[ high (>20KV) back EMF suppression ] - Page 1
Forum section: electronics / beginners

author: texane
author_rank: Regular Contributor
timestamp: April 03, 2012, 05:43:24 am
content: Hi,I am working on a project where I load capa then discharge them rapidlyin a coil using a MOSFET to produce a high magnetic field. To be efficient, oneconstraint is to make the field as short (in time) as possible. Thus, di/dt isvery high, and the back EMF can reach up to 20kV... and I did not find any'flyback' diode that can handle such a voltage. I found some monster diodesthat can handle 6kV, but they burned up during the first test ... :/I try to change the switching MOSFET by an IBGT, but same problem asabove.Maybe there are some well known techniques to address this issue, and yourhelp would be very appreciatedCheers,Fabien.

author: Psi
author_rank: Super Contributor
timestamp: April 03, 2012, 06:35:45 am
content: Maybe look at TVS diodes or MOVs

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 03, 2012, 09:01:18 am
content: Your diode will try to clamp the voltage at about a volt so the problem is really the current. More precisely, the power that's dissipated by the diode.You can try a diode with a higher power rating.And/or you can clamp at a higher voltage by putting several diodes in series. This will spread the power dissipation equally across the diodes.A small value resistor in series with the diode is another option. This will limit the current but raise the voltage.

author: BravoV
author_rank: Super Contributor
timestamp: April 03, 2012, 09:10:39 am
content: Quote from: TerminalJack505 on April 03, 2012, 09:01:18 amWill a series of diodes also increase the total break down voltage proportionally ?

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 03, 2012, 09:19:14 am
content: Quote from: BravoV on April 03, 2012, 09:10:39 amAs in the reverse breakdown? That I don't know.It shouldn't problem in this particular case, though.

author: Neilm
author_rank: Super Contributor
timestamp: April 03, 2012, 06:32:40 pm
content: There are some high voltage diodes used for rectifying at high voltages but they are not that easily available. One I have seen is the SL500 that can take 5kV. I think that it does go up to 12kV. I don't know how much power they can take. I have seen diodes that go up higher than that (>20kV) but I don't know the part number for that. I also don't know where you could get it. I used these in a project and ended up speaking to specialist distributors.YoursNeil

author: IanB
author_rank: Super Contributor
timestamp: April 03, 2012, 06:52:22 pm
content: Do you have a circuit diagram? I don't understand where your diodes are placed and what function they are performing.(For example, back EMF suppression usually involves putting a diode in parallel with the coil to shunt the collapsing field current away from the rest of the circuit and prevent an EMF building up. What is the configuration here and why does the diode need 20 kV capability?)

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 03, 2012, 06:53:43 pm
content: I'm confused as to why everyone seems to think the diode needs to have a high voltage rating. The voltage rating of the diode just needs to be greater than the voltage that the capacitors are charged-up to before dumping their current into the inductor.The whole purpose of the flyback diode is to prevent the EMF voltage spiking up to, in this case, 20kV. It does this by clamping the voltage at its forward voltage, Vf. This will just be a volt or two in this particular case.So I'm pretty sure what is killing the diode is the heat dissipated during flyback. If the current through the inductor was 10A when the MOSFET was turned off then this 10A is going to flow through the diode. 10A * 2V is 20W of power, which the diode turns into heat.

author: texane
author_rank: Regular Contributor
timestamp: April 06, 2012, 11:12:03 am
content: Hi,Thanks for your replies.I join you a partial schema, but I have to take a look at theactual capa value so that I can compute the actual currentand the power that needs to be dissipated by the diode, asTerminalJack505 suggests.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 06, 2012, 12:47:44 pm
content: Assuming the capacitor and inductor are lossless (i.e., worst case), the peak current through the diode is:Imax = V * sqrt(C/L)The peak power (given the diode forward drop is Vf) is:Pmax = Imax * Vfand the total energy per pulse absorbed by the diode is:Qd = 0.5 * C * V^2And the decay time of the field is:t_decay = sqrt(L*C) * V/Vf