# Zerocross test issue

#### Benengineer

Joined Feb 6, 2016
133
Ok, I undertand it now. I will apply the circuit you recommended in next version. But now I still use the old pcb and just asked them to change the resister in solving not enough current issue. One more question, is the data sheet showing the saturation current at the current rate?
No, the output of the opto is not always higher than the input. It depends upon the current transfer ratio as shown in post #50.
For some versions, such as the EL817, you can see that the minimum current gain gives an output current that is 1/2 the input.
Why is that hard to understand? You seem to be focusing on the details without understand the concept.

I believe your confusion is believing that the output always has to be a fixed relation to the input.
That's true when operating in the linear mode but not in the switching mode as is the case here.
In the switching mode you want to overdrive the input by at least a factor of 2 or 3 above the value required by the minimum current gain so that the output is fully saturated in the ON state.
That's true of any bipolar transistor also.
So the output resistor is selected to give a current that meets this criteria.
In my circuit the input current is about 1.7mA and the output current is about 0.6mA through the 5kΩ, insuring that the output is well in saturation when ON.
That's probably more overdrive than is needed by the current gain of the opto you are using but it doesn't hurt anything to do that.
One more question. Since our unit is EL817, why can't choose 50% for Ii/Io ratio rather than choosing 300% ratio?
Thanks

#### Benengineer

Joined Feb 6, 2016
133
The current gain shown is for linear operation where the transistor collector is biased at some voltage above the emitter, not for the saturation condition where the transistor is fully turned on and Vce is at a minimum.
So in this case, Vce= 0.1V in the data sheet?

#### Benengineer

Joined Feb 6, 2016
133
Since I am a bignner, I don't really undertand the data sheet. There is lot of things to learn.

#### crutschow

Joined Mar 14, 2008
26,866
One more question. Since our unit is EL817, why can't choose 50% for Ii/Io ratio rather than choosing 300% ratio?
Thanks
The ratio variation is from 50% to 300% unit-to-unit.
You don't get to pick the value, only that the device with the stated suffix will have a unit-to-unit variation as shown in the data sheet for that suffix.
So in this case, Vce= 0.1V in the data sheet?
Yes, as show in Figure 5 of the data sheet.

Data sheets can be intimidating and somewhat cryptic at times.
You just have to take some time and try to understand what all the parameters mean, sort of like learning a new language.

#### Benengineer

Joined Feb 6, 2016
133
The ratio variation is from 50% to 300% unit-to-unit.
You don't get to pick the value, only that the device with the stated suffix will have a unit-to-unit variation as shown in the data sheet for that suffix.
Yes, as show in Figure 5 of the data sheet.

Data sheets can be intimidating and somewhat cryptic at times.
You just have to take some time and try to understand what all the parameters mean, sort of like learning a new language.
Yes, I agree with you and i have to spend time to learn those parameters mean. But it is more important to meet a good advisor even though i have passion on it. I try to use conversion functionality to invite you. But I don't know why it does work. Don't you allow people to invite you?
My problem is that I am not good at applying what I learned from the school when a circuit is complicated. for example, I try to debugg the circuit in attachment.
I can see that as below:

R48 and C22 are voltage divider V1 = VLF1* (1/2pifC22)/(R48+1/2pifC22); C73|| R21 and C78||C79||R23 + C77 are second voltage divider. V2 = V1* C78||C79||R23 + C77/(C73|| R21 + C78||C79||R23 + C77). C74 blocks DC. At 3 terminal of the amp, voltage is V2/2pi*f, current I = Vcc/2-V2/2pi*f/R24. it is very small. At feedback loop, R53//C30. C28 and C82 + R22 are other voltage divider V3 = V2/2pi*f.*(R32+C82/R32+C82+C28). V3/R32 = (Vo-V3)/R53. I can caculate very voltage. But I need people explain what the circuit is doing. I also compare a good unit voltate to the bad unit voltage with data.
Good unit Bad unit
TP45 528mV P-P 80mV p-p
TP46 2V 1.04V
TP47 0.68V 1.04V
TP48 1.04V 0.96V
TP49 1.32V 1.32V
TP50 1.32V 1.32V
TP51 2.72V 1.36V

Thanks

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