OPA549 single supplied?

Robert,

Almost all op amps can be used in single or dual configuration and OPA549 is no exception.  However, since most of the specifications like input common-mode voltage or output voltage range are specified with respect to the supply rails (see table below), you have to make sure you do not violate such conditions. 

Looking at the Output Voltage Swing vs Output Current graph below, you may expect around 3.5V swing to positive rail while sourcing 5A and 2.5V swing to negative rail while sinking 5A.

Therefore, using +30V and -5V supplies you mentioned, OPA549 minimum output voltage swing under Iout=5A load would be from:  (-5V+2.5V)=-2.5V to maximum of: (30V-3.5V)=26.5V 

Hi Mark,

Thank you for your answer. So, to summarise: in the application like programmable power supply (datasheet page 13, figure 10) where the OPA549 is used as a power source (only source, not sink) and output power requirements are: voltage range 0...25V, maximum output current 5A, I CAN use following power sources for supplying the OPA549:

V+ -> 30V/5A
V- -> -5V/50mA

Right? And you are absolutely sure that the OPA549 will not "take" from the negative power source more than 35mA in any conditions (taking into accounts that input voltage is always >=0)? So, if it is right I would use simple single power supply (for positive rail) and charge pump for negative rail as shown below (attached file). Right?

So, there is still one problem unsolved. Look at the figure 10 (datasheet, page 13). The application uses single power supply with V+=30V and V-=0V...and the Texas wrote: Vo = 1V to 25V and Io = 0 to 10A! Output voltage started from 1V??? Why not from 3.5 or even from 4.5V (output swing shown on charts). Anyway, could you look at the schematic above? Best regards... Robert

Robert,

The ultimate test is to breadboard the circuit and confirm that it works as envisioned before going into production. A negative power supply should only need to sink DC quiescent current of OPA549 and, in case of sudden lowering of the OUTPUT voltage, also AC transient current coming from the load. For that reason, there will be some delay in the OUTPUT settling time when the negative supply’s max 50mA sinking capability is exceeded by the transient loading demand of dt=~C*dV/Iout.

There is no inconsistencies in the Fig 10 of the datasheet; if you look at the Output Voltage Swing vs Output Current graph above, you will see that under unloaded (sinking) conditions the output can swing within less than 1V above negative rail while under maximum 10A (sourcing) load it can get only within 5V below positive rail. Therefore, on 0 to 30V supplies, the OUTPUT can swing from 1V (Vo=0V+1V, Iout=0) to 25V (Vo=30V-5V, Iout=10A). The 3.5V swing from negative rail applies only under 5A sinking current.

Hi Mark,
So, now I'm not sure if charge pump working as negative rail power source shown above is good choice in this application since OPA549 can sink more than 50mA from this source :-(. I'm not able to estimate if 100mA or more is sufficient in any conditions to have stable output power supply. What do you think about it? Robert

Mark,

As the circuit will act as arbitrary power supply (that is programmable) the output voltage can be set via remote controller every 1ms (for every 1ms the difference between previous and the next output voltage can be 25V!). So, I'm not sure if the negative rail power source shown above is sufficient in this case. The alternative is to use additional small power transformer wind (for example 300-400mA) and 7905 to supply negative rail...Robert

Robert,

I did not realize that you plan to reprogram the power supply every 1ms. However, with 100nF load you show on your schematic, this still would be feasible since it would take at most 0.167ms of slewing time to bring the Output to a new output setting: dt = C*dV/Iout = 100E-9*25/(50mA-35mA) = 0.000167 seconds.

Marek

Robert,

Yes, that is correct - you have to know TOTAL active load in order to calculate the OUTPUT settling time with the very limited negative supply sinking current capability. Using LM7905 with its peak current of 2.2A should greatly mitigate the problem BUT since it is nothing other than a negative voltage LDO, it requires a negative supply of  -7V to -20V to provide the fixed -5V rail – see PDS table below. Thus your power transformer would also have to include a rectifier to supply the required negative voltage of at least -7V.

Marek

Marek,

OK. I know that I should use typical application provided for the 7905 to provide the fixed -5V rail but since the settling time depends on the actual OPA's negative rail current (that is when the negative supply’s max sinking capability is exceeded by the transient loading demand) the question is how to estimate this current in any conditions? In other words, how to estimate the maximum negative rail capability for the worst case (worst load type)? Best regards...Robert