Generally what I see on 'spec' sheets for voltage drop is: "max 1 volt@1 amp".
In my circuit running at 1.2vdc, 50ma, I measure a votage drop of about 0.2v. Is this correct? Does voltage drop change with current?

 

Generally what I see on 'spec' sheets for voltage drop is: "max 1 volt@1 amp".
In my circuit running at 1.2vdc, 50ma, I measure a votage drop of about 0.2v. Is this correct? Does voltage drop change with current?

Yes, there is a V/I curve for diodes. Look at the datasheet. it is not linear.

 

Thanks GopherT. I'll try to find a data sheet. (when I said 'spec sheets' I meant the data that is listed on a site like digikey)
I looked at some data sheets and voltage drop seems to be about 0.6v @ 100ma. Would a measured voltage drop of 0.2v be reasonable for say 10ma?

 

I looked at some data sheets and voltage drop seems to be about 0.6v @ 100ma. Would a measured voltage drop of 0.2v be reasonable for say 10ma?

According to this from the attached datasheet; forward voltage would typically be 0.6V @ 10mA and 0.75V @ 100mA.

 

Post your circuit diagram.

 

Thanks GopherT. I'll try to find a data sheet. (when I said 'spec sheets' I meant the data that is listed on a site like digikey)
I looked at some data sheets and voltage drop seems to be about 0.6v @ 100ma. Would a measured voltage drop of 0.2v be reasonable for say 10ma?

No, that is not reasonable at all.

For the Fairchild 1N4002 (https://www.fairchildsemi.com/datasheets/1N/1N4002.pdf), you see that the typical (as opposed to max) voltage at 100 mA is about 750 mV.

For diodes, a good rule of thumb is that the current changes by a factor of ten for a change in forward voltage of about 120 mV (this corresponds to an ideality factor of 2 near room temperature). As you can see from Figure 2, the voltage at 10 mA is about 600 mV, which is in pretty decent agreement with this rule of thumb.

As you can see, the current-voltage characteristic (on a semi-log plot) is roughly linear for lower currents. At higher currents other factors, such as parasitic resistances, start playing a more pronounced role.

 

Thanks WBahn. I'll double check in what I am doing/measuring.