Light coupling PC817 was used for the first time, there was a question I'm not sure, as shown in figure: + 3.3 V input, + 5.5 V circuit conduction? If the pressure drop at the ends of the conduction R02 or R04 after close to 5 v? These two connection light coupling can be conducting?

CodePudding user response:

Can conduction! Specific voltage values need to be calculated according to the current transfer ratio, Vr4=5-10 k * (3.3 v to 0.7 v)/1 k * transfer ratio], but generally do switch signal to use, you can pick up in pc817 4 feet, but pay attention to the current limiting resistor!

CodePudding user response:

VR4 can't just use transmission ratio to calculate, because of the effect of input and output the current limiting resistor, actually is not that big, transmission electric current, should see is conduction voltage drop at this time,

CodePudding user response:

Should be no problem,
After you pick up good circuit, the actual measure R02 or 04 at both ends of the voltage, if close to 5 v is normal, otherwise, you need to reduce R01 or R03, or the resistance increase R02 or 04

CodePudding user response:

Light coupling start current is too low, you this figure light coupling pressure drop 1 V, 2.3 V/1000 ohms, the remaining 2.3 mA. Obviously not, now the product can't do this, to understand light coupling datasheet

CodePudding user response:

Can conduction, light coupling Vf=1.2 V, the If=(3.3 1.2)/1=2.1 mA, CTR=80 ~ 600%, and saturated turn-on corresponds to the Ie=1.6 mA ~ 12 mA, corresponding R02 and R04 voltage V=* Ie 10 k=16 V ~ 120 V & gt; 5.5 V, consider VCE (sat)=0.1 V, all R02 and R04 voltage V=5.5-0.1=5.4 V,

CodePudding user response:

reference 5 floor weixin_44616774 reply:

can conduction, light coupling Vf=1.2 V, the If=(3.3 1.2)/1=2.1 mA, CTR=80 ~ 600%, and saturated turn-on corresponds to the Ie=1.6 mA ~ 12 mA, corresponding R02 and R04 voltage V=* Ie 10 k=16 V ~ 120 V & gt; 5.5 V, consider VCE (sat)=0.1 V, all R02 and R04 voltage, V=5.5-0.1=5.4 V

Excuse me, warrior, saturated conductivity when Ie=1.6 ma - 12 ma, this number is not absolute, and according to the actual calculation, the result of the Vce=0.1, so R02, partial pressure on R04 is 5.4, so we can obtain Ie=5.4/10=0.54 k ma, right, sent several times, is this normal? How do you explain?

CodePudding user response:

CTR just shows light coupling ability of amplification, the final size is decided by the load current, you can put the light coupling is simple to understand for triode, reminder, light coupling of CTR to consider derating, related to the use of time and temperature are, also need to consider when light load of dark current situation,

CodePudding user response:

reference 4 floor qq_39542548 response:

light coupling start current is too low, you this figure light coupling pressure drop 1 V, 2.3 V/1000 ohms, the remaining 2.3 mA. Obviously not, now the product can't do this, to understand light coupling datasheet

I also think this analysis is correct, you should use the following the connection the R03, resistance R04 also should be 5 v series network, otherwise when leds catheter, 5 v network won't conduction,

CodePudding user response:

Optical coupling in some cases can see approximate triode (just different amplification ratio), as long as the voltage at the threshold, is the current scaling relationship between the input and output, a similar triode,