IL 300 driving

[Tabs]

Just having a look at this app note now and looks like Fig 18 on page 9 describes the use of the PNP and Fig 21 is where the circuit snippet actually comes from and may be the differential amp referred to earlier.

The PNP (if selected and biased properly) compensates for the temperature drift of the IL300 forward diode. There are explanations of using a diode in the emitter for this reason (Page 9), and a full explanation of the circuit is on page 11.

The application note says nothing of a sort:

... The output current capability of the OP-07 is extended by including a buffer transistor between the output of U1 and the LED. The buffer transistor minimizes thermal drift by reducing the OP-07 internal power dissipation if it were to drive the LED directly. ...

The forward voltage drop of the diode is irrelevant because it is inside the feedback loop.  Its input referred effect will be on the order of nanovolts or less and the Vbe of the transistor will double it instead of compensating for it.

The app note section that you quoted is discussing circuit as shown in figure 18. This shows the output of the opamp connected to the base of the pnp transistor. This effectively grounds the cathode of the input diode. The current through this diode is limited by the 100k so the opamp is never going to lose that fight (especially when backed up by the pnp - as the app note states). This represents what is going on in the upper circuit.

That entire section of the app note does not apply to the lower circuit opamp (the circuit snippet of the OP). This is because the lower opamp output is connected the base of the lower pnp and pin 3 of the lower IL300. This connection does not appear in the upper opamp or in the app note. The connection to pin 3 causes the opamp to directly drive one of the IL300 outputs. These outputs look like the attached drawing (app not 55). The current boosted photdiode is essentially an npn in parrallel with the output (this is the current source in the schematic symbol). When the photodiode turns the npn on, the Vce will go to Vce(sat). The npn is saturated on and effectively a short to ground with no current limit resistor. The lower opamp is where the current is coming from and this will be limited by the output impedance of the opamp. Hence the lower opamp drives into a short circuit and there is negative feedback. In fact theres no feedback because the opamp output is grounded.

correct me if I'm wrong f5r5e5d, but I believe this is what you were talking about.

[David Hess]

That entire section of the app note does not apply to the lower circuit opamp (the circuit snippet of the OP). This is because the lower opamp output is connected the base of the lower pnp and pin 3 of the lower IL300. This connection does not appear in the upper opamp or in the app note. The connection to pin 3 causes the opamp to directly drive one of the IL300 outputs. These outputs look like the attached drawing (app not 55). The current boosted photdiode is essentially an npn in parrallel with the output (this is the current source in the schematic symbol). When the photodiode turns the npn on, the Vce will go to Vce(sat). The npn is saturated on and effectively a short to ground with no current limit resistor. The lower opamp is where the current is coming from and this will be limited by the output impedance of the opamp. Hence the lower opamp drives into a short circuit and there is negative feedback. In fact theres no feedback because the opamp output is grounded.

correct me if I'm wrong f5r5e5d, but I believe this is what you were talking about.

That is an error in the schematic shown in figure 21 of application note 50.  Pin 3 of the optocoupler should connect directly to the inverting input of the operational amplifier and not the output as shown.

The IL300 outputs use only photodiodes and not phototransistors.

[Tabs]

Apologies,

Got app note 55 in my head (phototransistor output)

I think were all agreed then. The schematic is wrong because the inverting input is not balanced. Mirroring the top half fixes the problem.

[sandy]

its not "a different circuit" its just wrong

Yes , As I was not following it so I didnt bothered to look more into it. You have shown HCNR 200 in the LTspice , are the two optos electrically similar can it complement IL 300.

[f5r5e5d]

you have to carefully read the datasheets to see if the HCNR is a sub in your circuit - most cases it should be a functional equivalent with slightly different typical specs, and spreads - such as a much tighter k3 spec

understanding the circuit operation really means understanding the op amp and LED/IP1 servo circuit - no detailed analysis of the incorrect IP1 photocurrent is needed to see the lower incorrect drawing of the circuit connection doesn't give a feedback loop

I put the pieces needed together for LTspice sim: