Help on the circuit with Opamp

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manikandan123

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hi,
I used LM358 op amp. In my application I need voltage follower. the output is forward biased with diode.. So need to compensate the voltage drop on the diode. SO I want to get output, input voltage + voltage drop across the diode. so only we get same input as output after the diode. How do get it? when the input changes, the same output should come after the diode. Please suggest some ideas. awaiting for your response...
 

Dear manikandan123
Hi
If you attach your schematic i can help you .
Best Wishes
Goldsmith
 

Humm , as i understood you want design a precise rectifier !
follow below circuit :

You should use your diode at feed back loop !
Best Wishes
Goldsmith
 
BTW : the circuit that i attached will work fine , but not as well as possible ! do you want that i attach the best circuit for that ?
 

goldsmith said:
BTW : the circuit that i attached will work fine , but not as well as possible ! do you want that i attach the best circuit for that ?
Click to expand...
Already i know about precision rectifier.It good. but I need to put the diode after feedback for prevention purpose. It act as blocking diode. is any other ideas.... I want to get Vout of opamp is input+drop voltage.so only it followed same input voltage after the diode... did u got it?
 

If you use Capacitor loading not resistor, you can get Vin = Vout
 

Are you sure ? how it is possible to compensate the drop diode of automatically without diode at feed back loop ?
What do you want to do exactly ! i confused . say your question more clear.
Respect
Goldsmith
 

If you use Capacitor loading not resistor, you can get Vin = Vout. See the attached image.

For dc forward current diode will provide forward voltage drop, but if capacitor loading is there, forward voltage drop can be removed.
 

Dear varunkant2k
Hi
As you probably know the gama voltage of silicon diode is about 0.6 volt . and thus you'll have this voltage across your diode . and thus your out put will be lower than your input . you can do it , just if you use that in an active feed back loop .
Respect
Goldsmith
 

you told correct Goldsmith. The capacitive load does not avoid the forward voltage drop. Precision rectifier is also good . thanks for all...
 

Are you sure ? how it is possible to compensate the drop diode of automatically without diode at feed back loop ?
Click to expand...
Of course it's not possible.

The problem is, that the original poster decided to place a diode without thinking about a circuit specification (or at least didn't tell it). Connecting the feedback loop on the load side of the diode is the only way to get an exact diode voltage compensation. Input protection for the OP can be provided by other means. But you need to know details about the intended "blocking" behaviour.

A non-feedback compensation could be implemented unter specific conditions, but can't achieve the same performance.
 

I guess something I missed in the discussion.
I have simulated the same response, Capacitor stores the average dc value of VIN
 

Capacitive load is just one of several options, but hasn't been mentioned by the original poster at all. Without a resistive load component, respectively a certain sink current, Vout is more or less undefined, for both feedback configurations.

Capacitor stores the average dc value of VIN.
Click to expand...
No. It stores Vpeak - 0.6 V.
 

Capacitor stores the average dc value of VIN.
No. It stores Vpeak - 0.6 V.
Click to expand...
In post #14, simulation result shows for dc as well as sin wave input, but output shows the dc value of input.
A non-feedback compensation could be implemented unter specific conditions, but can't achieve the same performance.
Click to expand...
Yes I agree to this point, performance may be affected at resistive loading.
 

In post #14, simulation result shows for dc as well as sin wave input, but output shows the dc value of input.
Click to expand...
That's an example how you can easily get unrealistic results in simulation. During bias point calculation, capacitors are ignored, so Vout can rise to Vin with effectively zero (or fA) current. The capacitor will be charged through the diode leakage resistance.
 

Dear varunkant2k
Hi
is it possible that you explain how it is possible , please ? ( as i saw your simulation and i re simulated this , it seems correct , but how ? your circuit destroyed some principles of diode ! ;-)
Thanks
Goldsmith
 

Thanks FvM for your attention . but as i understood until now , the conduction of each silicon diode will be maximum at 0.65 volts ( approx ) and it will start to conduction at 0.35 volts ( approx and at pretty logarithmic region ) .
But here how the out put wave can be 1 volt if vin is one volt ??!! and i think if vin = 1 volts , the out put should be 0.4 volts ( approx ) because diode need 0.6 volts at all the time !
Is it possible that you elaborately explain it , how it is possible , please ?
Thanks
Goldsmith
 

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