How can I adjust output voltage from pulse delay generator without loading effect.

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
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?
 

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
What is the load impedance?
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.
 

#12

Joined Nov 30, 2010
18,210
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.
 
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Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
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.
Thanks for replying

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?
 

#12

Joined Nov 30, 2010
18,210
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.
 

Attachments

ScottWang

Joined Aug 23, 2012
6,881
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?
 

crutschow

Joined Mar 14, 2008
24,704
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.
 

#12

Joined Nov 30, 2010
18,210
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.
Thanks. I might have been unable to address the audience correctly. Still not sure.
 

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
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?
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.
 

Attachments

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
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.
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?
 

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
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.
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?
 

#12

Joined Nov 30, 2010
18,210
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.
 

crutschow

Joined Mar 14, 2008
24,704
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.
 

ScottWang

Joined Aug 23, 2012
6,881
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.

DG535_Pulse_Dong-gyu Jang_ScottWang.gif
 

#12

Joined Nov 30, 2010
18,210
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. :(
 

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
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.

View attachment 88664
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?
 
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