Choice of resistor value for unit gain amp affecting output

Thread Starter

samihawasli

Joined Oct 19, 2020
8
Hello,

I am trying to create a steady DC output to drive some pHEMTS, HMC7357. To control the gates I need to produce -1.4, and -0.8V and supply about 0.5-1 mA. To do so, I was planning on using a filtered PWM output from a microcontroller, followed by a -1 inverting op-amp circuit. I was playing around in LT-spice and i got some results I didn't expect. Perhaps someone can help me understand what is happening and why.
Capture.PNG
Above is a screen capture of my schematic and simple transient analysis. I am using a op-amp model from the TI website for the LM358. The input is a 500 Hz 50% duty cycle square wave with a magnitude of 5V. I am not plotting the input on the graph because the X axis time scale is long. The circuit acts as one expects, the PWM signal is filtered at each stage, V1,V2 and the output, Vout, is a steady -2.5V.
Capture2.PNG
However, when i change the value of the feedback resistors, now 1K instead of 1 Meg, the value of the output changes drastically. From what I understand about IDEAL op-amps it shouldn't really matter what value of feedback resistors I use? clearly I am missing something fundamental to op-amps.

BONUS question: I tested the first circuit, with 1Meg feedback, on a breadboard. Without the pHEMT connected, I got the -1.4 and -0.8V as simulated and expected. Connecting the circuit to the pHEMT's gate:
With no driving power to the pHEMT, meaning virtually no current DC to the gate, the outputs work as expected.
When I apply a RF power, ~20 dBm causing at most 1mA gate current, the PWM output from the AVG drops, subsequently dropping both of my voltages, expected -0.8V to -0.7V and -1.4 to -1.3! None of my circuit elements should be drawing much, if ANY current, does anyone have any clues?

Happy to provide and more context,
Thanks for any and all help,
Sami
 

AlbertHall

Joined Jun 4, 2014
10,558
Because of the negative feedback, the inverting input of the amplifier is a virtual groud - at 0V.

In the first circuit you have 2x 10k resistors and then a 1M resistor connected to this 0V point.
In the second circuit the 2 x 10k resistors have just 1k connected to 0V. This severely loads down the signal from the input resistors.
 

Thread Starter

samihawasli

Joined Oct 19, 2020
8
Because of the negative feedback, the inverting input of the amplifier is a virtual groud - at 0V.

In the first circuit you have 2x 10k resistors and then a 1M resistor connected to this 0V point.
In the second circuit the 2 x 10k resistors have just 1k connected to 0V. This severely loads down the signal from the input resistors.
Was my understanding that the inputs to op-amps draw virtually no current? With no current no voltage drop?
 

AlbertHall

Joined Jun 4, 2014
10,558
The amplifier inputs draw near zero current but are we agreed that the inverting amplifier input is at zero volts?
See https://www.electronics-tutorials.ws/opamp/opamp_2.html
1603141846550.png
And that current i must flow through the 10k input resistors and will drop voltage across those resistors.
Or to put it another way, Rin in the above diagram consists of 2x 10K + 1k = 21k. The feedback resistor is 1k. There fore the gain is -1/21.
 

Audioguru again

Joined Oct 21, 2019
2,061
The input two 10k resistors feeding your 1k opamp input resistor attenuate the input level to 1/21 times which is 4.76%.
The inputs of opamps draw virtually no current, but the 1k resistor does.
 

Ian0

Joined Aug 7, 2020
750
The input doesn't draw any (significant amount of) current.
If V2 is at 1.4V and the output is at -1.4V, then a current will flow between the two points, equal to 2.8V divided by 2k.
The inverting input is therefore halfway between 1.4V and -1.4V, which is 0V. It's at 0V even though it's not connected to 0V. That's why it's called a "virtual earth".
 

Ian0

Joined Aug 7, 2020
750
Hello,

I am trying to create a steady DC output to drive some pHEMTS, HMC7357. To control the gates I need to produce -1.4, and -0.8V and supply about 0.5-1 mA. To do so, I was planning on using a filtered PWM output from a microcontroller, followed by a -1 inverting op-amp circuit. I was playing around in LT-spice and i got some results I didn't expect. Perhaps someone can help me understand what is happening and why.
I'd suggest you use a MFB filter.
GO here: http://sim.okawa-denshi.jp/en/OPtazyuLowkeisan.htm and scroll down to Calculate the R and C values for the multiple feedback filter at a given frequency and Q factor. Type in about 50Hz in and leave the Q at 0.7.
 

Thread Starter

samihawasli

Joined Oct 19, 2020
8
I'd suggest you use a MFB filter.
GO here: http://sim.okawa-denshi.jp/en/OPtazyuLowkeisan.htm and scroll down to Calculate the R and C values for the multiple feedback filter at a given frequency and Q factor. Type in about 50Hz in and leave the Q at 0.7.
Thanks, that is a good solution to my problem. One more follow up question if you don't mind.
Capture.PNG
Using components I had available, I built and simulated the circuit above. It worked as expected, but when I connected it to the gates of the amplifier, something rather strange happened. For reference, Vdrain for the amplifier is 6V.
1) Controlling the amp's gate using a benchtop power supply, applying a gate voltage Vg = -1.4 yields a drain current Id = 13mA. Vg = -1.1 yields Id = 239mA, and Vg = -0.8 yields Id = 513mA
2) Controlling the amp's gates using the above circuit and PWM inputs; output voltages similar to voltage levels in step 1 yields Id = 20mA, 427mA, and 1093mA, almost twice the current at each current level!
I hope that makes sense, any clue as to why this might be happening?
Thanks in advance
 

Ian0

Joined Aug 7, 2020
750
I don't know anything much about HEMT amplifiers.
All I can suggest is:
1) there might be some ripple remaining from the PWM, in which case you could use a higher PWM frequency or a lower frequency filter
2) 1k is quite a large load for a microcontroller output, which has an output resistance of around 40 ohms. You could increase the impedances in the filter by multiplying all the Rs by 10 and dividing all the Cs by 10. That leaves the filter frequency the same but increases the input impedance to 10k. (But then the 100k resistor might cause offset voltage due to the op-amp bias currents - that depends on the op-amp)
 
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