Let's suppose a Rectifier with Vd,on=0.7 V is connected to a sin-voltage source with Vamp=10 V. When in this case of forward bias can I ignore the voltage drop of the diode? Or is it only possible if the input voltage has a very high amplitude? I've heard that you can do in for example 5000 V, is it right and why is that? In this case I haven't ignored the voltage drop and the output voltage is Vpeak-2Vd,on.
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\$\begingroup\$ Did you have anything in particular in mind? Are you asking only about forward-biased diodes? \$\endgroup\$– Andrew MortonCommented Oct 28, 2020 at 9:23
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1\$\begingroup\$ A warm welcome to the site. Please note that it's not a homework-answering service or on-line technical encyclopaedia, copied out to you on demand. People will help you take the next step if your question shows you've already done as much as you possibly could - which yours doesn't, I'm afraid. Please edit your question and greatly improve it. Show your own work and own findings in considerable detail with any schematic. The schematic tool here is easy to use. The better the quality of your question, the better the quality of the answers it will attract. Again, welcome. \$\endgroup\$– TonyMCommented Oct 28, 2020 at 9:24
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\$\begingroup\$ Simply you can't. It is not just the diode has a voltage drop it regulates current as well. If you put a battery connected to a diode in a circuit without resistors the diode will prevent current from reaching infinity. \$\endgroup\$– Se1fieCommented Oct 28, 2020 at 9:24
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1\$\begingroup\$ Yes, there are such situations. The forward voltage drop across a diode is in the order of 0.6 V to 1 V. If you're dealing with 5 V that drop is significant. If you're dealing with 5000 V that drop might not be significant. \$\endgroup\$– BimpelrekkieCommented Oct 28, 2020 at 9:25
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1\$\begingroup\$ I think it's bad practice to assume the forward voltage drop on a diode is fixed. I recall an Analog Devices app note which used a diode for its "0.7V" drop. The diode fed MOSFET gates and the authors didn't even put a load resistor to force a current through the diode, so the suggested circuit in the app note didn't even work as intended. In this day and age, it's better to use a SPICE simulator and have it compute the actual voltage based on its exponential I-V relationship. \$\endgroup\$– Ste KulovCommented Oct 28, 2020 at 9:49
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3 Answers
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It is depend on which diode model you are using to make calculations. There are three common diode models.
Exponential Mode:
Constant-Voltage Mode:
Ideal Mode:
So, as you see like all figures voltage drop on the diode is not just related to voltage. Current effects the diode too. If you want to answer when will you neglect the diode voltage drop, you should first select the which diode model you are going to use.
Figures are from: Microelectronic Circuits Adel S. Sedra & Kenneth C. Smith (Table 3.1)
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\$\begingroup\$ Nice answer showing the relation between model choice and voltage drop. Regarding the last sentence, I'd say you should probably first determine whether you can neglect the voltage drop in order to decide which model to use though. (Unless you are in a class and your teacher tells you which model to use.) \$\endgroup\$– JustinCommented Oct 28, 2020 at 13:34
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\$\begingroup\$ Could you please add citations or links to the source of the graphics you copied into your answer? We want to be sure to give credit to the original creator. \$\endgroup\$ Commented Oct 28, 2020 at 14:10
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\$\begingroup\$ @ElliotAlderson Sure, I edited my answer \$\endgroup\$– ratatoskCommented Oct 29, 2020 at 20:10
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Neglecting a diode's voltage drop depends on the circuit it is operating in. You can neglect it all the time, and all may be fine, but this is not good practice as it can be significant like you say if the operating voltage is only a few times higher than the diode drop. For an operating voltage, orders of magnitude higher it may be OK to assume the diode drop is insignificant, but this depends on how much insignificance you want to assign to the diode in your expectation of how the circuit behaves. Diode voltages are also a function of current flow and so may increase or decrease with current passing through, and this is very predicatble. It all boils down to your concern of how much importance you want to assign to the diode's voltage drop and if you decide to ignore it how this may impact on your expectation of circuit behavior otherwise ... If this won't impact on how you predict the circuit will function, it may be ok, otherwise if you ignore it but actually does have an impact on overall circuit operation, then this wouldn't be ok. Generally you are right in some regard that you can assume "small-signal" effects are second-order when they become so, and may be "rounded-off" to simplify the prediction of circuit operation etc. Just like rounding-off decimal points when the precision does not call for it ...
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1\$\begingroup\$ Use 2 x <Enter> for paragraph breaks. \$\endgroup\$ Commented Oct 28, 2020 at 13:43
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At the risk of stating the obvious, you can neglect it if the voltage drop is too small to matter.
If you put a 0.7V diode into a circuit that runs at 5000V, then you will have 4999.3V left. That's a trivial difference.
If you put a 0.7V diode into a circuit running from a 1.5V alkaline cell, that leaves you with only 0.8V to power the rest of the circuit. That's a huge difference.
