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高頻變壓器參數(shù)該如何計(jì)算</h3>
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<span>文章出處??a href=""><a >深圳市晨飛電子有限公??/a></a></span><span>網(wǎng)責(zé)任編輯：<a >NSW</a></span><span>作者：<a >NSW</a></span><span>人氣??span
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<p> <a href="http://m.linhang.com.cn/product/product-0001,0015.shtml">高頻變壓??/a>一直被大家所熟悉，那么今天就由深圳晨飛電子的小編來給大家講一??a href="http://m.linhang.com.cn">高頻變壓器的參數(shù)</a>是怎么計(jì)算的：首先，我們應(yīng)該知道高頻變壓器的參數(shù)有以下幾點(diǎn)??、高頻變壓器輸入輸出要求 2、確定初次級匝數(shù)??3、計(jì)算功率場效應(yīng)管的最高反峰電??4、計(jì)算PWM占空??5、算變壓器初級電感量 6、計(jì)算初級線圈和次級線圈的匝??7、計(jì)算次級線圈的峰值電??8、計(jì)算激勵繞??也叫輔助繞組)的匝數(shù)，下面就來看看它具體是怎么計(jì)算出來的：<br />
1??nbsp;高頻變壓器輸入輸出要求：<br />
輸入直流電壓??00--- 340 V<br />
輸出直流電壓??3.5V <br />
輸出電流??.5A * 2<br />
輸出總功率：117.5W<br />
2??nbsp;確定初次級匝數(shù)比??br />
次級整流管選用VRRM =100V正向電流(10A)的肖特基二極管兩個，若初次級匝數(shù)比大則功率所承受的反壓高匝數(shù)比小則功率管反低,這樣就有下式??br />
N1/N2 = VIN(max) / (VRRM * k / 2) <br />
N1 ----- 初級匝數(shù) VIN(max) ------ 最大輸入電??nbsp; k ----- 安全系數(shù)<br />
N2 ----- 次級匝數(shù) Vrrm ------ 整流管最大反向耐壓<br />
這里安全系數(shù)??.9<br />
由此可得匝數(shù)比N1/N2 = 340/(100*0.9/2) ??7.6<br />
3??nbsp;計(jì)算功率場效應(yīng)管的最高反峰電??<br />
Vmax = Vin(max) + (Vo+Vd)/ N2/ N1 ??br />
Vin(max) ----- 輸入電壓最大??nbsp; Vo ----- 輸出電壓 <br />
Vd ----- 整流管正向電??br />
Vmax = 340+(23.5+0.89)/(1/7.6) <br />
由此可計(jì)算功率管承受的最大電?? Vmax ??525.36(V)<br />
4??nbsp;計(jì)算PWM占空比：<br />
由⑽式變形可得：<br />
D = (N1/N2)*E2/(E1+(N1 /N2*E2) <br />
D=(N1/N2)*(Vo+Vd)/Vin(min)+N1/N2*(Vo+Vd) ⒀<br />
D=7.6*(23.5+0.89)/200+7.6*(23.5+0.89)<br />
由些可計(jì)算得到占空比 D??0.481<br />
5??nbsp;算變壓器初級電感量：<br />
為計(jì)算方便假定變壓器初級電流為鋸齒波，也就是電流變化量等于電流的峰??也就是理想的認(rèn)為輸出管在導(dǎo)通期間儲存的能量在截止期間全部消耗完。那么計(jì)算初級電感量就可以只以PWM的一個周期來分析，這時可由⑼式可以有如下推導(dǎo)過程??br />
（P/η?? f = 1/2 * I2 * L ??br />
P ------- 電源輸出功率 (瓦特) η ---- 能量轉(zhuǎn)換效率 f ---- PWM開關(guān)頻率將⑺式代入⒁式??br />
（P/η?? f = 1/2 * (EL * ⊿t / L)2 * L ??br />
⊿t = D / f (D ----- PWM占空?? <br />
將此算式代入⒂式變形可得??br />
L = E2 * D2 *η/ ( 2 * f * P ) ??<br />
這里取效率為85%??PWM開關(guān)頻率為60KHz.<br />
在輸入電壓最小的電感量為??br />
L=2002* 0.4812 * 0.85 / 2 * 60000 * 117.5<br />
計(jì)算初級電感量為: L1 ??558(uH)<br />
計(jì)算初級峰值電流：<br />
由⑺式可得：<br />
⊿i = EL * ⊿t / L = 200 * (0.481/60000 )/ (558*10-6)<br />
計(jì)算初級電流的峰值為: Ipp ??2.87(A)<br />
初級平均電流為： I1 = Ipp/2/(1/D) = 0.690235(A)<br />
6??nbsp;計(jì)算初級線圈和次級線圈的匝數(shù)??br />
磁芯選擇為EE-42(截面??.76mm2)磁通密度為防治飽和取值為2500高斯也即0.25特斯?? 這樣由⑹式可得初級電感的匝數(shù)??<br />
N1= ⊿i * L / ( B * S ) = 2.87 * (0.558*10-3)/0.25*(1.76*10-4)<br />
計(jì)算初級電感匝數(shù): N1 ??36 (??<br />
同時可計(jì)算次級匝數(shù)： N2 ??5 (??<br />
<br />
7??nbsp;計(jì)算次級線圈的峰值電流：<br />
根據(jù)能量守恒定律當(dāng)初級電感在功率管導(dǎo)通時儲存的能量在截止時在次級線圈上全部釋放可以有下式??br />
由⑻⑼式可以得到??br />
Ipp2=N1/N2* Ipp ??br />
Ipp2 = 7.6*2.87 <br />
由此可計(jì)算次級峰值電流為：Ipp2 = 21.812(A)<br />
次級平均值電流為I2=Ipp2/2/(1/(1-D))= 5.7(A)<br />
8.計(jì)算激勵繞??也叫輔助繞組)的匝數(shù)：<br />
因?yàn)榇渭壿敵鲭妷??3.5V，激勵繞組電壓取12V，所以為次級電壓的一??br />
由此可計(jì)算激勵繞組匝數(shù)為: N3 ??N2 / 2 ??3 (??<br />
激勵繞組的電流?? I3 = 0.1(A)</p>
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