I am using a high power op amp (PA194) in the non inverting manner. It has a gain of 70. I used it with a single feedback resistor, and it worked. However, i had to change it to multiple resistors in series to take care of the wattage issue when using just one resistor. When I make the change to multiple resistors in series, it begins to suck more current and the op amp just oscillates at around 1Mhz. It works, but the high frequency oscillation is present. When I change back to just 1 feedback resistor, the issue goes away and everything is happy again. I did some searching it signs point to parasitic capacitance....what else could it be?

 

There's a good chance that your research was correct!
If such a small change can tip the circuit into instability, then with the single resistor it's almost there.
Do you have enough power supply decoupling?

 

I am using a high power op amp (PA194) in the non inverting manner. It has a gain of 70. I used it with a single feedback resistor, and it worked. However, i had to change it to multiple resistors in series to take care of the wattage issue when using just one resistor. When I make the change to multiple resistors in series, it begins to suck more current and the op amp just oscillates at around 1Mhz. It works, but the high frequency oscillation is present. When I change back to just 1 feedback resistor, the issue goes away and everything is happy again. I did some searching it signs point to parasitic capacitance....what else could it be?

Three words; schematic, schematic, schematic.

 

When I make the change to multiple resistors in series to take care of the wattage issue

Three words; schematic, schematic, schematic.

If your concern is that there is too many watts for one resistor, then you should use higher resistance resistors in parallel such that their equivalent resistance is what your circuit needs.

Your circuit needs a 200Ω resistor (for example only). Consider two 100Ω resistors in series. Each one will have be rated the same wattage as one 200Ω resistor because the same current flows in series through both resistors.

Now, two 400Ω resistors in parallel is equivalent to 200Ωs. But they only need to be rated at half the wattage than a single resistor. Because the current splits between the two parallel paths.

I may not completely understand where you are putting multiple resistors. That I’d why I quoted Papabravo, “Schematic, schematic, schematic.”

 

If the opamp circuit oscillates when a string of resistors adding up to the same resistance is used, then the problem is not simple at all. But first, why is there a lot of current in the resistors? I am not familiar with the PA194 device but it does not seem like an opamp should require that much input current. It seems that 200 ohms is a very low feedback resistance, so I am asking about what else is connected to the inverting input?
It may be that the physical arrangement of a string of resistors is providing excessive capacitance to the non-inverting input. or, as already stated, bypass capacitors between the power supply terminals and the system common may not be adequate.
So, if not an actual circuit schematic, a description of the circuit and the various supply voltages. Unlike some others, I can visualize a circuit from the detailed description fairly well.

 

If the opamp circuit oscillates when a string of resistors adding up to the same resistance is used, then the problem is not simple at all. But first, why is there a lot of current in the resistors? I am not familiar with the PA194 device but it does not seem like an opamp should require that much input current. It seems that 200 ohms is a very low feedback resistance, so I am asking about what else is connected to the inverting input?
It may be that the physical arrangement of a string of resistors is providing excessive capacitance to the non-inverting input. or, as already stated, bypass capacitors between the power supply terminals and the system common may not be adequate.
So, if not an actual circuit schematic, a description of the circuit and the various supply voltages. Unlike some others, I can visualize a circuit from the detailed description fairly well.

It's high voltage. That is why. It has a gain of close to 70 so a 8V signal will output over 500V, so you need a high watt resistor. The feedback resistor needs to be able to handle the over 500V signal to the 8V signal on the inverting terminal. Not sure where the 200 ohms is coming from. My feedback is 65k, then I tried doubling and changing the resistor to ground to compensate. I suppose I could try 120k, but I do not have a high wattage resistors that go that high. It is just a non inverting opamp with a 700V postive supply and a -100V negative supply.

 

Not sure where the 200 ohms is coming from.

My post explicitly stated that I was using 200Ω for an example only. That’s where it came from.

 

OK, clearly this is not a simple small op amp. Thus the difficulty increases with the voltage.
And still the scheme is voltage divider feedback. So possibly much of that current is not into the inverting input, but rather into the second resistor.
One suggestion is to check with the manufacturer and their applications information person, because they may have a solution.
Is the controlling signal fed into the inverting input, or the non-inverting input?

 

Do you use a compensation capacitor on the Cc pin? A schematic would really help.

 

Your circuit needs a 200Ω resistor (for example only). Consider two 100Ω resistors in series. Each one will have be rated the same wattage as one 200Ω resistor because the same current flows in series through both resistors.

Now, two 400Ω resistors in parallel is equivalent to 200Ωs. But they only need to be rated at half the wattage than a single resistor. Because the current splits between the two parallel paths.

In both cases, serial and parallel, it's 1/2 the wattage. In the first case because V is halved across each resistor and in the second case because I is halved through each resistor.

 

evidently there is a lot more to the problem than we see directly, because an opamp will not oscillate unless the phase margin is wrong. That part will not change simply because of the physical number of resistances in the feedback portion.. Perhaps one of the resistors is wirewound, and the inductance is having an effect. I have seen many cautions about avoiding wirewound resistors because of their inductance.
So possibly the presence or absence of a wirewound resistor is the cause of the problem. That would be a logical reason, given the small amount of information that we have. And it may possibly be due to one of the series devices not being exactly as marked.

and still my question stands: Why use a voltage divider for the feedback portion, when certainly the part of the heating current in the feedback resistor is due to the resistor from the inverting input to the common?

 

As requested, please post a schematic. It can be on typing paper drawn with a crayon, we just need to see the circuit in as fine a detail as you can show.

 

Even at 65K your feedback resistor may be too small. with an output if 500V that would be 7mA of current. The current in the divider only needs to be large compared to the the input bias current, which is hopefully in the nanoamp region.

So, in addition to the schematic, we need the datasheet of your opamp.

 

Even at 65K your feedback resistor may be too small. with an output if 500V that would be 7mA of current. The current in the divider only needs to be large compared to the the input bias current, which is hopefully in the nanoamp region.

So, in addition to the schematic, we need the datasheet of your opamp.

The 65K value can be used because the feedback is a voltage divider, not just a simple resistance back to the inverting input. AND that is probably why the heating happens.

 

The 65K value can be used because the feedback is a voltage divider, not just a simple resistance back to the inverting input. AND that is probably why the heating happens.

And a voltage divider can use higher value resistors, depending in the current that will be drawn from the junction. Unless, of course, the other resistor cannot be changed..

 

There was a statement about an "eight volt input", which somehow affected the resistor value, and so once again there is not enough information to adequately evaluate what is happening. And eight now I am guessing that if one resistor is wirewound that inductance could cause a problem with stability.