3 gm error edited Apr 7 '17 at 15:04 analogsystemsrf 18.2k22 gold badges88 silver badges2323 bronze badges Q2 has to drive the speaker. Assume that speaker looks like ONE OHM. At the base of Q2, that ONE OHM becomes (maybe 100 Ohms ------ Q2 beta * Z(speaker)). Now the base of Q2 is the AC load on collecter of Q1. The gain of Q1 is gm*Rload. Rload is 220K||10K||4.4K||100_Ohms, or approximately 100_Ohms. What is the gm (transconductance) of Q1? Ie_ma/26. Ie = Ic = (+5v-Vce)/10K. Given the ratio of Rc to Rb (10K to 220K), 1:22 or a lot lower than BETA, I expect Vc of Q1 to be much lower than VDD/2. In fact, Vc may be only 1 volt, and that transistor is biased poorly for gain. By operating the bipolar on the far left, we bring the very low Rout of the collector (normally predicted well as Vearly/Ic) into very inaccurate value that also is a serious load upon the gm behavior. Right now with 4 volts across 10Kohm, you have Ic = 0.4mA, and gm = 0.4/26 or 0.01;016; the total gain is 0.01016 * 100 = 1.6x. Solution? Make that 220Kohm be 1MegaOhm; and convert the Q2 into a Darlington. EDIT>>>> make that output cap Z(1KHz) be 1_ohm. To achieve that, use 160UF. For some base signal, use 1,000uF. EDIT>>>> corrected gm math error Q2 has to drive the speaker. Assume that speaker looks like ONE OHM. At the base of Q2, that ONE OHM becomes (maybe 100 Ohms ------ Q2 beta * Z(speaker)). Now the base of Q2 is the AC load on collecter of Q1. The gain of Q1 is gm*Rload. Rload is 220K||10K||4.4K||100_Ohms, or approximately 100_Ohms. What is the gm (transconductance) of Q1? Ie_ma/26. Ie = Ic = (+5v-Vce)/10K. Given the ratio of Rc to Rb (10K to 220K), 1:22 or a lot lower than BETA, I expect Vc of Q1 to be much lower than VDD/2. In fact, Vc may be only 1 volt, and that transistor is biased poorly for gain. By operating the bipolar on the far left, we bring the very low Rout of the collector (normally predicted well as Vearly/Ic) into very inaccurate value that also is a serious load upon the gm behavior. Right now with 4 volts across 10Kohm, you have Ic = 0.4mA, and gm = 0.4/26 or 0.01; the total gain is 0.01 * 100 = 1. Solution? Make that 220Kohm be 1MegaOhm; and convert the Q2 into a Darlington. EDIT>>>> make that output cap Z(1KHz) be 1_ohm. To achieve that, use 160UF. For some base signal, use 1,000uF. Q2 has to drive the speaker. Assume that speaker looks like ONE OHM. At the base of Q2, that ONE OHM becomes (maybe 100 Ohms ------ Q2 beta * Z(speaker)). Now the base of Q2 is the AC load on collecter of Q1. The gain of Q1 is gm*Rload. Rload is 220K||10K||4.4K||100_Ohms, or approximately 100_Ohms. What is the gm (transconductance) of Q1? Ie_ma/26. Ie = Ic = (+5v-Vce)/10K. Given the ratio of Rc to Rb (10K to 220K), 1:22 or a lot lower than BETA, I expect Vc of Q1 to be much lower than VDD/2. In fact, Vc may be only 1 volt, and that transistor is biased poorly for gain. By operating the bipolar on the far left, we bring the very low Rout of the collector (normally predicted well as Vearly/Ic) into very inaccurate value that also is a serious load upon the gm behavior. Right now with 4 volts across 10Kohm, you have Ic = 0.4mA, and gm = 0.4/26 or 0.016; the total gain is 0.016 * 100 = 1.6x. Solution? Make that 220Kohm be 1MegaOhm; and convert the Q2 into a Darlington. EDIT>>>> make that output cap Z(1KHz) be 1_ohm. To achieve that, use 160UF. For some base signal, use 1,000uF. EDIT>>>> corrected gm math error 2 added 114 characters in body edited Apr 7 '17 at 3:05 analogsystemsrf 18.2k22 gold badges88 silver badges2323 bronze badges Q2 has to drive the speaker. Assume that speaker looks like ONE OHM. At the base of Q2, that ONE OHM becomes (maybe 100 Ohms ------ Q2 beta * Z(speaker)). Now the base of Q2 is the AC load on collecter of Q1. The gain of Q1 is gm*Rload. Rload is 220K||10K||4.4K||100_Ohms, or approximately 100_Ohms. What is the gm (transconductance) of Q1? Ie_ma/26. Ie = Ic = (+5v-Vce)/10K. Given the ratio of Rc to Rb (10K to 220K), 1:22 or a lot lower than BETA, I expect Vc of Q1 to be much lower than VDD/2. In fact, Vc may be only 1 volt, and that transistor is biased poorly for gain. By operating the bipolar on the far left, we bring the very low Rout of the collector (normally predicted well as Vearly/Ic) into very inaccurate value that also is a serious load upon the gm behavior. Right now with 4 volts across 10Kohm, you have Ic = 0.4mA, and gm = 0.4/26 or 0.01; the total gain is 0.01 * 100 = 1. Solution? Make that 220Kohm be 1MegaOhm; and convert the Q2 into a Darlington. EDIT>>>> make that output cap Z(1KHz) be 1_ohm. To achieve that, use 160UF. For some base signal, use 1,000uF. Q2 has to drive the speaker. Assume that speaker looks like ONE OHM. At the base of Q2, that ONE OHM becomes (maybe 100 Ohms ------ Q2 beta * Z(speaker)). Now the base of Q2 is the AC load on collecter of Q1. The gain of Q1 is gm*Rload. Rload is 220K||10K||4.4K||100_Ohms, or approximately 100_Ohms. What is the gm (transconductance) of Q1? Ie_ma/26. Ie = Ic = (+5v-Vce)/10K. Given the ratio of Rc to Rb (10K to 220K), 1:22 or a lot lower than BETA, I expect Vc of Q1 to be much lower than VDD/2. In fact, Vc may be only 1 volt, and that transistor is biased poorly for gain. By operating the bipolar on the far left, we bring the very low Rout of the collector (normally predicted well as Vearly/Ic) into very inaccurate value that also is a serious load upon the gm behavior. Right now with 4 volts across 10Kohm, you have Ic = 0.4mA, and gm = 0.4/26 or 0.01; the total gain is 0.01 * 100 = 1. Solution? Make that 220Kohm be 1MegaOhm; and convert the Q2 into a Darlington. Q2 has to drive the speaker. Assume that speaker looks like ONE OHM. At the base of Q2, that ONE OHM becomes (maybe 100 Ohms ------ Q2 beta * Z(speaker)). Now the base of Q2 is the AC load on collecter of Q1. The gain of Q1 is gm*Rload. Rload is 220K||10K||4.4K||100_Ohms, or approximately 100_Ohms. What is the gm (transconductance) of Q1? Ie_ma/26. Ie = Ic = (+5v-Vce)/10K. Given the ratio of Rc to Rb (10K to 220K), 1:22 or a lot lower than BETA, I expect Vc of Q1 to be much lower than VDD/2. In fact, Vc may be only 1 volt, and that transistor is biased poorly for gain. By operating the bipolar on the far left, we bring the very low Rout of the collector (normally predicted well as Vearly/Ic) into very inaccurate value that also is a serious load upon the gm behavior. Right now with 4 volts across 10Kohm, you have Ic = 0.4mA, and gm = 0.4/26 or 0.01; the total gain is 0.01 * 100 = 1. Solution? Make that 220Kohm be 1MegaOhm; and convert the Q2 into a Darlington. EDIT>>>> make that output cap Z(1KHz) be 1_ohm. To achieve that, use 160UF. For some base signal, use 1,000uF. 1 answered Apr 7 '17 at 3:00 analogsystemsrf 18.2k22 gold badges88 silver badges2323 bronze badges Q2 has to drive the speaker. Assume that speaker looks like ONE OHM. At the base of Q2, that ONE OHM becomes (maybe 100 Ohms ------ Q2 beta * Z(speaker)). Now the base of Q2 is the AC load on collecter of Q1. The gain of Q1 is gm*Rload. Rload is 220K||10K||4.4K||100_Ohms, or approximately 100_Ohms. What is the gm (transconductance) of Q1? Ie_ma/26. Ie = Ic = (+5v-Vce)/10K. Given the ratio of Rc to Rb (10K to 220K), 1:22 or a lot lower than BETA, I expect Vc of Q1 to be much lower than VDD/2. In fact, Vc may be only 1 volt, and that transistor is biased poorly for gain. By operating the bipolar on the far left, we bring the very low Rout of the collector (normally predicted well as Vearly/Ic) into very inaccurate value that also is a serious load upon the gm behavior. Right now with 4 volts across 10Kohm, you have Ic = 0.4mA, and gm = 0.4/26 or 0.01; the total gain is 0.01 * 100 = 1. Solution? Make that 220Kohm be 1MegaOhm; and convert the Q2 into a Darlington.