resistor values? Because the load I am driving is 2Kohms, a bit unclear on that. Jul 3, 2020 at 10:45
• Top left graph is for a 2 kohm load and bear in mind the feedback resistor makes the effective load smaller than 2 kohms. I'm sure you don't need me to hold your hand while looking at graphs. Jul 3, 2020 at 10:51
• I see. However I m still not sure how to approach choosing the resistor values Jul 3, 2020 at 10:59
• There are no hard and fast rules for any op-amp. You choose Rf to be as high as possible without causing the op-amp high frequency response to be poor and this depends on the actual op-amp BUT, older devices (like the really old but good LF351) don't give this info because they are targeted at audio applications. Folk use simulators now and these things can usually be determined with an accurate op-amp model but, given the pensioner status of the 351, it's unlikely models will be accurate enough. Jul 3, 2020 at 11:02
• Ahhh, also I see a lot of videos online that actually have a resistor connected to the non-inverting input, is it necessary and what does it do? Jul 3, 2020 at 11:43

I by default use 10,000 ohm resistor as basic value; thus 100Kohm would be the Rfeedback value for inverting gain_of_10. With 10 volt output, only 0.1mA (100uA) output current is needed. And the total resistive_contribution noise, assuming 1MHz UGBW and thus 100,000Hz opamp+resistors bandwidth because of gain of 10X, will be (approximately)

• 4 nanoVolt/rtHz * sqrt(10Kohm/1Kohm) * sqrt(100,000Hz BW) * Av=10

or

• 120nanoVolt * sqrt(100,000) = 120nV * 316 = 36 microVolts RMS

Thus the opamp is comfortable with an easy load. And the output noise is 36uV rms, spread over the entire bandwidth of 100,000 Hz.

The picking of resistor values is constrained by

• Operational Amplifier output current and heating; many opamps have a SHORT CIRCUIT protection feature, usually about 10mA to 20mA; read the datasheet

• random NOISE generated inside the resistors you choose; the wider the bandwidth, the more noise, which increases as the square_root of the bandwidth. The noise_power increases linearly with frequency; the noise_voltage increases as the square_root; remember Power = Voltage^2 / Resistance.

• the PCB parasitic capacitance and the OpAmp input capacitance will interact with the resistors you pick, to limit the bandwidth and to cause phase shifts that may cause ringing and poor settling and even oscillation.

==========================================

High resistor values introduce more feedback phase shift, where the C_input_differential and PCB parasitic capacitance will matter.

If frequency response peaking, and time_response ringing, and possibility of oscillation, do not make you nervous, then you can ignore this.

Solutions:

• An opamp with higher phase margin may be your friend.

• remove any ground plane under or near the summing node PCB foil; that node is the Vin- of the opamp

• move all components of summing node very close to Vin-, so minimal PCB area is used

• enlarge the solder pads of the Rfeedback, and have them just barely separated (almost shorting), so the electric fields across the Rfeedback are maximum

• have a tiny capacitor across Rfeedback (some people use trimmer-caps; you could use a "gimmick" cap made of 2 wires twisted together (insulated from each other)

• replace the Rfeedback with a "T" attenuator of much lower Rvalues

• pick an opamp with lower input capacitance (some opamp datasheets do not provide this information)