how does it work?

supertone · 10-07-2008, 09:01 PM

Who can explane me how the channel switching circuit of a fender blues deville actually works, wat does the 4560D do? ect....

Thanks

Michel

techineer · 10-07-2009, 06:43 AM

The 4560 op-amps are configured as comparators with hysteresis, which means that they have an electrical 'snap action' to make them less sensitive to noise. When an input signal changes above or below the reference voltage at the pin 3 input, the output snaps to the opposite polarity maximum value.

The amp feeds an AC voltage through a resistor to the footswitch jack. The footswitch buttons are each connected in series with a diode to ground. The diode for one switch is reversed relative to the other switch.

The detector circuitry looks for the presence or absence of the positive and/or negative alternation of the AC signal to determine which channel is to be active and whether or not to enable reverb.

When both switches are open, both alternations of the AC signal are present from the jack. This signal is fed to two half-wave rectifier circuits (CR7 & CR8) feeding the 4560 U3A and U4A. This produces a positive DC voltage at pin 2 of U3A, and a negative output at pin 1. This does not activate RY2, but it gets inverted by U3B to create a high voltage level at pin 7, which will turn on RY1 to engage the clean channel relay. A negative DC voltage is produced at pin 2 of U4A, which tells it to turn the reverb on.

When the Channel switch is pressed, the diode in the switch shorts the positive alternation to ground (actually 0.6V) so no voltage passes through the rectifier CR7 to U3A. The op amp output goes high, turning on RY2 and the Drive channel, while the inversion of U3B turns off RY1.

When the reverb switch is pressed, a similar event happens, but of opposite polarity. The negative alternations are clamped to -0.6v, which cannot pass through CR8. Since pin 2 is less than pin 3, the output switches to ~+15V, which controls the reverb circuit to turn it off.

It's a slick way to get 4 switch states across a single wire with respect to ground.

10-07-2008, 09:01 PM	#1
supertone Junior Member Join Date: Oct 2008 Location: Loon op zand, the Netherlands Posts: 11	how does it work? Who can explane me how the channel switching circuit of a fender blues deville actually works, wat does the 4560D do? ect.... Thanks Michel

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10-07-2009, 06:43 AM	#2
techineer Member Join Date: Jun 2009 Posts: 60	The 4560 op-amps are configured as comparators with hysteresis, which means that they have an electrical 'snap action' to make them less sensitive to noise. When an input signal changes above or below the reference voltage at the pin 3 input, the output snaps to the opposite polarity maximum value. The amp feeds an AC voltage through a resistor to the footswitch jack. The footswitch buttons are each connected in series with a diode to ground. The diode for one switch is reversed relative to the other switch. The detector circuitry looks for the presence or absence of the positive and/or negative alternation of the AC signal to determine which channel is to be active and whether or not to enable reverb. When both switches are open, both alternations of the AC signal are present from the jack. This signal is fed to two half-wave rectifier circuits (CR7 & CR8) feeding the 4560 U3A and U4A. This produces a positive DC voltage at pin 2 of U3A, and a negative output at pin 1. This does not activate RY2, but it gets inverted by U3B to create a high voltage level at pin 7, which will turn on RY1 to engage the clean channel relay. A negative DC voltage is produced at pin 2 of U4A, which tells it to turn the reverb on. When the Channel switch is pressed, the diode in the switch shorts the positive alternation to ground (actually 0.6V) so no voltage passes through the rectifier CR7 to U3A. The op amp output goes high, turning on RY2 and the Drive channel, while the inversion of U3B turns off RY1. When the reverb switch is pressed, a similar event happens, but of opposite polarity. The negative alternations are clamped to -0.6v, which cannot pass through CR8. Since pin 2 is less than pin 3, the output switches to ~+15V, which controls the reverb circuit to turn it off. It's a slick way to get 4 switch states across a single wire with respect to ground.

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