LogicLimited-DM101-tt-sch 维修电路原理图.pdf
-
资源ID:164158
资源大小:196.93KB
全文页数:3页
- 资源格式: PDF
下载积分:8积分
会员登录下载
三方登录下载:
友情提示
2、PDF文件下载后,可能会被浏览器默认打开,此种情况可以点击浏览器菜单,保存网页到桌面,就可以正常下载了。
3、本站不支持迅雷下载,请使用电脑自带的IE浏览器,或者360浏览器、谷歌浏览器下载即可。
4、本站资源下载后的文档和图纸-无水印,预览文档经过压缩,下载后原文更清晰。
5、试题试卷类文档,如果标题没有明确说明有答案则都视为没有答案,请知晓。
|
LogicLimited-DM101-tt-sch 维修电路原理图.pdf
Logic DM101 Electronic Reverse engineered to fix mine by; Kelvin Unsworth 21st October 2006 Note: Some component values may be wrong due to colour fade. This is how I think the circuit works. There are NO guarantees. Part numbers/values and circuit connections could be wrong. Use at your own risk. 240 VAC enters via a 100ma anti surge fuse (not shown on scematic) then passes through a 100 ohm resistor. There is then an anti surge varistor to prevent mains spikes. The 240 VAC is then recified and filtered with two 22uF electrolytic capacitors giving approx +150 and -150 VDC with a 0V in between. On the -150VDC side the -ve voltage is supplied to the -ve out put transistors Q5 and Q6. It also supplies a 15k wire wound resistor that feeds on to two 7.5v zenner diodes. These make the -15v supply rail. On the +150VDC side the +ve voltage is supplied to the +ve out put transistors Q3 and Q4. It also supplies a 15k wire wound resistor that feeds the LED circuitry. Current will flow through either Q1 or LED1 depending on the state of the two flip flops in IC2. If the 45RPM button has been pressed then out puts QA (IC2a) and QB(IC2b) will be on ( at +15v) and D1 and D2 will not conduct and Q1 will turn off. The current will then flow through LED1 lighting it. I either QA or QB are off then Q1 will be turned on and there will not be enough voltage across LED1 to light it. The current flow will continue through to Q2. Current will flow through either Q2 or LED2 depending on the state of the two flip flops in IC2. If the 33RPM button has been pressed then out puts NOTQA (IC2a) and QB(IC2b) will be on ( at +15v) and D3 and D4 will not conduct and Q2 will turn off. The current will then flow through LED2 lighting it. I either NOTQA or QB are off then Q2 will be turned on and there will not be enough voltage across LED2 to light it. QB = turn motor on QA = 45RPM NOTQA = 33RPM Pressing SW3 (45RPM button) puts 15v onto the set input of flip flop a and b latching QA and QB on via D9 and D8. Pressing SW1 (33RPM button) puts 15v onto the set input of flip flop b via D7 and reset flipflop b latching NOTQA and QB on. Pressing SW2 resets flipflop b turning QB off. The current flowing via Q1/LED1 Q2/LED2 feeds through to two 7.5 v zenner diodes making the +15v supply. The flipflip a out put QA is used to operate three solid state switches in DG308ACJ these short out R29,R30,R26 and R25 in the oscilator circuit changing its frequency from 50 Hz (33RPM) to 67.5 Hz (45RPM) Flipflop b out put QB is used to switch the oscilator out put from IC3b into the drive opamp IC1. The DG308ACJ ic is only shown as the four solid state switches on the scematic. IC3a/b is an LF353N my CAD program did not have this IC so it is shown as an LM348N. RadioFans.CN 收音机爱 好者资料库 The LF353N is the 50/67.5 Hz oscilator which to me looks like it gives out a clipped sawtooth wave form. ZD9 and ZD10 appear to limit the output to + - 6v. Q9 seems to pull the solid state switches off when the 45RPM button is pressed ?! The output from the oscilator is feed to the 741 opamp which will amplify the signal and feed it into a 1k resistor R16. This signal does not go any further. The only way I can see that this circuit works is via the supplies to the 741. Q7 and Q8 form two regulator circuits for the 741 supply. Looking at Q7, +150v is fed to Q4 and Q3 this is tapped off via D5 and R12 to bias the base. This base signal is also fed via ZD7 a 75V zenner to Q7. Q7s base is at +15 v thus giving 15v from its emmitter supplying the 741 +ve rail. When the 741s out put swings high it draws current form the +ve rail this takes current from Q7 and all the way back to the biasing resistor thus drawing current from the base and reducing the out put from Q3/Q4. Q5/Q6 and Q8 will work the same but on the negative side. Thus Q3/Q4 give the +ve half of the output and Q5/Q6 give the -ve. R8,R9,R10 and R11 appear to limit the current through Q3-Q6. Why the out put transistors are doubled up I dont know. The part numbers for Q5,Q6 and Q8 are calculated. The parts are not marked. Q9 is marked 238 I assume BC238. C3 is to alter the phase of the out put to one of the motor coils to ensure it goes the right way. C4 and R15 form a feed back loop to the 741. I do not know the value of R17 but I guess some where between 10k and 50k. The motor looks very similar to Radio Spares () stock no 441-0423. 110vac-50Hz C 0.22uF 250RPM 117vac-60Hz C 0.18uF 300RPM 1.8W 18mA 24 pole. http:/www.mclennan.co.uk/datasheets/european/synchronousdata/990411131813.pdf RadioFans.CN 收音机爱 好者资料库