Replacing UCC37321 with IRF2110 in a DRSSTC - Page 1
Forum section: electronics / projects

author: MrOmnos
author_rank: Frequent Contributor
timestamp: January 12, 2017, 05:13:26 pm
content: HiI am working on a tesla coil with can generate musical notes. I am basing my project on oneTesla DRSSTC design.http://onetesla.com/tutorials/how-a-tesla-coil-worksHere's the schematic of the controller:The problem is, I can't find any UCC37321 but the IRF2110 is readily available. So, I have decided to use the IRF2110.As you can see in the schematic above, the UCC37321 has a EN and IN for the input.Now the tutorial says:The inverting gate driver turns on when IN is high and EN is low. The noninverting gate driver turns on when IN is high and EN is high.I don't know much about about both UCC37321 or IRF2110 but I was thinking of ORing the IN and EN signal for Hin of the IRF2110 and ANDing the IN and EN signal for the Lin of the IRF. So, when IN is high and EN is low Highside will be driven and when both IN is high and EN is high Lowside will be driven.Will this work? What else should I taken in consideration?

author: Arjan Emm
author_rank: Regular Contributor
timestamp: January 13, 2017, 09:07:47 am
content: Quote from: MrOmnos on January 12, 2017, 05:13:26 pmThat is not correct. It should be:The inverting gate driver turns on when IN is low and EN is high. The noninverting gate driver turns on when IN is high and EN is high.When EN is low outputs are always low.Looking at the datasheet for the irf2110, it has a shutdown function SD instead of an enable function. So that input needs to be inverted. I guess connecting to Q pin 5 instead of /Q pin 6 should do the job. HIN and LIN can be connected together and work the same way as the two IN's of the ucc's connected together.The ucc's are available at farnell.

author: bktemp
author_rank: Super Contributor
timestamp: January 13, 2017, 09:27:18 am
content: Quote from: Arjan Emm on January 13, 2017, 09:07:47 amBoth are active high. So one signal must be inverted.But I doubt you can easily replace the UCCs with a IR2110:UCC is much faster (35/25ns instead of 120/94ns for IR2110). It can also ouput much more current (+/-9A vs +/-2A for 2110).

author: Arjan Emm
author_rank: Regular Contributor
timestamp: January 13, 2017, 09:34:54 am
content: Quote from: bktemp on January 13, 2017, 09:27:18 amYup, correct on all points, rushed through the IR datasheet too fast. The UCC is really fast, 1ns with a scope as a load. Really nice driver.

author: MrOmnos
author_rank: Frequent Contributor
timestamp: January 13, 2017, 10:21:44 am
content: Quote from: bktemp on January 13, 2017, 09:27:18 amHi! Can you please brief me on how the performance will be affected if I use IRF instead of UCC.

author: bktemp
author_rank: Super Contributor
timestamp: January 13, 2017, 11:38:02 am
content: Higher propagation delay -> more phase shift -> IGBTs don't operate at zero crossing -> they have more switching lossesLower drive current -> lower switching speed -> IGBTs spend more time in linear operation -> more switching lossesAnd you want the lowest switching losses possible, because the IGTs already work at a much higher frequency than usual.

author: MrOmnos
author_rank: Frequent Contributor
timestamp: January 13, 2017, 12:26:57 pm
content: Quote from: bktemp on January 13, 2017, 11:38:02 amI get what you are saying.The current transformer will detect the zero crossing but it will take time for the IRF to switch the IGBT, by that time IGBT will be no longer at zero crossing. I am I understanding this correct? So, If I were to use UCC, can I use both UCC37321 instead of one UCC37321 and another UCC37322? I don't have access to UCC her so I might have to get it from outside the country. Also IGBTi am using is Fairchild's 25n120. Will this IGBT do the job?

author: bktemp
author_rank: Super Contributor
timestamp: January 13, 2017, 12:58:52 pm
content: Quote from: MrOmnos on January 13, 2017, 12:26:57 pmYes.QuoteUCC37321 is inverting, UCC37322 is non inverting. Both form a H-bridge driving the gate drive transformer at +/-15V (when using 15V supply voltage for both UCCs).So if you use only UCC37321 or UCC37322, the signal going to one IC needs to be inverted.Since there are already inverters in the circuit, you can simply use the signal at the input of G2 for one driver and the signal at the output for the other one.

author: MrOmnos
author_rank: Frequent Contributor
timestamp: January 13, 2017, 04:02:47 pm
content: Quote from: bktemp on January 13, 2017, 12:58:52 pmSo, if I use Ir2110 what kind of output should I expect? If it were only a SSTC not DRSSTC would it make any difference?

author: Arjan Emm
author_rank: Regular Contributor
timestamp: January 13, 2017, 04:40:34 pm
content: The output is mostly dependent on the coil design, the resonant frequency and choice of igbt's. The driver is just there to keep the igbt's happy. They are expensive to replace...

author: bktemp
author_rank: Super Contributor
timestamp: January 13, 2017, 04:47:01 pm
content: It depends on how fast you need to switch:Your IGBT has a gate charge of about 200ns. The gate charge of both IGBTs is maybe 300nC total.Let's assume you want to switch in 100ns, then you need 300nC/100ns=3A gate drive.IR2110 can drive 2-3A, so its output resistance is probably somewhere around 5ohms. Since the gate drive transformer is driven using a H-bridge, there are two driver in series. So you have around 10 ohms series resistance, limiting the output current to less than 2A at 15V, that's less than 1A per IGBT.UCC3732x instead of IR2110 uses mosfets + BJTs for its output stage, therefore it has much less resistance and can drive much larger currents because of its fixed voltage drop.IR2110 could work, but it is probably less reliable than when using the more powerful driver.Maybe you could add an additional driver stage between IR2110 and gate drive transformer using BJTs. But this will add an additional delay.