26-06-2015, 03:53 PM
Although the circuit has been reliable, I'll expand on my thinking in the hope of giving an insight into the way I think about these things. Hopefully it'll be of interest to someone, somewhere 
Let's take the worst-case of 5A across the diodes. At that current, the BY550 could drop 0.9V at 5A. So that's 2.7V in total.
This is being reduced by D5 by 0.3V approximately*, so there is 2.4V available for the LED...
The forward voltage of the LED is 1.2 to 1.4V. At a forward current of 50mA, it might be as high as 1.5V. Trying to put more than 1.5V across it would clearly result in more current, but the graph stops at 50mA (which is the absolute maximum rating for the LED). In other words, if you try to "force" 2.4V across it, it wouldn't be terribly happy!
OK, the above is slightly simplified, but hopefully you can see my concerns
*In practice, the BAT81 has a very "soft" knee, which implies a high slope resistance (not untypical with a signal diode). In reality, it's protecting the LED, and if anything, this diode might be more vulnerable than the LED, given that it's only rated for 30mA....
Of course, the trouble with developing a circuit like this is the difficulty of measuring the current in the LED without changing it. The best I can think of - in lieu of a genuine Hall-effect current probe - would be a 1 ohm sense resistor and a differential 'scope probe (and an isolation transformer powering everything, of course!). A lot of DMMs would present too much "burden voltage" and reduce the current while making the reading. The EEVBlog "Microcurrent" adaptor would do a good job here, probably.
Having had a closer look at the datasheets for the diodes and SSR, I reckon that my earlier wild guess of 100 ohms should be more like half that value. Let's see how I got to that:
You need about 8mA in the LED, though more won't hurt. But let's assume 8mA for starters.
At low currents, the BY550 has a Vf of just under 0.8V (hence ~2.4V for 3 of them).
At 8mA, the BAT81 has a Vf of nearly 0.8V.
At 8mA, the LED has a Vf of about 1.2V.
So, 2.4V minus 0.8V minus 1.2V is 0.4V. Hence, we need a resistor of 50 ohms to get to 8mA. Call it 47 ohms.
How does that change at high currents?
With 2.7V across the BY550s, you'd have around 14 or 15mA in the LED if nothing else changed. In practice, the voltage lost across the BAT81 will increase slightly, but either way, ~14mA is perfectly safe for the LED and the BAT81.
As I said before, with no resistor, I can't put exact numbers on the current in the LED, but instinctively, I am sufficiently worried about it that I'd want to measure it before deciding that it's OK. Remember; any more than 30mA will endanger the BAT81, and any more than 50mA is doing the same for the LED. The 3 BV550s form a "stiff" voltage source. The BAT81 is providing some "softness". I haven't began to think about how the capacitor is influencing things (other than the fact that it will greatly increase peak currents seen by the BAT81). And because we're seeing half-wave current peaks, we might need to reduce the series R further to achieve an average of 8mA in the LED - but at least it's easy to measure the average DC voltage across the capacitor and calculate the required resistance. Surprisingly complex, these simple circuits!
Anyway, I hope this helps,
Mark
Let's take the worst-case of 5A across the diodes. At that current, the BY550 could drop 0.9V at 5A. So that's 2.7V in total.
This is being reduced by D5 by 0.3V approximately*, so there is 2.4V available for the LED...
The forward voltage of the LED is 1.2 to 1.4V. At a forward current of 50mA, it might be as high as 1.5V. Trying to put more than 1.5V across it would clearly result in more current, but the graph stops at 50mA (which is the absolute maximum rating for the LED). In other words, if you try to "force" 2.4V across it, it wouldn't be terribly happy!
OK, the above is slightly simplified, but hopefully you can see my concerns
*In practice, the BAT81 has a very "soft" knee, which implies a high slope resistance (not untypical with a signal diode). In reality, it's protecting the LED, and if anything, this diode might be more vulnerable than the LED, given that it's only rated for 30mA....
Of course, the trouble with developing a circuit like this is the difficulty of measuring the current in the LED without changing it. The best I can think of - in lieu of a genuine Hall-effect current probe - would be a 1 ohm sense resistor and a differential 'scope probe (and an isolation transformer powering everything, of course!). A lot of DMMs would present too much "burden voltage" and reduce the current while making the reading. The EEVBlog "Microcurrent" adaptor would do a good job here, probably.
Having had a closer look at the datasheets for the diodes and SSR, I reckon that my earlier wild guess of 100 ohms should be more like half that value. Let's see how I got to that:
You need about 8mA in the LED, though more won't hurt. But let's assume 8mA for starters.
At low currents, the BY550 has a Vf of just under 0.8V (hence ~2.4V for 3 of them).
At 8mA, the BAT81 has a Vf of nearly 0.8V.
At 8mA, the LED has a Vf of about 1.2V.
So, 2.4V minus 0.8V minus 1.2V is 0.4V. Hence, we need a resistor of 50 ohms to get to 8mA. Call it 47 ohms.
How does that change at high currents?
With 2.7V across the BY550s, you'd have around 14 or 15mA in the LED if nothing else changed. In practice, the voltage lost across the BAT81 will increase slightly, but either way, ~14mA is perfectly safe for the LED and the BAT81.
As I said before, with no resistor, I can't put exact numbers on the current in the LED, but instinctively, I am sufficiently worried about it that I'd want to measure it before deciding that it's OK. Remember; any more than 30mA will endanger the BAT81, and any more than 50mA is doing the same for the LED. The 3 BV550s form a "stiff" voltage source. The BAT81 is providing some "softness". I haven't began to think about how the capacitor is influencing things (other than the fact that it will greatly increase peak currents seen by the BAT81). And because we're seeing half-wave current peaks, we might need to reduce the series R further to achieve an average of 8mA in the LED - but at least it's easy to measure the average DC voltage across the capacitor and calculate the required resistance. Surprisingly complex, these simple circuits!
Anyway, I hope this helps,
Mark
