A very basic question about diode voltage drop (Solved) - Page 1
Forum section: electronics / beginners

author: sam1275
author_rank: Frequent Contributor
timestamp: March 01, 2017, 12:57:31 pm
content: Hello.I did play a lot with electronics but this is the first time I use a diode for voltage drop purpose, on a expensive device, so I want to make sure it won't break anything...The device require anything between 1.8-3.2v to operate, and the power require will be various depend on device state, up to about 1.9 watts. I want to use aregulated 3.3v sourceLi-ion battery, can I just puta2 diodes in serial, and expect it to drop0.7v1.4v, thus the device will never receive higher than 3.2v at ANY case?Thanks.

author: Brumby
author_rank: Unknown
timestamp: March 01, 2017, 01:13:17 pm
content: Not sure what diode family you are looking at - but here is the relevant chart for the 1N4004:As you can see the actual voltage drop is dependent on current and temperature.

author: capt bullshot
author_rank: Super Contributor
timestamp: March 01, 2017, 01:20:15 pm
content: From your description of the constraints, I'd say: yes it'll work. From my experience: don't do it. Use a LDO. I've made a circuit like yours many years ago, only to discover that the diode drop supplied part of the circuit wouldn't start early in the morning. (BTW: reason was room temperature, not earliness in the morning)There's no diode in this world that will drop 0.7V. Voltage drop of a silicon diode varies from below 0.4V to above 1V (depending on temperature, diode size and current)

author: S13
author_rank: Regular Contributor
timestamp: March 01, 2017, 01:32:10 pm
content: If its an expensive device then chances are it wont work as expected. Yes you get a voltage drop, but the amount of drop depends on temperature and the current drawn as mentioned before. Especially if your device has erratic current draw (like digital switching circuitry for example) then you get a very unstable power supply which could lead to circuitry instability. Perhaps something you dont want with an expensive device like you described?Also you should guarantee a minimum voltage drop on the output of your diode so when the attached device draws no current at all you still dont wind up with too little voltage drop and thus an overvoltage situation. You can take care of this by placing a dummy load (a small value resistor load) that guarantees a certain amount of voltage drop.Otherwise I agree with other comments here. Use a true regulator if possible, especially when trying to power complex circuits this way. If its just a light bulb, led or something equally simple that you want to power then dont bother with the LDO and go for the quick and dirty diode solution.

author: sam1275
author_rank: Frequent Contributor
timestamp: March 01, 2017, 01:54:56 pm
content: Thank you everyone, I just edited OP because that 3.3v source is not enough for the power draw...That is a special camera with lots of ADC, DAC, processors, motors... and will be damaged if exceed 3.2v according to manual.Maybe I would consider something else as all of you suggest me to not do it

author: S13
author_rank: Regular Contributor
timestamp: March 01, 2017, 02:04:02 pm
content: If its a camera that you are trying to power from a li-ion then please go for a true regulator solution!An LDO should be sufficient in your case, but keep in mind that cold temperatures and batteries dont mix well together and can cause unexpected low output voltages. With the addition of an LDO-drop you might not make it to the minimum 1.8V, but that is something for you to verify.

author: sam1275
author_rank: Frequent Contributor
timestamp: March 01, 2017, 02:31:18 pm
content: Thank you everyone!

author: KL27x
author_rank: Super Contributor
timestamp: March 01, 2017, 09:13:25 pm
content: I have done this to put li ion battery (but with a single diode) for many 3V devices. This is somewhat rolling the dice. If it is digital and it is expensive, don't do it! Some - well, all - of my devices are still working to this day. in a digital device, the IC is drawing close to nothing when it's in standby. So, technically voltage is slightly too high at standby - the resting voltage of the li ion battery can be as high as 4.2, and the diode drops something something closer to 0.3 or 0.4V (for very, very sensitive diodes, you can technically record voltage drop at low as 245mV at very tiny draw). Now, there are several reasons this slightly high standby voltage is probably not an issue, but it is still guessing, unless you really know the device in question. 2 diode drops would be safer, but at high current you are now way too low. This is a hack job and should be left where a hack job is sufficient.

author: sam1275
author_rank: Frequent Contributor
timestamp: March 02, 2017, 02:41:16 pm
content: Quote from: KL27x on March 01, 2017, 09:13:25 pmThank you, I'm back here to add my findings and saw you said exactly what I'm going to say.The diode's(in4007) voltage drop is as low as ~0.3v when only load in serial is a volt meter, so it may be dangerous for a sensitive device when not on heavy load.