Do differential i/p need to be clamped?

[dknguyen]

Hi. I have a differential amplfiier here that has an input common range of +/-2.5V (half the supply voltage). I was wondering if I need to clamp the inverting input like in the attached schematic since the transformer's outputs are floating differential outputs (as opposed to the differential outputs that aren't floating from something like an op-amp where the power driving the differential outputs come from the power rails and are therefore inherently have a bounded common-mode range.

I will be clamping at least the non-inverting input for sure since abnormal operating conditions might cause the voltage to exceed the maximum.

1. But I was wondering if I should clamp the non-inverting input to make sure that the common-mode stays within the maximum limits of the amplifier?'

2. Is there any mechanism on the amplifier's differential inputs that would try and pull the common-mode range of a floating differential outputs of a source (the transformer) to within the amp's common mode input range?

3. If there is no such mechanism like in #2 , then I guess I must tie the inverting input to grounds. But if there is, should I still just tie the inverting input to ground? Because if I don't, it seems like the common-mode signal might float away beyond the limits of the amp. If I tied one input to ground it would ensure that the signal would stay within the common mode input range of the amp.

4. And is it pointless to use the amp in differential mode since it's already being fed from a transformer which kind of makes it differential already?

 

[audioguru]

Your circuit must keep the opamp's inputs in their operating common-mode voltage range or the opamp won't work properly. Some opamps suddenly invert the output when an input goes lower than its allowed negative common-mode voltage range.

You can use the transformer's secondary as a single-ended voltage source instead of being a differential.

 

[Roff]

You have no source of input bias current for your VCA except the diodes to +5V. I'm assuming you are still using VCA810. The bias current is -10uA. The two input signals will have their positive half cycles clipped at +5.6V.
For any differential amp, if the internal input circuit does not provide a path to ground, you must provide one. Here's a quote from the datasheet:

To preserve termination options, no internal
circuitry connects to the input bases of this differential
stage. For this reason, the user must provide DC paths for
the input base currents from a signal source, either through
a grounded termination resistor or by a direct connection
to ground.

A center-tapped transformer, with the center tap grounded, is one solution. The other is resistors to GND from each input. Just keep in mind that 100k would give you a nominal 1V common-mode offset, and will load your inputs, which won't happen with the center tapped version.
Or, you can go single-ended, as AG suggested.

 

[dknguyen]

Okay, thanks. New question now...I am thinking cascading two of these VCAs for a max gain of ~80dB (I don't really know how much I Need, but that's what the maximum adjustable gain is between the two), except that each amp has a maximum offset of 8mV. Now this 8mV after it comes out of the first amp is going to get amplified right along with everything else, but tHe simulation shows that if I place a DC block capacitor between the two amps then the net output ends up being attenuating rather than amplifying.

Is there a way to get around this? It seems I need a way to AC couple the two amps without messing around with impedances.

 

[Roff]

Okay, thanks. New question now...I am thinking cascading two of these VCAs for a max gain of ~80dB (I don't really know how much I Need, but that's what the maximum adjustable gain is between the two), except that each amp has a maximum offset of 8mV. Now this 8mV after it comes out of the first amp is going to get amplified right along with everything else, but tHe simulation shows that if I place a DC block capacitor between the two amps then the net output ends up being attenuating rather than amplifying.

Is there a way to get around this? It seems I need a way to AC couple the two amps without messing around with impedances.

Why would a blocking capacitor cause attenuation? Did you provide a path for bias current?

 

[dknguyen]

Why would a blocking capacitor cause attenuation? Did you provide a path for bias current?

Is that what it is? Right now the cap just directly connects the output of one op-amp to the non-inverting input of the other op-amp (the inverting input is grounded). So you are saying that a large resistor to ground between the capacitor and input of the next amp would solve the problem? It has happened with every other op-amp circuit I simulated so far so perhaps that is it.

 

[Roff]

Is that what it is? Right now the cap just directly connects the output of one op-amp to the non-inverting input of the other op-amp (the inverting input is grounded). So you are saying that a large resistor to ground between the capacitor and input of the next amp would solve the problem? It has happened with every other op-amp circuit I simulated so far so perhaps that is it.

Did you read both of my most recent posts?

 

[dknguyen]

Yeah, I thought I remember people just sticking a capacitor on the non-inverting terminal of op amps in a non-inverting amplifier and having it work just fine. I'm not sure why it's not working here. There is a path for the current to ground isn't there since it's in between op-amps and the output of an op-amp has a path to ground?

THis is the circuit currently (except without those two resistors on either side of the DC block capacitor, I added those in as I was messing around with it).

 

[Roff]

Yeah, I thought I remember people just sticking a capacitor on the non-inverting terminal of op amps in a non-inverting amplifier and having it work just fine. I'm not sure why it's not working here. There is a path for the current to ground isn't there since it's in between op-amps and the output of an op-amp has a path to ground?

THis is the circuit currently (except without those two resistors on either side of the DC block capacitor, I added those in as I was messing around with it).

it's in between op-amps

That's the problem. It's a DC blocking capacitor. No DC current can get through it.
In your schematic, you don't need a resistor to GND on the output of the first amp, but you should change the one on the input to 10k or less, and add an identical one between the other input and GND. Otherwise, as you can see, you get a big DC input offset. The offset current is typically +/- 600nA, so 10k resistors would give you +/-6mV input offset.
Do you know what you are going to do on the transformer-coupled input?

 

[dknguyen]

I was trying values of 250K to 1M but I guess those were far too large (I seem to remember trying 500R, 1K, and 5K too but for some reason I didn't notice a difference). Even 10K seems to be too high. It seems to need around 1-2K.

Why wouldn't 250K work though? 5V/250K = 20uA > 12uA (specified as bias current in the datasheet)? Or is something wrong behind that reasoning?

Maybe I wasn't paying enough attention to the graph because on an axis scaled to the maximum gain which is a hundreds of times larger than your smallest gain, a gain of 1 look almost like a gain of 100. But yeah, it seems fixed up now.

For the transformer coupled amp, this is what I have right now because it easily balances the impedances on the both outputs. But I have to sit down and think about how the voltage is going to behave there and may have to add another rail clamp. THe thing I like about grounding the inverting terminal and one terminal fo the transformer to make it single ended is how the AC voltage is predictably centered around GND which ensures the input is in the optimal range for the amp.

EDIT: Seems that I will stick with this setup and add another rail clamp. If I thought it through properly, the transformer secondary voltage will still center around zero, except that both pins will do it unlike the single-ended way where just one pin jumps up and down centered around ground.

 

[dknguyen]

Here is the whole circuit right now. The output to this circuit goes through that AC block capacitor all the way onto the right into a MAX913 comparator. So if the comparator also has no connections to ground then I also need to provide a resistor for the DC bias currents then again?

Clarification- do I need to add that resistor when AC coupling two op-amps/comparators together every singe time? Or is it done internally on some of them?

 

[Roff]

The signal will be centered around ground and symmetrical with the center-tapped transformer.
I don't understand why, in your cascaded amp sim, you show -71mV on the input bias resistor, when the datasheet says the current is -10uA typical. I ran a sim on the TI model, and the results are weird. With 100k on one side and zero ohms on the other, like your schematic, I get -71mV, as you did. Then I put in 100k on each side, and I got about -600mV on each input. To further confuse things, the datasheet says the bias current is negative, but the spice model says the input transistors are NPN. Their LM324 datasheet also says the bias current is negative, and the input transistors are PNP.
I wouldn't trust the results of VCA810 simulations.

 

[Roff]

Here is the whole circuit right now. The output to this circuit goes through that AC block capacitor all the way onto the right into a MAX913 comparator. So if the comparator also has no connections to ground then I also need to provide a resistor for the DC bias currents then again?

Clarification- do I need to add that resistor when AC coupling two op-amps/comparators together every singe time? Or is it done internally on some of them?

Yes, you need the resistor for all op amps. Most op amps have much lower bias current than the VCA810 (which is not an op amp). The bias current in CMOS amps is typically picoamps, but you still need a path for DC current. The resistor for a CMOS amp can be megohms.

 

[dknguyen]

I'm slightly confused now because I ran two simulations for cascaded AC coupled non-inverting amps and cascaded AC coupled buffers both with and without those resistors on the AC coupled input of the second amp and it says it works? I ran them using an actual op-amp, an OPA743. So I was about to come to the conclusion that it was dependent on the op-amp.

 

[Roff]

I'm slightly confused now because I ran two simulations for cascaded AC coupled non-inverting amps and cascaded AC coupled buffers both with and without those resistors on the AC coupled input of the second amp and it says it works? I ran them using an actual op-amp, an OPA743. So I was about to come to the conclusion that it was dependent on the op-amp.

If your sim started out with zero charge on the cap, it could run for a long time without showing a problem.

 

[dknguyen]

Ah. I see.

I've just been muddlign around with it and it seems that the problem I had above is only recognizable when the amps are set for large gains. If they are set for small or fractional (attenuating) gains it doesn't show that there is anythign wrong. I guess that also has to do with the capacitor charging.

EDIT: And now that I upped the gains on the cascaded non-inverting amp circuit by a lot, it also exhibits the same problem without the resistors.

Thanks for your help. I'd give you more rep, but the system has not been letting me give out rep.

*yay*
https://www.analog.com/library/analogdialogue/archives/41-08/amplifier_circuits.html

 

[marksanders]

Differential System

Firstly a differential system using a transformer provides good common mode isolation and unless the cost or size/weight is restrictive then keep it as part of the design. The overall common mode rejection ratio CMRR that is the amount of rejection of common mode signal that is rejected compared to the differential that is required should be high as the common mode impedance from either input transformer connection to 0V is high, lots of Megohms when compared to the differential impedance of the transformer input winding - that is mathematically related to the secondary impedance and turns ratio.

Secondly you will want to insure that the opamp works properly. This is normally done by providing a path for the amplifier input biasing currents to the 0V line. Therefore you will need to provide some 100k resistors from the + and - pins to 0V for the amplifier to work properly. If these resistors are not added then the capacitor across the transformer output will slowly charge up to either the + or - rail and damage the amplifier inputs.

Thirdly you will be able to measure some ac voltages on either side of the output transformer referred to 0V. Looking on an oscilloscope and assuming a sinusoidal input then one transformer side will be positive going as the other side is negative going with respect to 0V. There will be voltage gain or loss according to the turns ratio. If you usea signal generator a frequency sweep would confirm the bandwidth of your system.
What you really need to do here is determine is the maximium likely voltage swing to the amplifier + and - inputs - this could damage the amplifier inputs.
If the differential voltage is then outside the amplifier differential spec then you will have to provide voltage excess protection systems using diodes to the supply rails or another voltage reference rail, to 0V or also fitted across the + and - input pins - perhaps back to backdiodes. Opamp data sheets may show some useful varieties. ( Some opamps may need different circuits )

I hope that this approach is at least logical and straight forward

 

[dknguyen]

Firstly a differential system using a transformer provides good common mode isolation and unless the cost or size/weight is restrictive then keep it as part of the design. The overall common mode rejection ratio CMRR that is the amount of rejection of common mode signal that is rejected compared to the differential that is required should be high as the common mode impedance from either input transformer connection to 0V is high, lots of Megohms when compared to the differential impedance of the transformer input winding - that is mathematically related to the secondary impedance and turns ratio.

It isn't. I'm actually using the transformer to isolate the silicon from the high-voltage bias for the transducer on the primary side of the transformer. My theory is, that any failures are high-voltage DC failures rather than AC ones so that if something fails the transformer should be able to block the high-voltage (otherwise I would need more isolators on the output side of the analog receiver logic- which is what this circuit is).

Secondly you will want to insure that the opamp works properly. This is normally done by providing a path for the amplifier input biasing currents to the 0V line. Therefore you will need to provide some 100k resistors from the + and - pins to 0V for the amplifier to work properly. If these resistors are not added then the capacitor across the transformer output will slowly charge up to either the + or - rail and damage the amplifier inputs.

THat capacitor is there as an impedance matching thing (the transducer is capacitive and supposedly has a capacitance of 400-500pF. It never occured to me it might charge up. I'll add something (I guess I shouldn't rely on the amp's differential 1M input impedance huh?). BUt won't the capacitor discharge itself through the transformer winding (maybe LC oscillating as it does so?)

Thirdly you will be able to measure some ac voltages on either side of the output transformer referred to 0V. Looking on an oscilloscope and assuming a sinusoidal input then one transformer side will be positive going as the other side is negative going with respect to 0V. There will be voltage gain or loss according to the turns ratio. If you usea signal generator a frequency sweep would confirm the bandwidth of your system.
What you really need to do here is determine is the maximium likely voltage swing to the amplifier + and - inputs - this could damage the amplifier inputs.
If the differential voltage is then outside the amplifier differential spec then you will have to provide voltage excess protection systems using diodes to the supply rails or another voltage reference rail, to 0V or also fitted across the + and - input pins - perhaps back to backdiodes. Opamp data sheets may show some useful varieties. ( Some opamps may need different circuits )

I hope that this approach is at least logical and straight forward

The turns ratio is 1:1. My useful bandwidth is about 40-70kHz (and 10kHz-200kHz, but only if I use really good $300 (vs $30) transducers. The transformer has a bandwidth of 125MHz and the amp has a bandwidth of at least a few MHz at maximum gain so I think I should be okay there.

THe regular AC voltage from the transducer (I was told) is on the order of uV-mV. THe excess voltages I am more worried about are those that come from the transmitter as it's transmitting which might be really close to the receiving transducer. Which is for those two pairs of diode rail clamps I have in there.

 

[marksanders]

Differential Amplifier Bias Currents and Input Protection

Just a few articles that you may find useful.
Basically the bias currents from both the + and - amplifier inputs must flow to 0V through resistors, these resistors can be quite high value, say 100k, as the bias currents should be quite low.
The diode protection circuits will increase the differential amplifiers life in cases where either the common mode or differential voltages applied to either the + or - amplifier inputs exceeds the manufacturers specifications.
I hope you really enjoy electronics - remember quite often it is when things dont go as expected that you really learn more although it can be very frustrating at the time ! The web is great for a reality check and general information and engineers are a helpful bunch so if you ask you can normally expect a reasonable answer.


Mark

 

[marksanders]

Differential System

You are correct that the C cannot charge up due to the low resistance of the secondary winding that would always discharge it. This is a mistake on my part.

However any DC bias currents flowing in or out of the + or - amplifier inputs have no route to 0V. All of the protection diodes would be normally be off, apart from leakage currents, at the normal voltage levels at the + and - pins of the differential amplifier around about 0V.
The capacitor will provide some filtering of the transformer output signal and reduce the bandwidth at some point.