Hello,

I have been reading various threads in this forum for at least a year. I replied to one thread once. I'm not sure exactly how to format my post. I hope some kind person here may be able to help me.

I have built this circuit:



When nothing is connected to JP1 the output of IC1A oscillates. The output is approximately a square wave with a frequency of 7 kHz to 10 kHz depending on e.g. op amp type.

There is a lot more information about what I am trying to do and the details of the problem I am having in the attached pdf. Sections 1 and 2 are background information. Section 3 is about the current problem I am having.

I would like to know why it oscillates and how to prevent it doing so.

Thanks in advance for any help.
Dave

 

Your logbook shows that many opamp part numbers oscillate almost the same.
Maybe because your + and - power supply connections on the perf board are not decoupled to ground?
Note that most opamps oscillate if the capacitance of a shielded output cable is connected to their output.

 

Your logbook shows that many opamp part numbers oscillate almost the same.
Maybe because your + and - power supply connections on the perf board are not decoupled to ground?
Note that most opamps oscillate if the capacitance of a shielded output cable is connected to their output.

Thanks for the quick reply Audioguru. I'll try changing the decoupling so that both rails are decoupled to ground. I think I should also include a resistor between the output pin of the op amp and output from the board. I will do some reading to check what value it should be. It may be a week or two before I can do more work on this as I have 3 grandkids coming to stay tomorrow for a week.

Thanks again for your help.
Dave

 

Don't the opamp inputs need a DC path to GND? The 2M resistor R2 looks more like an open for small currents.

 

Don't the opamp inputs need a DC path to GND? The 2M resistor R2 looks more like an open for small currents.

Hi Papabravo,
Thanks for your reply. It's 2M because I want a high impedance input similar to an oscilloscope. I figure that the input impedance will be approx 1M with R1 and R2 in parallel (ignoring the op amp input impedance). The imbalance in the impedances for bias currents to + and - inputs caused me to add C2. Without C2 the op amp output goes to about -13.6V with the input (JP1) shorted. With C2 fitted the offset at the output is about -0.4V which I think is acceptable other than it would cause asymmetrical clipping if I wanted to have such high levels (I don't need more than about 5V RMS output).
Best regards
Dave

 

Hi Papabravo,
Thanks for your reply. It's 2M because I want a high impedance input similar to an oscilloscope. I figure that the input impedance will be approx 1M with R1 and R2 in parallel (ignoring the op amp input impedance). The imbalance in the impedances for bias currents to + and - inputs caused me to add C2. Without C2 the op amp output goes to about -13.6V with the input (JP1) shorted. With C2 fitted the offset at the output is about -0.4V which I think is acceptable other than it would cause asymmetrical clipping if I wanted to have such high levels (I don't need more than about 5V RMS output).
Best regards
Dave

@Dave Lowther
All the components (and their leads) connected to pin3 of the opamp make an excellent antenna for receiving a signal sent by pin1 of the opamp (whenever it happens to swing from low-to-high or high-to-low) and this could easily lead to oscillation. Keep all leads as short as possible and keep components connected to pin1 well separated/shielded from components connected to pin3. It's possible that a small capacitance (10pF - 1nf) from pin2 to pin1 could be useful. However, the larger the capacitance, the less high frequency response will be provided by the opamp.

With JP1 shorted and C1 not added, you see saturation at the output of the opamp? The input bias current is less than 100nA and thus the drop across R3 (due to bias current) is less than 0.001V. With C2 blocking current to ground, the DC closed loop gain of the opamp should be ~1. Even with the bias current flowing through the 10K feedback, the opamp output should be close to 0V. It seems that something else is going on. Board leakage, miswire, mismarked R, ?

Finally, be aware that zener diodes have a very large junction capacitance (in this case from the inverting input to ground). Zeners also have large leakage currents below their breakdown voltage. You might consider replacing the zeners with a low leakage signal diode from pin2 to +15V(cathode to +15V) and another such diode from pin2 to -15V (anode to -15V); equivalent protection with less capacitance and less leakage current (depending on which diode you choose).

 

@Dave Lowther
All the components (and their leads) connected to pin3 of the opamp make an excellent antenna for receiving a signal sent by pin1 of the opamp (whenever it happens to swing from low-to-high or high-to-low) and this could easily lead to oscillation. Keep all leads as short as possible and keep components connected to pin1 well separated/shielded from components connected to pin3. It's possible that a small capacitance (10pF - 1nf) from pin2 to pin1 could be useful. However, the larger the capacitance, the less high frequency response will be provided by the opamp.

Hi TeeKay6,
Thanks for your information. I'll try adding the small capacitance next time I can work on it (not next week). I'll see if I can shorten the leads, or if all else fails maybe I'll get a PCB made and shorten the tracks / shield pin 3 that way.
Best regards
Dave

 

Hi TeeKay6,
Thanks for your information. I'll try adding the small capacitance next time I can work on it (not next week). I'll see if I can shorten the leads, or if all else fails maybe I'll get a PCB made and shorten the tracks / shield pin 3 that way.
Best regards
Dave

@Dave Lowther
The issue of my second paragraph ("With JP1 shorted...) seems more urgent.

 

I was going to mention feedback in the tiny capacitance between the input and output but it would be at low radio frequencies, not at the audible frequency you have.

 

@Dave Lowther

With JP1 shorted and C1 not added, you see saturation at the output of the opamp? The input bias current is less than 100nA and thus the drop across R3 (due to bias current) is less than 0.001V. With C2 blocking current to ground, the DC closed loop gain of the opamp should be ~1. Even with the bias current flowing through the 10K feedback, the opamp output should be close to 0V. It seems that something else is going on. Board leakage, miswire, mismarked R, ?

Finally, be aware that zener diodes have a very large junction capacitance (in this case from the inverting input to ground). Zeners also have large leakage currents below their breakdown voltage. You might consider replacing the zeners with a low leakage signal diode from pin2 to +15V(cathode to +15V) and another such diode from pin2 to -15V (anode to -15V); equivalent protection with less capacitance and less leakage current (depending on which diode you choose).

Oh I'm sorry, I only saw the first paragraph previously when I replied. It's past my bed time and I'm getting a bit sleepy.

Do you mean "With JP1 shorted and C*2*" not added? I was talking about that in post #5. I don't think I've mentioned removing C1, and if it was removed then shorting JP1 would do nothing. I don't think it's a wiring mistake. I've built two versions of the board (as detailed in the file attached to my opening post). They are laid out quite differently and I checked both boards for expected continuity / resistance values, I also checked for likely shorted nets. 100nA through R2 1M would be -? 100 mV and with a gain of 100 that would be -10V at pin 1.

At the moment the Zener diodes are not connected and R3 is shorted out. Thanks for the tip about alternatives.

Best regards
Dave

 

Oh I'm sorry, I only saw the first paragraph previously when I replied. It's past my bed time and I'm getting a bit sleepy.

Do you mean "With JP1 shorted and C*2*" not added? I was talking about that in post #5. I don't think I've mentioned removing C1, and if it was removed then shorting JP1 would do nothing. I don't think it's a wiring mistake. I've built two versions of the board (as detailed in the file attached to my opening post). They are laid out quite differently and I checked both boards for expected continuity / resistance values, I also checked for likely shorted nets. 100nA through R2 1M would be -? 100 mV and with a gain of 100 that would be -10V at pin 1.

At the moment the Zener diodes are not connected and R3 is shorted out. Thanks for the tip about alternatives.

Best regards
Dave

I apologize. I did misread "C2" as "C1" in your comment: "Without C2 the op amp output goes to about -13.6V with the input (JP1) shorted. With C2 fitted the offset at the output is about -0.4V." Of course I had no knowledge that you had not installed zeners as shown in your schematic, nor that you had shorted out R3.

 

I apologize. I did misread "C2" as "C1" in your comment: "Without C2 the op amp output goes to about -13.6V with the input (JP1) shorted. With C2 fitted the offset at the output is about -0.4V." Of course I had no knowledge that you had not installed zeners as shown in your schematic, nor that you had shorted out R3.

I mean JP2 is fitted with a shorting jumper and JP3 is open circuit. Sorry, I should have mentioned this earlier.
Back to your comment about zeners. I was concerned about their capacitance varying with voltage and thus causing some distortion of the signal. When I did have them connected I saw no noticeable increase in THD from the usual few ppm. I was only putting 30 mV RMS across them. I expect there may be some degradation if the amplifier had a lower gain / higher input level. I need to have a more careful look to confirm this. I had considered the diodes to rail method but I wasn't sure if it was safe to take the inputs to rail plus a diode drop. This is why I decided to try the zeners initially and there's a jumper to isolate them to check their affect on the signal. When I've solved the oscillation problem and got a satisfactory THD/SNR I'll take another look at the protection and may well switch methods.
Thanks again for your comments.
Best regards
Dave

 

I mean JP2 is fitted with a shorting jumper and JP3 is open circuit. Sorry, I should have mentioned this earlier.
Back to your comment about zeners. I was concerned about their capacitance varying with voltage and thus causing some distortion of the signal. When I did have them connected I saw no noticeable increase in THD from the usual few ppm. I was only putting 30 mV RMS across them. I expect there may be some degradation if the amplifier had a lower gain / higher input level. I need to have a more careful look to confirm this. I had considered the diodes to rail method but I wasn't sure if it was safe to take the inputs to rail plus a diode drop. This is why I decided to try the zeners initially and there's a jumper to isolate them to check their affect on the signal. When I've solved the oscillation problem and got a satisfactory THD/SNR I'll take another look at the protection and may well switch methods.
Thanks again for your comments.
Best regards
Dave

The data sheet for the LM4562 shows that inputs may exceed the rails by 0.7V. Your concern is justified.

Here is an excerpt from an old (2003) 1N52xx zeners datasheet. You can see that the (typical is shown) capacitance varies greatly with voltage rating and with applied voltage, with the capacitance being greatest at 0V bias across the junction.

There is another class of diodes known as TVS (transient surge suppressors) that will likely have lower capacitance.


View attachment 184199

 

Signal Input → .1uF → 100Ω → OpAmp Input
. . . . . . . . . . ↓ . . . . . ↓ . . . . . .↓↓
. . . . . . . . . 2MΩ . . 2MΩ . . 10kΩ Bi-Di-Zener
------------------------ Signal GND ------------------

 

Signal Input → .1uF → 100Ω → OpAmp Input
. . . . . . . . . . ↓ . . . . . ↓ . . . . . .↓↓
. . . . . . . . . 2MΩ . . 2MΩ . . 10kΩ Bi-Di-Zener
------------------------ Signal GND ------------------

Hi ci139,
Thanks for your comment. Unfortunately I don't understand what point you are making. Can you elaborate? In the signal input path you have 100Ω. Should that be R3 10kΩ?
Best regards
Dave

 

The data sheet for the LM4562 shows that inputs may exceed the rails by 0.7V. Your concern is justified.

I'm not sure whether you mean 'justified' or maybe you missed a 'not' before it?
I was aware of the abs max rating you quoted from the data sheet. My concern was that a regular silicon diode drop may be too close for comfort to (or exceed) the abs max rating. I had considered using schottky diodes. Like I said earlier the zeners are just an experiment and I'll take another look at the protection when it's working ok without protection.
Best regards
Dave

 

I'm not sure whether you mean 'justified' or maybe you missed a 'not' before it?
I was aware of the abs max rating you quoted from the data sheet. My concern was that a regular silicon diode drop may be too close for comfort to (or exceed) the abs max rating. I had considered using schottky diodes. Like I said earlier the zeners are just an experiment and I'll take another look at the protection when it's working ok without protection.
Best regards
Dave

@Dave Lowther
Responding to your post#16.
I did mean justified for the reason you noted. 0.7V is not hard to achieve with a forward biased silicon diode and would put the input at its "abs max" limit; that is not usually a good idea. Schottky diodes have lower forward voltages, but higher leakage currents; it may be hard to find a suitable compromise. As I suggested, there are other diodes specifically designed to have low capacitance (although I do not know about their leakage currents) and are used like zeners (i.e. they have a specified breakdown rating): TVS diodes. Beyond that, there are several other ways to protect circuitry that can minimize the diode defects...at the cost of increased circuit complexity. Bear in mind that bandwidth of the amplifier may well involve the protection circuitry. Your use of 12V zeners indicates some leeway in what protection is provided and that makes a solution more likely. If you ultimately decide that you cannot satisfy your over-voltage protection need, start a new thread with that specific topic.

 

C2 in your circuit makes no sense to me. There is no DC path to GND on the inverting input.

 

C2 in your circuit makes no sense to me. There is no DC path to GND on the inverting input.

Hi Mr Chips,
I think the inverting input gets its bias current from the op amp output via R5. A simple voltage follower (like the unused half of the op amp in the circuit in my opening post) also doesn't have a path to GND. I think of C2 as effectively making the DC gain 1 rather than 101. Adding C2 definitely reduces the offset at the output from -13.6V to -0.4V.
Best regards
Dave

 

Thanks. That makes sense now. I need to get some sleep.