Help|what i'm doing wrong?

Thread Starter

AndrewAK

Joined May 12, 2019
12
Hello,
i'm using opa650 (http://rtellason.com/chipdata/opa650.pdf) as unity gain amp with +- 5v supply.
i'm supplying 5MHZ square wave with 2vp-p amp and offset of 1v to the non inverting input and the output follows the input but the problem is
if i disable the input of the op amp the output goes to -3.2v and i dont understand why!
can someone pelase explaine what i'm doing wrong?
 

Zeeus

Joined Apr 17, 2019
615
Hello,
i'm using opa650 (http://rtellason.com/chipdata/opa650.pdf) as unity gain amp with +- 5v supply.
i'm supplying 5MHZ square wave with 2vp-p amp and offset of 1v to the non inverting input and the output follows the input but the problem is
if i disable the input of the op amp the output goes to -3.2v and i dont understand why!
can someone pelase explaine what i'm doing wrong?
Also, any reason why to use offset given the +-5 source?
 

Thread Starter

AndrewAK

Joined May 12, 2019
12
hi,
You must not leave the Non Inverting input floating, add say a 100K from the NI to 0V.
The OPA must have a DC current path to 0v on that NI input.
E
View attachment 178882
It worked! thank you!
the output now is -0.328v.
can you please explain me why 100KOHM resistor solved this problem?
is there soming i can do to make the output closer to 0v?
 

Thread Starter

AndrewAK

Joined May 12, 2019
12
hi A,
What is the OPA output load.?
You could try a 10k in place of the 100K.
E[/QUOT
Hi Eric,
op amp will drive 100 ohm load with 20MHZ square wave.
eric can you please explain me the solution?or refer me to subject that i can learn and understand it deeply?please i wana learn!
 

Sensacell

Joined Jun 19, 2012
2,718
The opamp always needs to have the inputs connected so that bias current can flow, otherwise, it will malfunction.
Leaving it unconnected makes it go nuts.

The resistor to ground provides a default path for the input bias current.
When the generator is connected, the bias current flows via the generator output impedance.

The 0 offset voltage is going to be the bias current X the resistor value, plus any input offset in the amp itself.

5 uA X 100K ~ 0.5V - use a lower value resistor if this offset is problematic.

Read the data sheet.
 

Thread Starter

AndrewAK

Joined May 12, 2019
12
hi,
Is the 100R load connected when you measure the -0.328v.
Look thru this OPA PDF.
E
after connecting lower value resistor the offset goes nearly 0v thank you!
The opamp always needs to have the inputs connected so that bias current can flow, otherwise, it will malfunction.
Leaving it unconnected makes it go nuts.

The resistor to ground provides a default path for the input bias current.
When the generator is connected, the bias current flows via the generator output impedance.

The 0 offset voltage is going to be the bias current X the resistor value, plus any input offset in the amp itself.

5 uA X 100K ~ 0.5V - use a lower value resistor if this offset is problematic.

Read the data sheet.
thank you!
 

danadak

Joined Mar 10, 2018
4,057
Pay careful attention to C loading. Eg. if you use coax cabling on
output or actual C in load. This is a two pole response OpAmp
and its phase margin is of concern. The datasheet discusses this.

Also when bypassing supplies, because OpAmp is so fast, ceramic
as well as bulk caps should be used. Low ESR is focus over frequency.
Not all caps, same C value, perform well -




Regards, Dana.
 
I have another comment that really got me once too. The input of an OP-amp cannot see an effective 0 ohms resistance (Very small). It to, has Ib across it. Ib needs a place to drop across.
 

crutschow

Joined Mar 14, 2008
26,365
I have another comment that really got me once too. The input of an OP-amp cannot see an effective 0 ohms resistance (Very small). It to, has Ib across it. Ib needs a place to drop across.
What? o_O
What do you mean "Ib needs a place to drop across."?
An op amp will work normally with zero impedance for Ib.
The plus input of an op amp, connected as an inverting amp, is typically connected directly to ground with no added resistance.

The only reason you would want some resistance is to cancel the offset voltage from Ib through the other input equivalent resistance (if that voltage is significant in your application).
 
I'd have to scan the circuit,

But I had sense (10) and Vout(9) of a AD524 conected together +RESISTOR. it went to an LM711(-) and then to the input of a AD524(2)

(+) of the AD711 would either be connected to ground or a voltage source through mercury wetted relay.
There was a 470 pf cap between (-) and out of the LM711.

So, you would get a unity gain or an open loop feedback tot he voltage source.

I had to add a 10K RESISTOR after the (9)(10) pin junction of the first AD524 or really bad things happened.

The circuit was wierd. It was basically a garded differential measuring circuit with gaurd drive. It could be placed in "measure mode" or inside a loop with an I-V converter that maintained a given voltage across the device. So, active of passive feedback. Currents were 100 mA max at +-0.6 V. DUT was a meter away.
 
The uploaded file should be close to the issue I had. There is whole bunch of other stuff including multiple power supplies, current boosters, Zero resistance ammeters etc. that I left out.

Connect AG4 to AG1
Put some DUT (10K Resistor) between pins 1 and 2 of IC4
Put a 1 ohm resistor between pin 10 of IC2 and pin 1 of IC1 (simulate lead resistance)
Put a 1 ohm resistor between pin 1 of IC2 and AG4 (simulate lead resistance)

MV (Measured Value) is a Voltmeter.
SP (SetPoint is a voltage source)

In 2T mode, the ckt worked fine.

In 4T mode, the 10K resistor was required. It could be almost anything. The real output is capable of 100 mA, but this excerpt is not, so in 4T mode, the lead resistance is compensated for.





``
 

Attachments

crutschow

Joined Mar 14, 2008
26,365
Sorry, but I'm still confused.
I'll agree, the circuit is weird.
The output of an instrument amp goes to a comparator, whose output goes to another instrumentation amp, which makes no sense.
The output of a comparator is a digital high or low signal, so there's no reason to amplify that.
And what all that has to do with the 10k resistor is beyond my understanding. :confused:
 
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