BJT with really tight tolerance on hfe?

[crutschow]

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I will have to think if their is a gain now to include due to R(LED) and the 931 ohm pullup. I believe there is but I will have to check.

"R(LED)", as you know, is the resistor in series with the opto diode.

The 931 ohm is part of the current gain circuit.
The circuit just adds a current gain of 7 to the opto output, so should have no significant effect on whatever the opto gain is due to R(LED) (other than the frequency response of the gain circuit, already discussed).

 

[treez]

I am sure you realise the brilliance of your circuit in that it takes away the variance in gains in the feedback loop, which is more crucial for voltage mode supplies because these are more likely to go unstable with gain variation than current mode designs. The Darlington would otherwise have been needed for high vout flybacks, because the high control current and operating current effectively flows in the secondary side opto diode when topswitch is used, and if vout is 48v, then that is a too-high loss.
Power Integrations have a lot to thank you for here.

 

[crutschow]

............
The only problem I can see is the sensed voltages being close to the rail, but that can be solved by just putting in an extra resistor, or indeed supplying the opamp from a bit higher voltage.
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I just realized I incorrectly replied to this in my post #39.
I still don't see how an extra resistor would help, but you could use a standard (non-rail-rail) op amp if the op amp supply voltage was increased so that the opamp's maximum common-mode input voltage and output voltage is greater than the supply voltage to R2 and R3.

Also the current gain of the circuit is R2/R3, not R3/R2 as I originally stated.
The accuracy of this gain is largely determined by the tolerance on the ratio of these two resistor.

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Thanks Crutschow, I see your point, though I am not sure if ltspice correctly simulates the base-collector capacitance of optocouplers, and I fear it will miss the pole due to that and the pullup resistor (the 931 ohm).
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It does simulate that capacitance, but the 4N25 model I used may have a different capacitance than the opto you are using.

 

[Easy peasy]

Double sigh.
You can degenerate the voltage gain of a transistor but not the current gain.

Triple sigh, if the load impedance is known, then the voltage gain becomes the current gain...

 

[crutschow]

Triple sigh, if the load impedance is known, then the voltage gain becomes the current gain...

Quadruple sigh.
I can't argue with that, but that doesn't apply here.
You haven't been paying attention to the OP's circuit requirements.
There is no specified load impedance.
It needs a current not a voltage into the control IC input.
The circuit used a transistor in a Darlington configuration to amplify that current from the opto output, and the current gain is determined by the transistor Beta.
There's not a practical way to control that.

 

[Easy peasy]

On the topswitch to load presented by the pin is approx 5.2V/ 3mA = 1700 ohms at the midpoint,

5k ohm at light load and ~ 5.2/4.8mA = 1100 ohm worst case at full load, before loss of regulation, so limiting the volt gain to a known maximum, should suffice reasonably well, and gives a simper cheaper circuit...

 

[crutschow]

On the topswitch to load presented by the pin is approx 5.2V/ 3mA = 1700 ohms at the midpoint,

5k ohm at light load and ~ 5.2/4.8mA = 1100 ohm worst case at full load, before loss of regulation, so limiting the volt gain to a known maximum, should suffice reasonably well, and gives a simper cheaper circuit...

The Zc dynamic impedance (normal operating point) of the C input is stated to be from 13Ω to 25Ω once the 5.8V input threshold voltage is reached (page 25 Table and Figure 33 in the data sheet) which is much lower than the values you are using.
This wide tolerance, low impedance input coupled with the 5.8V threshold voltage would make it difficult to control the gain as your proposed, at the low mA operating current levels for the C input.

 

[treez]

As you know, Power Integrations also serves the market of "companies who want a power supply but can't afford to permanently employ a sophisticated PSU designer forever"

...what I like about Crutschow's circuit is that there is no doubt whatsoever what is the current gain, and its easy for anyone to work out. And it has the advantage (also as does the single BJT solution) of drastically reducing standby losses especially when the vout of the flyback is high.
Though Crutschow's cct has well tolerance gain, unlike the single BJT method, which even if there is ways of reducing it, its still not defineable, at least not yet....and Voltage mode control dynamics are such that wide tolerances of gains in the feedback loop are not wanted....especially when many of the Power Integ. users are not dedicated PSU people.
The problem is deriving the small signal transfer function of Crutschow's circuit. Once that is done, I believe Crutschow's circuit will go into Power Integrations Folklore.

As far as I can see , Crutschow's cct has the same features as others with a grounded opto-emitter, ie there is the gain term of the 931 ohm divided by the resistor in series with the opto...and then its just the gain of seven.......so I think that's it, then its all the same as the original one on AN-57. But I would have to check.

If I were Crutschow i'd get a patent on that circuit, as it is a cracker to make up as a module and flog with the topswitch as a method to reduce standby losses with high vout flybacks using topswitch, which is the best selling offline monolithic flyback controller in the world

 

[crutschow]

I have no interest in patenting the circuit and I don't think I could even if I did (it's just a variation of a common constant current circuit), but if you want to contact Power Integrations about it or otherwise use the circuit, be my guest. :thumbsup: