Inductive Kickback

[gyulasun]

Hi Gyula
Yes,the measurements are from the scope shot below.
If we look at the input side of thing's,it is saying our inductor has a resistance of around 1890 ohms at that frequency when we subtract the 10 ohm CVR. The reason i asked that question is that the LED dose not need that amount of current to get the 2.85 volts across it at that 70% duty cycle. So with this circuit,how do we measure the power being dissipated by the LED?.-->circuit below.

Hi Brad,

I can agree with the 4.8mA current (from the scope) for the LED when the voltage across it is 2.85 V, (this 13.68mW is the mean or average comsumption of the LED)  and this could be checked with brightness comparison in your "light box": just feed the same pure DC power from your power supply to the same LED to get the same DC output voltage across the box output resistor what the circuit provides to it from the flyback energy.  (Be careful to the LED when changing the supply output voltage, better to use a series few hundred Ohm resistor to limit current and measure the voltage across LED with a DMM, of course the current is to be checked too.)

I think the dissipation in the 10 Ohm CVR comes first from the input current of your function generator during the ON times and to this we need to add the dissipation caused by the 4.8 mA current during the OFF times.
The wire resistance of the coil also dissipates during the ON time and the OFF times like the 10 Ohm, while the LED dissipates only during the OFF times, this latter is the 13.68 mW.

EDIT:  I think the duty cycle is to be considered for the LED too because the 4.8 mA current through the 10 Ohm includes both the ON and the OFF time currents and I based the LED dissipation on that.   

Gyula

[tinman]

Hi Brad,

and this could be checked with brightness comparison in your "light box": just feed the same pure DC power from your power supply to the same LED to get the same DC output voltage across the box output resistor what the circuit provides to it from the flyback energy.  (Be careful to the LED when changing the supply output voltage, better to use a series few hundred Ohm resistor to limit current and measure the voltage across LED with a DMM, of course the current is to be checked too.)

I think the dissipation in the 10 Ohm CVR comes first from the input current of your function generator during the ON times and to this we need to add the dissipation caused by the 4.8 mA current during the OFF times.
The wire resistance of the coil also dissipates during the ON time and the OFF times like the 10 Ohm, while the LED dissipates only during the OFF times, this latter is the 13.68 mW.

Gyula

I can agree with the 4.8mA current (from the scope) for the LED when the voltage across it is 2.85 V, (this 13.68mW is the mean or average comsumption of the LED)

This was the very reason i asked the question Gyula,as it dose not take 4.8mA to get 2.85v across the LED-->it only takes around 1.2mA to get the 2.85v across the LED. This i checked simply by using my FG,the LED and CVR. With the FG set to a 70% duty cycle,and set so as i have the 2.85v across the LED,i only had 12mV across the 10 ohm CVR. So it seems that the power dissipated by the LED cannot be measured using the voltage across it,and current flowing through the CVR when in the flyback circuit,as it gives us 4x the actual power required to obtain the 2.85v at a 70% duty cycle-by way of the FG.

Brad

[gyulasun]

Hi Brad,

Yes and in the meantime I wrote an addition to my previous post.

The CVR should be placed directly in series with the LED and not in series with the coil.

Thanks,  Gyula

[tinman]

Hi Brad,

Yes and in the meantime I wrote an addition to my previous post.

The CVR should be placed directly in series with the LED and not in series with the coil.

Thanks,  Gyula

Gyula
The CVR is directly in series with the LED-see circuit below.
As there is only 3 components in the circuit during the flyback stage,how can they not all be in series?.

[gyulasun]

Yes it is in series but it is also included in the input circuit current and the scope shot includes that too, no?

EDIT  This is why I goofed by saying the 13.68 mW LED dissipation.

Gyula