Load sharing concept of op amps

Hello Spanika,

Can you post a link to the load sharing app note? What is the final circuit or system you are designing? Why do you need to load share?

Do you need to use terminated transmission line?

The LMH6552 is not really a 5V amplifier. If you want to run on 5V the LMH6554 would be a better choice.

Power = V^2 / R Your maximum V is set by the +- 2.5V power supplies, hence the maximum power in the 75 Ohm load is not going to require very much current. I really don't see any need to parallel amplifiers in this application.

Regards,
Loren

hello loren,

this is the schematic i would like to develop. My target is to achieve a Maximum power at my load end.I want to achieve a wide bandwidth indeed 220MegHertz.

when asked what is the set of 3 op amps in my third stage, I was said that a concept from TI - load sharing  is used and the purpose of using it is to maximise the Amplitude and to see that max power is obtained at Output.

but when i read the application note, other than load current sharing, I cant understand how it is reducing the power distortions. I am unable to understand exactly why a load sharing concept is used here.

some explaination on how load sharing would effect the power at Output would be a great help.

Hello Spanika,

Unless you need very high linearity (-80dBc or better), I don't think you need to use load sharing for this application. The LMH6552 can drive a 75 Ohm matched load (150 Ohms total). If you increase the supply voltage of the LMH6552 to +-5V you would have more power available than if you use a +-2.5V supply. The LMH6552 works best on a 10V power supply, so this is a double benefit.

Are you aware that on a +-2.5V supply that most amplifiers will only swing +-1.5V? If you require more voltage swing using power sharing won't work, it only works for current and even then with limits.

Power sharing works for linearity because sharing current will increase linearity for a given power level. This is because amplifiers give better linearity into a higher impedance load (less current).

Regards,
Loren

Hello Spanika,

Here is a good app note for load sharing of amplifiers.

www.ti.com/.../TIDA-00023


Regards,
Loren

hello loren,

I have already read the app note u sent.

I would like to make my question simpler. I am trying to understand the benfits of paralleling of op amps first, rather than the load sharing concept.

putting my entire schematic and my purpose of developing this schematic aside for a Moment, all I know about paralleling concept is the below three Statements.

1. With  amplifiers in parallel, the output drive capability increases. In general, the drive capability increases by N (where N is the number of amplifiers in parallel).

2. If x is the rms power of a single op amp, then adding a second amplifier in parallel increases the RMS signal power by 2x

3. Paralleling the amplifiers reduces the power dissipation of a single amplifier by sharing the load with multiple amplifiers. The current load on each amplifier is reduced by 1/N where N is number of amplifiers.

 I could prove the 3rd Statement right. I am unable to Show the 2nd Statement (most important for my application the reason for which i want to use the op Amps in parallel). Here is a simple schematic by which I am trying to understand the things.

 4454.Parallel.TSC

 

Hello Spanika,

"Why is the power always seen same irrespective of the number of op amps in parallel.
 Can you explain me this."

You are driving the same load with the same voltage, hence your power is the same.  This is basic Ohms law, it won't change based on how many amplifiers you use. 

In order to benefit from parallel amplifiers you need a load resistance that is so low that one amplifier cannot supply adequate current.  You do not have those conditions, hence you DO NOT NEED to parallel amplifiers in your application. 

Try the schematic I attached. 

Regards,

Loren

2678.4454.Parallel.TSC

Thank you loren. Now I understand the actual purpose of paralleling op amps. I was trying to understand this concept from the view Point of why it is used in schematic given to me, thats why i was confused.

I have a question,

In the load sharing concept it is stated that -------if Rl is load then, series reistors (Rs) for placing 2 op amps in parallel is 2*Rl .,to match the load.

but in the sample you sent me, Rs is Rl/2, what is the correct method for selecting the value for Rseries....

just a last question to know if I understood it right..

In the example you sent me, since the Output current of THS3091 is 350mA(as per data sheet), and the load we are driving here is too small,

we increase the op amps in parallel so that each op amp meets the condition of not exceeding 350mA.

____Output current specified in the datasheet is the Maximum current that an op amp can drive. The current measured at Output of op amp can be less than this but not more, otherwise it damages the op amp.___is this Statement right.

than what is the Minimum current required below which i cannot drive the op amp. Can i find that value on datasheet:

Hello Spanika,

 "what is the correct method for selecting the value for Rseries"

Rseries should be large enough to ensure that every op amp delivers the same current to the load and small enough that the voltage drop across it does not reduce your system performance.  The exact value will depend on your circuit. 

"than what is the Minimum current required below which i cannot drive the op amp. Can i find that value on datasheet:"

There is no minimum current. 

"____Output current specified in the datasheet is the Maximum current that an op amp can drive. The current measured at Output of op amp can be less than this but not more, otherwise it damages the op amp.___is this Statement right."

The absolute maximum current for the THS3091 is 350mA.  This is the current that will damage the amplifier.  Look on page 7 of the datasheet.  You will see that the amplifier can source or sink around 250mA at room temperature, but we can only guarantee 175mA over temperature and process variations.

Based on this information you would really only want to plan on a maximum of 150mA per amplfier.  If you were using the maximum voltage swing on a +-15V power supply(13Vpp) then the smallest load you could drive is an 87 Ohm load. 

Can you again confirm the supply voltage you are using?  If you stepped up to a +-15V power supply and used more gain in your differential amplifier you could get enough output voltage swing to require using multiple amplifiers to drive your 75 Ohm load. 

More voltage will give you the potential to deliver more power. 

Regards,

Loren

hello loren,
can you suggest a better single ended to differential op amp,which has a wideband more than 300MHz, for which I can give an Input of 5V. currently I am using LMH6552, I am giving +-5V supply voltage and Maximum input i could give is only 3.5V and the max gain I can achieve is 3. If I am increasing my gain or Input beyond this Limits my waveform is distorted.

I want my single ended to differential op amp to take an Input of 5V and deliver at least 5 Volts or even more at each differential Output.

as you said I am driving My parallel op amps with this single ended to differential op amps and want to achieve a Maximum voltage possible at the Output of parallel op amps.

Hello Spanika,

The other way to get more voltage at your load is to use a higher turns ratio transformer. Can you use a 1:4 or 1:8 transformer?

Have you contacted your transformer vendor for assistance? The power levels you are looking for may require special transformers.
Regards,
Loren

hello loren,
i am using a 1:5 transformer.
i want to use 2 transformers in parallel