Riccardo Posted November 25, 2009 Report Share Posted November 25, 2009 Hi,I'm looking for an integrated H-Bridge IC like those used in motor control applications. I've seen the VNH3SP30-E (http://www.st.com/stonline/products/literature/ds/12688.pdf) but it is only rated for 36V. The current rating it has of 30A is just enough but ideally I would like something higher.This IC looks just what I need except for the voltage rating (I need at least 50V). I've been searching for hours but can't seem to find what I need. Does anyone know of an equivalent for this? I must be using the wrong search terms because I'm sure something like this will exist.Cheers Quote Link to comment Share on other sites More sharing options...
Hero999 Posted November 25, 2009 Report Share Posted November 25, 2009 50V@30A.I think you'll need to build the h-bridge from scratch from MOSFETs.How fast does it need to switch? Quote Link to comment Share on other sites More sharing options...
Riccardo Posted November 25, 2009 Author Report Share Posted November 25, 2009 Hmm. I was hoping to avoid that as I'd need to make a high and low side driver too. Those IC's also have fault indicators and other useful stuffIt only needs to go at 50Hz or less. Quote Link to comment Share on other sites More sharing options...
Hero999 Posted November 26, 2009 Report Share Posted November 26, 2009 50Hz is fairly low so it should be easy to level shift the h-side P-channel MOSFETs using BJTs and a potential divider or zener to prevent the gate voltage from being exceeded. At higher frequencies this sort of thing gets harder as the resistors slow the switching speed too much.I'll post a schematic if you don't get it. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted November 26, 2009 Author Report Share Posted November 26, 2009 Yes, a schematic would be great thanks. Do you also know of a control IC that can be used to simplify the driving if it. e.g, prevents shoot through or other fault conditons? Quote Link to comment Share on other sites More sharing options...
Hero999 Posted November 26, 2009 Report Share Posted November 26, 2009 Here's a schematic, no control circuitry is required assuming the lower MOSFETs are logic level.The diodes prevent both the high and low side MOSFETs from being on simultaniously. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted November 27, 2009 Author Report Share Posted November 27, 2009 Wow, you made that diagram just for me! Thanks! I have a few more questions ::)Will I need to change those resistor values if the supply voltage drops to 25V ?Just to confirm: The high site MOSFETs are ordinatry P-Type, and the low side ones are logic level N-Type ? They must not have reverse diodes built in?Can I use IGBT's instead of MOSFET's? Quote Link to comment Share on other sites More sharing options...
Hero999 Posted November 27, 2009 Report Share Posted November 27, 2009 R1 and R2, and R3 and R4 form potential dividers which limit the gate voltage to the high side MOSFETs to a safe level, whether their values need to be changed for operation at 25V depends on the drain current at the lower gate voltage.The MOSFETs are normal enhancement MOSFETs with built-in diodes, if you're using this to power a large motor it's a good idea to add extra diodes because the internal diodes might not be fast enough.I don't know why you'd want to use IGBTs, at low voltages the on voltage loss is higher than MOSFETs, IGBTs are only any good for higher voltages >400V or so. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted December 1, 2009 Author Report Share Posted December 1, 2009 It will be to drive coils / electromagnets so I was thinking that an IGBT would be better for 2 reasons.1. They can have a high voltage rating and be more tolerant of voltage spikes.2. They don't get as hot for a given current.The supply voltage to the load will be variable (although there will be a constant 12V source available for the control electronics). Different coils would also be used so the output current could vary quite a lot.For those potential dividers, would it be ok to just add some back to back zenner diodes to prevent over voltage on the gates? Quote Link to comment Share on other sites More sharing options...
Hero999 Posted December 1, 2009 Report Share Posted December 1, 2009 2. They don't get as hot for a given current.That's not always true.It depends on the MOSFET and the IGBT, as I said before, for devices with higher voltage ratings IGBTs are best but for low voltage devices MOSFETs are much better.For those potential dividers, would it be ok to just add some back to back zenner diodes to prevent over voltage on the gates?If you're going to use zeners then you might as well replace R1 and R3 with zener diodes to limit the gate voltages to a safe level. It's still a good idea to put a resistor in parallel with the zeners, to alow the gate capacitance to discharge, but it's value is much less critical than for a potential divider. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted December 1, 2009 Author Report Share Posted December 1, 2009 Thanks again. I really appreciate your help.Does this diagram look ok?I've not found an appropriate pair of mosfets yet. I find one type, but then can't find its equivalent of opposite polarity. I'll update when I find them. Quote Link to comment Share on other sites More sharing options...
Hero999 Posted December 1, 2009 Report Share Posted December 1, 2009 No, you clearly don't understand how the circuit works.What voltage is driving the low site MOSFETs?I assume it's 5V so why do you need 10V zeners to protect it?Why do the zeners need to be bidirectional when the gate voltage can only be one polarity?The high side MOSFETs are connected backwards. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted December 1, 2009 Author Report Share Posted December 1, 2009 :-[I thought 5V would be too low to switch the mosfet properly. I was thinking of using some comparators to convert the voltage level to around 12V.I chose 10V zeners because that is high enough to allow the mosfets to switch on but low enough to protect them against any transients.I made them bidirectional because I'm concerned that transients of either polarity could occur since the load would be inductive and the current is being alternated. Is that wrong?I just connected the mosfets like in your diagram, but yes, I see now that they (and the diodes I added) should be the other way around. Don't know how I missed that! Quote Link to comment Share on other sites More sharing options...
Hero999 Posted December 1, 2009 Report Share Posted December 1, 2009 If you use logic level MOSFETs for the low side then you won't need to perform any level shifting because they will be able to conduct the full current with 5V at the gate. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted December 9, 2009 Author Report Share Posted December 9, 2009 I've got some STP40NF10L, and IRF5210PBF MOSFETs handy so I think I will use those as they are N and P type respectively and both rated for 100V and 40A.I also have some comparators, so I will just use those for the level shifting.Thanks Quote Link to comment Share on other sites More sharing options...
Hero999 Posted December 9, 2009 Report Share Posted December 9, 2009 Those should be fine but you'll need to connect two in parallel for 50A.Looking at the datasheet, the STP40NF10L should be able to pass 25A with a gate voltage of 5V so you shouldn't need any level switching. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted December 9, 2009 Author Report Share Posted December 9, 2009 Ok, Thanks for your help :) Quote Link to comment Share on other sites More sharing options...
Riccardo Posted February 3, 2010 Author Report Share Posted February 3, 2010 Hi there,I have another question about the diagram posted by Hero999.On the gates for the high side transistors, there is a pair of resistors (R1, R2). When input A goes high, this should activate Q1 and pull down the gate voltage of Q5. My question is, How did you choose those resistor values? As far as I can tell, with Q1 activated, the gate of Q5 will be at 32V (relative to ground).I've tried this in a test circuit with 9V input and a small solenoid as a load, but it wont switch it. Just wondering if the high side gate is not being pulled low enough.Cheers! Quote Link to comment Share on other sites More sharing options...
Hero999 Posted February 3, 2010 Report Share Posted February 3, 2010 The resistors form a potential divider which keeps the gate voltage to the top MOSFETs below the maximum rating (normally 20) but high enough to turn them fully on.V = 4.7*50/(10+4.7) = 16VIf the supply voltage is too low the gate voltage will be too low to enable the MOSFETs to carry the required current. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted February 5, 2010 Author Report Share Posted February 5, 2010 Great, thanks. I tested with 18V input and its working fine :) Quote Link to comment Share on other sites More sharing options...
Riccardo Posted February 10, 2010 Author Report Share Posted February 10, 2010 Ok, I melted it! (not so suprising :-[)Wen I tested with 18V, it was simply a pair of 9V PP3 batteries in series.The test load was a little solenoid with a resistance of 54 ohms.When I replaced the 18V with a 24V SLA battery, the MOSFETs (rated for 40A) blew immediatley. I had a 30A car fuse installed but I guess it was too slow acting.It seems like there must have been a shoot through condition, but I don't know why.I replaced the MOSFETs and then tested again at 18V from a 5A PSU. This again seems to be working fine, even with lower resistance loads. I do notice though that the heat sink holding the high side MOSFETs gets quite warm, while the low side one is totally cold.I noticed that you mentioned I could replace R1 and R3 with zeners. Is something like this going to make this work better with the different possible voltages/currents? Quote Link to comment Share on other sites More sharing options...
Hero999 Posted February 11, 2010 Report Share Posted February 11, 2010 The high side MOSFETs will dissipate more power because they are P-channel and will have a higher on resistance.two 9V batteries in series driving a 54R load is not 18V because the batteries are overloaded the voltage will drop to around 14V.I don't know why the MOSFETs blew when it was connected to a 24V supply. Replacing the resistors with zeners will enable the circuit to work from any voltage above the required gate voltage for the high side to pass the desired current, up to the maximum voltage rating of the MOSFET and the load. Just make sure that the power dissipation of the zeners and resistors is not exceeded. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted February 16, 2010 Author Report Share Posted February 16, 2010 Yes, that makes sense, they have twice the resistance (0.06 vs the 0.033 of the n types). I'm using IRF5210PBF and STP40NF10L.With the current limited to 5A, I can turn up the voltage without it blowing, but I do notice that if I feel the case of the p-types, they are very hot within seconds while the n-types are cold (this explains why the blew when using the 24V SLA).The load (a hollow coil with a screwdriver in it so I can hear it vibrate) sounds like it is working with the pulses I send from a PIC.So do you think the p-types are not switching on, or maybe off properly? How could I test this?I've attached my current circuit. Quote Link to comment Share on other sites More sharing options...
Hero999 Posted February 16, 2010 Report Share Posted February 16, 2010 Only D3 and D5 are required.Think about it.The gate voltage for the lower MOSFETs comes from a logic circuit so there's no chance of it exceeding 5V, let alone 20V.The voltage is also only one polarity so there's no need to connect the zeners back-to-back which is only required for AC.D3 and D5 are also in completely the wrong place: they need to be connected between the gate and source.The circuit should be the same as my previous circuit but with R1 and R3 replaced with zener diodes. D1 and D2 don't need to be Schottky barriers, ordinary silicon diodes will do. Quote Link to comment Share on other sites More sharing options...
Riccardo Posted February 16, 2010 Author Report Share Posted February 16, 2010 I see, so those zeners being in the wrong place would be why the p-types were not switching as I expected.I had the other excess zeners like that just as some overkill really. I can see they are not needed, but there's no harm in leaving them there is there (they are already soldered to my board)?The attached diagram is better? Quote Link to comment Share on other sites More sharing options...
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