Discussion in 'General Electronics Chat' started by Dong-gyu Jang, Jul 16, 2015.

1. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Hello.

I'm now stuck with how to adjust output voltage of given pulse delay generator.

Let's say the generator outputs 0 to 4 V single square signal with 10 us pulse width. Its raising time maybe within 200 ns.

4V is overkill to load so I need to reduce its amplitude to 1.5 V as recommended from load specification. I'd thought voltage divider is simple way for this job, however, I found that it may get some loading effect in which voltage applying to the load is different from my intention due to load impedance.

I've considered the linear voltage regulator but...Its response time to input voltage looks too slow compared to rising time above. I want to maintain the original rising time as much as possible with reducing voltage.

Is there a way to achieve this job?

Mar 14, 2008
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3. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Do you mean actual value of the load impedance? The load is actually LED of the optocoupler and Dynamic resistance of this at 1.5 V is about few Ohm. Maybe few Ohm is so small that loading effect is ignorable. However, I would like to know general solution for this case.

4. ### #12 Expert

Nov 30, 2010
16,685
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The general solution is that LEDs do not respond to voltage changes, they respond to current changes. If your 4 volt signal has no impedance of its own, you use Ohm's Law.
4v - 1.5v = 2.5v
2.5v/0.02 amps = 125 ohms

If your signal generator has some significant impedance, like 50 ohms, you can subtract 50 ohms from 125 ohms.

From this point you can measure the actual voltage and current of the LED under operating conditions to refine your calculations, but this is almost never necessary.

Very small duty cycle operations can be adjusted to overdrive the LED because it will survive several times its rated continuous current when used part time. The datasheet for the optocoupler usually recommends where the limits are for this application.

Last edited: Jul 16, 2015
5. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4

First of all, I would like to know how could you get 0.02 A and why 50 ohm is subtracted from LED impedance of 125 ohm at 1.5 LED voltage? I've only seen addition or resistance, not subtraction.

And more important, is there any other method like voltage regulation but also having fast response time?

6. ### #12 Expert

Nov 30, 2010
16,685
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I used 0.02 amps because most optocouplers use an LED with that specification. LED does not have impedance of 125 ohms. A resistor between 4 volts and an LED would have 125 ohms.

There are many fast methods of voltage regulation, like using a resistor, but controlling voltage is not how to drive an LED. Until you learn that fact, you will not be successful. An LED is not a high impedance amplifier that responds to a voltage level and drives a light bulb. It is a semiconductor that responds to current changes, and can be very fast, but the voltage is variable according to current and temperature so voltage is not the way to achieve good results.

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7. ### ScottWang Moderator

Aug 23, 2012
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1. How is the current of generator?
2. How is the spec of optocoupler, any datasheet?
3. Are you trying to using 10 us pulse to light up the led or can be converting the 10 us pulse to the DC to light up the led?

8. ### crutschow Expert

Mar 14, 2008
13,496
3,373
If the load is an optocoupler then just place a resistor in series with the opto input with a value to limit the current to the desired value based upon the source pulse voltage and impedance, and the opto LED voltage.
Since the LED load is low impedance, the opto LED response will be essentially as fast as the pulse.
You don't have to worry about impedances otherwise.

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9. ### #12 Expert

Nov 30, 2010
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Thanks. I might have been unable to address the audience correctly. Still not sure.

10. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Hello.

1. The current of generator is unknown. I was using DG535 pulse delay generator made from Standford Research Systems and I believe it is voltage source rather than current. As I-V curve of optocoupler tells 1.5 V is good to have operation current, 1.5 V is my target voltage.

2. Please see the attached document.

3. What do you mean "converting 10 us pulse to DC"? Do you mean pure DC signal of constant amplitude over very long time? For trigger purpose, I don't see it is useful.

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11. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Oh..thus voltage drop across the LED is to be set as 1.5 V and Delay generator gives me 4 V. It means I can use Ohm's law to generate right current to LED such as I_s = (4 - 1.5)/R where R is placed between generator output and input of the optocoupler. I guess it is good idea!

Just one more thing; Our delay generator requires 50 Ohm or high Z load. When I set 50 Ohm of impedance control for generator, How can I match this to load of R and optocoupler?

12. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Thanks for showing that figure.

Can I ask one different question? If I want to treat load having non-linear I-V curve such as LED here, How can I say its resistance? Let's say I want to know resistance of LED at 1.5 V then the way of getting resistance at that point is dynamic resistance, dV/dI = R_dyn?

13. ### #12 Expert

Nov 30, 2010
16,685
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Dynamic resistance can be found as you said, but it seems unimportant. You are not driving 0.25 ohms of dV/dI. You are driving current through the whole LED. If the LED uses 1.5 volts when 0.02 amps is applied, it might be called 75 ohms. 75 ohms plus a 125 ohm resistor makes 200 ohms and 4 volts driving into 200 ohms is 0.02 amps. If the LED plus resistor load appears to be 200 ohms and you want a 50 ohm load you can add a resistor from the generator output to ground of 67 ohms. 200 ohms in parallel with 66.6667 ohms is 50 ohms.

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14. ### crutschow Expert

Mar 14, 2008
13,496
3,373
A good part of engineering design is recognizing what parameters are important to the design and which have little or no effect. Otherwise you will waste time (and money) on the design effort with no practical improvement in the design, in which case your company may lose money and your boss won't be too happy with you.
In this case the dynamic resistance of the LED has little effect on the practical operation of the circuit so is normally ignored.

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15. ### ScottWang Moderator

Aug 23, 2012
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According to the datatsheet of DG535, the output current only 0.7mA, so the current is not enough to drive the led of optocoupler to light up to its degree, A driver for the optocoupler as below, you may adjust the VR1 to adjust the current of led, you should adjust the VR1 to the biggest values in the beginning, because the led needs to operate in 10 uS, so the current could be reach up to 5~10 times of normal current that says 50~100 mA, I choose it at 70 mA here.

When you adjust the current of led to suit your needs then you can using a higher values for R4.

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16. ### #12 Expert

Nov 30, 2010
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I did not notice the DG535. Thanks for catching that.

17. ### ScottWang Moderator

Aug 23, 2012
4,930
777
If we don't ask the datasheet or detail spec of parts then we can't get to the target, may guess this guess that.

18. ### #12 Expert

Nov 30, 2010
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I thought he had a normal signal generator with a 50 ohm output impedance (that you would plug into the wall), or at least he would know he doesn't have a normal signal generator. My bad.

19. ### Dong-gyu Jang Thread Starter Member

Jun 26, 2015
100
4
Thanks for very detailed support.

I've checked datasheet of the DG535 and 0.8 mA is not maximum current, but average current from real panel side when this optional output is available. I don't see any current limitation specification for front panel where I've mostly used.

But you made me lead to have one question of what happens when the load is inserted into outputs of the driver circuit which has current limiting function or feather. For example, open output voltage of the driver is 4V and its current limit is 0.7 mA. When the load of small impedance is connected to output of the driver and 4V/R_load exceeds 0.7 mA, then...the output voltage is automatically adjusted to the value where output current is 0.7 mA?

Last edited: Jul 17, 2015
20. ### #12 Expert

Nov 30, 2010
16,685
7,327
It is a 50 ohm output. ttl 0v-4v

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